Tüm alıştırma soruları

1526 soru

Soru 501Soru

If the expression 483+2+7232\frac{\sqrt{48}}{\sqrt{3} + \sqrt{2}} + \frac{\sqrt{72}}{\sqrt{3} - \sqrt{2}} is simplified into the form m+n6m + n\sqrt{6}, where mm and nn are integers, find the value of m+nm + n.

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Cevap: 26

Cevap

The simplified expression is 24+2624 + 2\sqrt{6}, giving m=24m = 24 and n=2n = 2, so m+n=26m + n = 26.
Simplifying 48\sqrt{48} to 434\sqrt{3} and 72\sqrt{72} to 626\sqrt{2} allows rationalization of each fraction by its conjugate. The first fraction becomes 124612 - 4\sqrt{6} and the second becomes 12+6612 + 6\sqrt{6}. Adding these expressions results in 24+2624 + 2\sqrt{6}, so m=24m = 24 and n=2n = 2, giving m+n=26m + n = 26.

Adım Adım Çözüm

1
Simplify the radical numerators
48=43\sqrt{48} = 4\sqrt{3} and 72=62\sqrt{72} = 6\sqrt{2}
Decomposing surds into perfect square factors simplifies subsequent algebraic expansion.
2
Rationalize the first term 433+2\frac{4\sqrt{3}}{\sqrt{3} + \sqrt{2}}
124612 - 4\sqrt{6}
Multiplying the numerator and denominator by the conjugate (32)(\sqrt{3} - \sqrt{2}) removes the surd from the denominator using the difference of squares (3)2(2)2=1(\sqrt{3})^2 - (\sqrt{2})^2 = 1.
3
Rationalize the second term 6232\frac{6\sqrt{2}}{\sqrt{3} - \sqrt{2}}
12+6612 + 6\sqrt{6}
Multiplying the numerator and denominator by the conjugate (3+2)(\sqrt{3} + \sqrt{2}) yields a rational denominator of 11.
4
Combine like surd terms
24+2624 + 2\sqrt{6}
Summing the rational components (12+12=24)(12 + 12 = 24) and combining similar surd terms (46+66=26)(-4\sqrt{6} + 6\sqrt{6} = 2\sqrt{6}).
5
Calculate the target sum m+nm + n
2626
Matching coefficients with m+n6m + n\sqrt{6} gives m=24m = 24 and n=2n = 2, yielding 24+2=2624 + 2 = 26.

Anahtar Kavram

Rationalization of Binomial Denominators using Conjugates
Soru 502Soru

When the polynomial P(x)=3x3+ax2+bx10P(x) = 3x^3 + ax^2 + bx - 10 is divided by (x2)(x - 2), the remainder is 1414, and when it is divided by (x+1)(x + 1), the remainder is 16-16. What is the value of a+ba + b?

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Cevap: 1

Cevap

The value of a+ba + b is 11.
According to the Remainder Theorem, dividing P(x)P(x) by (x2)(x - 2) and (x+1)(x + 1) gives P(2)=14P(2) = 14 and P(1)=16P(-1) = -16 respectively. Expanding these expressions forms two linear equations: 2a+b=02a + b = 0 and ab=3a - b = -3. Solving these simultaneously gives a=1a = -1 and b=2b = 2, so a+b=1a + b = 1.

Adım Adım Çözüm

1
Apply the Remainder Theorem for the first divisor (x2)(x - 2)
2a+b=02a + b = 0
By the Remainder Theorem, P(2)=14P(2) = 14. Substituting x=2x = 2 into P(x)P(x) yields 3(8)+4a+2b10=143(8) + 4a + 2b - 10 = 14, which simplifies to 2a+b=02a + b = 0.
2
Apply the Remainder Theorem for the second divisor (x+1)(x + 1)
ab=3a - b = -3
By the Remainder Theorem, P(1)=16P(-1) = -16. Substituting x=1x = -1 into P(x)P(x) yields 3(1)+ab10=163(-1) + a - b - 10 = -16, which simplifies to ab=3a - b = -3.
3
Solve the simultaneous equations for aa and bb
a=1a = -1 and b=2b = 2
Adding 2a+b=02a + b = 0 and ab=3a - b = -3 yields 3a=33a = -3, giving a=1a = -1. Substituting a=1a = -1 into 2a+b=02a + b = 0 gives b=2b = 2.
4
Calculate the value of a+ba + b
1
Adding the computed values yields a+b=1+2=1a + b = -1 + 2 = 1.

Anahtar Kavram

Polynomial Remainder Theorem and Systems of Linear Equations
Tahmini Süre:2m 0s
Soru 503Soru

In how many different ways can 55 boys and 33 girls be seated in a straight row such that all 33 girls must sit together?

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Cevap: 4320

Cevap

The total number of ways to seat 55 boys and 33 girls in a row such that all 33 girls sit together is 43204320.
To arrange 55 boys and 33 girls so that the girls are always together, treat the 33 girls as 11 single unit. Combined with the 55 boys, there are 66 units to arrange in a straight line, which can be done in 6!=7206! = 720 ways. Within their group, the 33 girls can be arranged in 3!=63! = 6 ways. By the multiplication principle, the total number of seating arrangements is 720×6=4320720 \times 6 = 4320.

Adım Adım Çözüm

1
Group the restricted items into a single block
The 33 girls form 11 unit. Combined with the 55 boys, there are 5+1=65 + 1 = 6 units to arrange.
Since all 33 girls must sit together, treating them as a single block ensures they are not separated.
2
Calculate the linear arrangements of the combined units
The 66 units can be arranged in 6!=7206! = 720 ways.
The number of distinct ways to arrange nn items in a line is n!n!.
3
Calculate internal arrangements of the girls' block
The 33 girls can be arranged among themselves in 3!=63! = 6 ways.
The 33 girls inside the block are distinct individuals and can swap positions.
4
Apply the fundamental counting principle
Total arrangements = 6!×3!=720×6=43206! \times 3! = 720 \times 6 = 4320.
The total number of arrangements is the product of external block arrangements and internal block arrangements.

Anahtar Kavram

Permutations with grouping constraints (string method)
Soru 504Soru

If xx is the smallest positive integer satisfying the modular congruence 2x7(mod11)2^x \equiv 7 \pmod{11}, what is the value of (3x24x+5)(mod11)(3x^2 - 4x + 5) \pmod{11} expressed in standard non-negative remainder form?

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Cevap: 3

Cevap

The smallest positive integer exponent satisfying 2x7(mod11)2^x \equiv 7 \pmod{11} is x=7x = 7. Substituting x=7x = 7 into 3x24x+53x^2 - 4x + 5 yields 124124, which simplifies to 3(mod11)3 \pmod{11}.
Evaluating powers of 2 modulo 11 shows that 27=1287(mod11)2^7 = 128 \equiv 7 \pmod{11}, giving x=7x = 7. Substituting x=7x = 7 into 3x24x+53x^2 - 4x + 5 gives 124124, which leaves a remainder of 33 when divided by 1111.

Adım Adım Çözüm

1
Find the smallest positive integer exponent xx satisfying 2x7(mod11)2^x \equiv 7 \pmod{11}
x=7x = 7
Evaluating consecutive powers of 2 modulo 11 shows 2122^1 \equiv 2, 2242^2 \equiv 4, 2382^3 \equiv 8, 2452^4 \equiv 5, 25102^5 \equiv 10, 2692^6 \equiv 9, and 2772^7 \equiv 7, making x=7x = 7 the smallest positive integer power.
2
Substitute x=7x = 7 into the expression 3x24x+53x^2 - 4x + 5
124
Direct substitution gives 3(7)24(7)+5=3(49)28+5=14728+5=1243(7)^2 - 4(7) + 5 = 3(49) - 28 + 5 = 147 - 28 + 5 = 124.
3
Reduce 124 modulo 11 to standard non-negative remainder form
3
Dividing 124 by 11 yields a quotient of 11 with a remainder of 3 (124=11×11+3124 = 11 \times 11 + 3).

Anahtar Kavram

Modular Exponentiation and Algebraic Evaluation in Modular Arithmetic
Soru 505Soru

The mean of five numbers arranged in ascending order is 2828. The mean of the first three numbers is 2222, while the mean of the last three numbers is 3636. Find the median of these five numbers.

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Cevap: 34

Cevap

The median of the five numbers is 3434.
For five numbers ordered from smallest to largest (x1,x2,x3,x4,x5x_1, x_2, x_3, x_4, x_5), the median is the middle value x3x_3. The sum of all five numbers is 5×28=1405 \times 28 = 140. The sum of the first three numbers is x1+x2+x3=3×22=66x_1 + x_2 + x_3 = 3 \times 22 = 66, and the sum of the last three numbers is x3+x4+x5=3×36=108x_3 + x_4 + x_5 = 3 \times 36 = 108. Adding these two partial sums gives (x1+x2+x3+x4+x5)+x3=66+108=174(x_1 + x_2 + x_3 + x_4 + x_5) + x_3 = 66 + 108 = 174. Substituting the overall sum of 140140 into the equation yields 140+x3=174140 + x_3 = 174, which simplifies to x3=34x_3 = 34.

Adım Adım Çözüm

1
Calculate the sum of all five numbers.
Sum of all 5 numbers is 5×28=1405 \times 28 = 140.
The total sum of a set of data equals the number of items multiplied by the mean.
2
Calculate the partial sums of the first three and last three numbers.
First three numbers sum to 3×22=663 \times 22 = 66; last three numbers sum to 3×36=1083 \times 36 = 108.
Multiplying each sub-group mean by the count of numbers in that sub-group yields the sub-group sum.
3
Set up an equation relating the partial sums to the total sum and the median.
Adding the partial sums counts the third number (median) twice: 66+108=140+median66 + 108 = 140 + \text{median}.
In an ordered set of 5 numbers, the 3rd term is the median and is shared by both the first three and last three elements.
4
Solve for the median.
Median =174140=34= 174 - 140 = 34.
Subtracting the total sum from the combined partial sums isolates the overlapping median value.

Anahtar Kavram

Measures of Central Tendency for Ungrouped Data (Relationship between sub-group means, total sum, and median in ordered data)
Soru 506Soru

For the matrix M=(k312k0152)M = \begin{pmatrix} k & 3 & 1 \\ 2 & k & 0 \\ 1 & 5 & 2 \end{pmatrix}, the determinant of MM is equal to 44. What is the positive value of kk?

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Cevap: 2

Cevap

The positive value of kk is 22.
Expanding the determinant along the second row yields det(M)=2(65)+k(2k1)=2k2k2\det(M) = -2(6-5) + k(2k-1) = 2k^2 - k - 2. Setting this equal to 44 yields 2k2k6=02k^2 - k - 6 = 0, which factors into (2k+3)(k2)=0(2k+3)(k-2)=0. The positive value is 22.

Adım Adım Çözüm

1
Evaluate the determinant of MM in terms of kk using row 2 cofactor expansion
\det(M) = 2k^2 - k - 2
Expanding along the second row gives 2(65)+k(2k1)0=2+2k2k-2(6 - 5) + k(2k - 1) - 0 = -2 + 2k^2 - k.
2
Set the determinant equal to the given value 4 and rearrange into standard quadratic form
2k^2 - k - 6 = 0
Subtracting 4 from both sides gives 2k2k6=02k^2 - k - 6 = 0.
3
Solve the quadratic equation by factorization
k = -1.5 \text{ or } k = 2
Factoring (2k+3)(k2)=0(2k + 3)(k - 2) = 0 gives roots k=1.5k = -1.5 and k=2k = 2.
4
Select the positive root
k = 2
The question specifically asks for the positive value of kk.

Anahtar Kavram

Evaluating a 3x3 matrix determinant using cofactor expansion and solving the resulting quadratic equation for an unknown parameter.
Soru 507Soru

A thermometer is calibrated on a custom scale, XX, where the ice point (0C0^\circ\text{C}) is marked as 10X-10^\circ\text{X} and the steam point (100C100^\circ\text{C}) is marked as 110X110^\circ\text{X}. What is the reading on this custom scale when a standard Celsius thermometer reads 35C35^\circ\text{C}?

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Cevap: 32

Cevap

32 °X
Using the linear temperature interpolation formula XLFPXUFPXLFPX=CLFPCUFPCLFPC\frac{X - \text{LFP}_X}{\text{UFP}_X - \text{LFP}_X} = \frac{C - \text{LFP}_C}{\text{UFP}_C - \text{LFP}_C}, substituting LFPX=10\text{LFP}_X = -10, UFPX=110\text{UFP}_X = 110, C=35C = 35, LFPC=0\text{LFP}_C = 0, and UFPC=100\text{UFP}_C = 100 gives X(10)110(10)=3501000\frac{X - (-10)}{110 - (-10)} = \frac{35 - 0}{100 - 0}. Simplifying gives X+10120=0.35\frac{X + 10}{120} = 0.35, leading to X+10=42X + 10 = 42, so X=32XX = 32^\circ\text{X}.

Adım Adım Çözüm

1
Determine fundamental intervals for both temperature scales
Celsius fundamental interval = 1000=100C100 - 0 = 100^\circ\text{C}; Custom scale fundamental interval = 110(10)=120X110 - (-10) = 120^\circ\text{X}
Linear temperature scale interpolation requires calculating the total interval between the lower fixed point (LFP) and upper fixed point (UFP).
2
Set up the ratio equation between the two thermometric scales
X(10)120=350100\frac{X - (-10)}{120} = \frac{35 - 0}{100}
The fractional position of any given temperature relative to its fixed points must be equal on all linear scales.
3
Solve the algebraic equation for XX
X+10=120×0.35=42    X=32XX + 10 = 120 \times 0.35 = 42 \implies X = 32^\circ\text{X}
Isolating XX gives the corresponding reading on the custom temperature scale.

Anahtar Kavram

Linear Temperature Scale Conversion and Interpolation
Soru 508Soru

Using differentiation from first principles, evaluate the value of the derivative dydx=limh0f(x+h)f(x)h\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} for the cubic function f(x)=2x39x2+12x5f(x) = 2x^3 - 9x^2 + 12x - 5 at the point x=3x = 3.

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Cevap: 12

Cevap

The derivative evaluated at x=3x = 3 is equal to 12.
Evaluating the definition of the derivative from first principles for f(x)=2x39x2+12x5f(x) = 2x^3 - 9x^2 + 12x - 5 yields limh0f(x+h)f(x)h=6x218x+12\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = 6x^2 - 18x + 12. Substituting x=3x = 3 gives 6(3)218(3)+12=5454+12=126(3)^2 - 18(3) + 12 = 54 - 54 + 12 = 12.

Adım Adım Çözüm

1
Set up the difference quotient definition from first principles
f(x)=limh0f(x+h)f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
Differentiation from first principles requires finding the limit of the average rate of change as the increment hh approaches zero.
2
Substitute (x+h)(x+h) into f(x)=2x39x2+12x5f(x) = 2x^3 - 9x^2 + 12x - 5 and expand
f(x+h)=2x3+6x2h+6xh2+2h39x218xh9h2+12x+12h5f(x+h) = 2x^3 + 6x^2h + 6xh^2 + 2h^3 - 9x^2 - 18xh - 9h^2 + 12x + 12h - 5
Expanding binomial terms (x+h)3(x+h)^3 and (x+h)2(x+h)^2 reveals all components involving hh.
3
Calculate f(x+h)f(x)f(x+h) - f(x) and factor out hh
f(x+h)f(x)=h(6x2+6xh+2h218x9h+12)f(x+h) - f(x) = h(6x^2 + 6xh + 2h^2 - 18x - 9h + 12)
Terms independent of hh cancel out completely, isolating hh as a common factor.
4
Divide by hh and evaluate the limit as h0h \to 0
f(x)=6x218x+12f'(x) = 6x^2 - 18x + 12
Canceling hh resolves the 00\frac{0}{0} indeterminate form, allowing direct substitution of h=0h=0.
5
Substitute x=3x = 3 into f(x)f'(x)
f(3)=6(3)218(3)+12=12f'(3) = 6(3)^2 - 18(3) + 12 = 12
Evaluating at x=3x = 3 gives the numerical value of the instantaneous rate of change at that specific point.

Anahtar Kavram

Differentiation from First Principles
Soru 509Soru

Given the function y=(x2+1)32x3y = \frac{(x^2 + 1)^3}{2x - 3}, what is the numerical value of dydx\frac{dy}{dx} evaluated at x=2x = 2?

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Cevap: 50

Cevap

The numerical value of dydx\frac{dy}{dx} at x=2x = 2 is 50.
Applying the Quotient Rule dydx=uvuvv2\frac{dy}{dx} = \frac{u'v - uv'}{v^2} along with the Chain Rule to find u(x)=6x(x2+1)2u'(x) = 6x(x^2 + 1)^2, and evaluating all components at x=2x = 2 gives u(2)=125u(2) = 125, u(2)=300u'(2) = 300, v(2)=1v(2) = 1, and v(2)=2v'(2) = 2. Substituting these values yields (300)(1)(125)(2)12=50\frac{(300)(1) - (125)(2)}{1^2} = 50.

Adım Adım Çözüm

1
Set up the Quotient Rule components
u(x)=(x2+1)3u(x) = (x^2 + 1)^3 and v(x)=2x3v(x) = 2x - 3
The given expression is a quotient of two functions requiring dydx=uvuvv2\frac{dy}{dx} = \frac{u'v - uv'}{v^2}.
2
Differentiate u(x)u(x) using the Chain Rule and v(x)v(x) using basic power rules
u(x)=6x(x2+1)2u'(x) = 6x(x^2 + 1)^2 and v(x)=2v'(x) = 2
Differentiating the outer power 3 gives 3(x2+1)23(x^2 + 1)^2, and multiplying by the derivative of the inner function (2x)(2x) gives 6x(x2+1)26x(x^2 + 1)^2.
3
Evaluate u(2)u(2), u(2)u'(2), v(2)v(2), and v(2)v'(2)
u(2)=125u(2) = 125, u(2)=300u'(2) = 300, v(2)=1v(2) = 1, and v(2)=2v'(2) = 2
Substituting x=2x = 2 into each function and derivative simplifies calculation of the overall derivative.
4
Substitute values into the Quotient Rule formula
dydxx=2=(300)(1)(125)(2)(1)2=50\left.\frac{dy}{dx}\right|_{x=2} = \frac{(300)(1) - (125)(2)}{(1)^2} = 50
Evaluating u(2)v(2)u(2)v(2)[v(2)]2\frac{u'(2)v(2) - u(2)v'(2)}{[v(2)]^2} yields the exact numerical result.

Anahtar Kavram

Combined Application of Quotient Rule and Chain Rule
Soru 510Soru

If 43x+56x=121x43_x + 56_x = 121_x, where xx represents a positive integer base, find the value of xx.

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Cevap: 8

Cevap

The value of the base xx is 8.
Expanding each base xx number into polynomial form gives (4x+3)+(5x+6)=x2+2x+1(4x + 3) + (5x + 6) = x^2 + 2x + 1. Simplifying yields the quadratic equation x27x8=0x^2 - 7x - 8 = 0, which factors as (x8)(x+1)=0(x - 8)(x + 1) = 0. Since a number base must be a positive integer greater than any digit present in the problem (x>6x > 6), x=8x = 8.

Adım Adım Çözüm

1
Convert all base xx numbers into base 10 algebraic expressions.
43x=4x+343_x = 4x + 3, 56x=5x+656_x = 5x + 6, and 121x=x2+2x+1121_x = x^2 + 2x + 1.
Place-value expansion expresses numbers in base xx as polynomials in xx.
2
Set up the algebraic equation corresponding to the addition.
(4x+3)+(5x+6)=x2+2x+1    9x+9=x2+2x+1(4x + 3) + (5x + 6) = x^2 + 2x + 1 \implies 9x + 9 = x^2 + 2x + 1.
The sum of the left-hand terms equals the right-hand term.
3
Rearrange into standard quadratic form and factor.
x27x8=0    (x8)(x+1)=0x^2 - 7x - 8 = 0 \implies (x - 8)(x + 1) = 0.
Moving all terms to one side allows solving for the roots of the quadratic equation.
4
Determine the valid base value.
x=8x = 8.
Number bases must be positive integers greater than all individual digits present in the expression (x>6x > 6).

Anahtar Kavram

Unknown base equations and expansion
Soru 511Soru

If the surd expression 7512+63\sqrt{75} - \sqrt{12} + \frac{6}{\sqrt{3}} is simplified to the form k3k\sqrt{3}, what is the value of kk?

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Cevap: 5

Cevap

The value of kk is 5.
Simplifying 75\sqrt{75} yields 535\sqrt{3}, simplifying 12\sqrt{12} yields 232\sqrt{3}, and rationalizing 63\frac{6}{\sqrt{3}} gives 232\sqrt{3}. Adding these together gives 5323+23=535\sqrt{3} - 2\sqrt{3} + 2\sqrt{3} = 5\sqrt{3}. Equating 535\sqrt{3} to k3k\sqrt{3} yields k=5k = 5.

Adım Adım Çözüm

1
Simplify the individual square roots.
75=25×3=53\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} and 12=4×3=23\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}.
Factoring out perfect square numbers allows surds to be written in basic radical form.
2
Rationalize the fractional surd term 63\frac{6}{\sqrt{3}}.
63×33=633=23\frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}.
Multiplying numerator and denominator by 3\sqrt{3} eliminates the radical from the denominator.
3
Combine like surd terms.
5323+23=535\sqrt{3} - 2\sqrt{3} + 2\sqrt{3} = 5\sqrt{3}.
Like surds share the same radical factor and can be added or subtracted algebraically.
4
Equate the simplified expression to k3k\sqrt{3}.
k=5k = 5.
Comparing coefficients of 3\sqrt{3} reveals the value of kk.

Anahtar Kavram

Surd Simplification and Rationalization of Denominators
Tahmini Süre:1m 30s
Soru 512Soru

Find the product of all real solutions to the exponential equation 9x+1283x+3=09^{x+1} - 28 \cdot 3^x + 3 = 0.

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Cevap: -2

Cevap

The product of all real solutions is -2.
Using the index laws 9x+1=9(3x)29^{x+1} = 9 \cdot (3^x)^2, let u=3xu = 3^x. The equation reduces to the quadratic 9u228u+3=09u^2 - 28u + 3 = 0, which factors as (9u1)(u3)=0(9u - 1)(u - 3) = 0. This gives u=1/9u = 1/9 or u=3u = 3. Solving 3x=1/93^x = 1/9 yields x=2x = -2, and solving 3x=33^x = 3 yields x=1x = 1. The product of these solutions is (2)×1=2(-2) \times 1 = -2.

Adım Adım Çözüm

1
Express 9x+19^{x+1} in terms of 3x3^x
9x+1=919x=9(32)x=9(3x)29^{x+1} = 9^1 \cdot 9^x = 9 \cdot (3^2)^x = 9 \cdot (3^x)^2
Apply index laws am+n=amana^{m+n} = a^m \cdot a^n and (am)n=(an)m(a^m)^n = (a^n)^m to establish a common base of 3.
2
Substitute u=3xu = 3^x to form a quadratic equation
9u228u+3=09u^2 - 28u + 3 = 0
Transform the exponential equation into a standard quadratic algebraic equation.
3
Solve the quadratic equation for uu
(9u1)(u3)=0    u=19(9u - 1)(u - 3) = 0 \implies u = \frac{1}{9} or u=3u = 3
Factorize the quadratic expression to determine its roots.
4
Substitute back u=3xu = 3^x to solve for xx
3x=32    x=23^x = 3^{-2} \implies x = -2, and 3x=31    x=13^x = 3^1 \implies x = 1
Equate exponents with matching bases to find all valid real solutions for xx.
5
Find the product of the two solutions
(2)×1=2(-2) \times 1 = -2
Calculate the required mathematical product of the solutions.

Anahtar Kavram

Solving exponential equations reducible to quadratic form using laws of indices
Soru 513Soru

Calcium hydride (CaH2\text{CaH}_2) reacts vigorously with water to produce calcium hydroxide and hydrogen gas. What volume of dry hydrogen gas, in dm3\text{dm}^3, measured at standard temperature and pressure (s.t.p.), is liberated when 10.5 g10.5\text{ g} of pure calcium hydride reacts completely with excess water?

[Relative atomic masses: Ca=40\text{Ca} = 40, H=1\text{H} = 1; Molar volume of gas at s.t.p. = 22.4 dm3 mol122.4\text{ dm}^3\text{ mol}^{-1}]

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Cevap: 11.2

Cevap

The volume of dry hydrogen gas liberated at s.t.p. is 11.2 dm³.
Calcium hydride reacts with water according to the reaction CaH₂ + 2H₂O → Ca(OH)₂ + 2H₂. Given 10.5 g of CaH₂ (molar mass 42 g/mol), there are 0.25 moles of CaH₂. Based on the 1:2 stoichiometric ratio, 0.50 moles of H₂ gas are generated. Multiplying by the molar volume at s.t.p. (22.4 dm³/mol) yields 11.2 dm³.

Adım Adım Çözüm

1
Write the balanced chemical equation for the reaction of calcium hydride with water
CaH₂ + 2H₂O → Ca(OH)₂ + 2H₂
Establishing the stoichiometric mole ratio between the reactant CaH₂ and the product H₂ gas.
2
Calculate the molar mass of CaH₂
42 g/mol
Molar mass is required to convert the given mass of CaH₂ into moles.
3
Determine the amount of CaH₂ in moles
0.25 mol
Moles = Mass / Molar mass = 10.5 g / 42 g/mol.
4
Calculate the moles of H₂ gas liberated using the 1:2 stoichiometric ratio
0.50 mol
1 mole of CaH₂ produces 2 moles of H₂ gas.
5
Calculate the volume of H₂ gas produced at standard temperature and pressure (s.t.p.)
11.2 dm³
Volume at s.t.p. = Moles × Molar volume at s.t.p. = 0.50 mol × 22.4 dm³/mol.

Anahtar Kavram

Laboratory and industrial preparation of hydrogen using metal hydrides and mole-volume stoichiometric calculations at s.t.p.
Tahmini Süre:2m 0s
Soru 514Soru

Calculate the solubility in mol/dm3\text{mol/dm}^3 of sodium nitrate (NaNO3\text{NaNO}_3) at 25C25^\circ\text{C}, if 17.0 g17.0\text{ g} of the salt dissolves in 100.0 g100.0\text{ g} of water to form a saturated solution. [Relative atomic masses: Na=23,N=14,O=16\text{Na} = 23, \text{N} = 14, \text{O} = 16; density of water =1.0 g/cm3= 1.0\text{ g/cm}^3]

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Cevap: 2

Cevap

The solubility of sodium nitrate at 25C25^\circ\text{C} is 2.0 mol/dm32.0\text{ mol/dm}^3.
To find solubility in mol/dm3\text{mol/dm}^3, first convert 17.0 g17.0\text{ g} of NaNO3\text{NaNO}_3 into moles by dividing by its molar mass (85.0 g/mol85.0\text{ g/mol}), obtaining 0.20 mol0.20\text{ mol}. Next, convert 100.0 g100.0\text{ g} of water into volume, which equals 0.100 dm30.100\text{ dm}^3. Dividing 0.20 mol0.20\text{ mol} by 0.100 dm30.100\text{ dm}^3 yields 2.0 mol/dm32.0\text{ mol/dm}^3.

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1
Determine the molar mass of the solute (NaNO3\text{NaNO}_3)
Molar mass of NaNO3=23+14+(3×16)=85.0 g/mol\text{NaNO}_3 = 23 + 14 + (3 \times 16) = 85.0\text{ g/mol}
Molar mass is required to convert mass of solute to amount in moles.
2
Calculate the amount of NaNO3\text{NaNO}_3 in moles
Moles=17.0 g85.0 g/mol=0.20 mol\text{Moles} = \frac{17.0\text{ g}}{85.0\text{ g/mol}} = 0.20\text{ mol}
Solubility in mol/dm3\text{mol/dm}^3 measures the amount of solute in moles per cubic decimeter of solvent.
3
Convert the mass of solvent (water) to volume in dm3\text{dm}^3
100.0 g of water=100.0 cm3=0.100 dm3100.0\text{ g of water} = 100.0\text{ cm}^3 = 0.100\text{ dm}^3
Density of water is 1.0 g/cm31.0\text{ g/cm}^3, and 1000 cm3=1 dm31000\text{ cm}^3 = 1\text{ dm}^3.
4
Calculate the solubility in mol/dm3\text{mol/dm}^3
Solubility=0.20 mol0.100 dm3=2.0 mol/dm3\text{Solubility} = \frac{0.20\text{ mol}}{0.100\text{ dm}^3} = 2.0\text{ mol/dm}^3
Dividing the moles of solute by the volume of solvent in dm3\text{dm}^3 gives the concentration in mol/dm3\text{mol/dm}^3.

Anahtar Kavram

Solubility expressed in molar concentration (mol/dm³)
Soru 515Soru

In a nuclear fusion reaction, two light nuclei fuse together to form a heavier nucleus. The total mass of the reactants before fusion is 4.028 u4.028\text{ u}, and the total mass of the products after fusion is 4.003 u4.003\text{ u}. Given that 1 u=931.5 MeV1\text{ u} = 931.5\text{ MeV}, what is the total energy released in this reaction in MeV\text{MeV}?

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Cevap: 23.2875

Cevap

The total energy released in the reaction is 23.2875 MeV.
The energy released in nuclear fusion is calculated using the mass defect \(\Delta m = m_{\text{reactants}} - m_{\text{products}}\). Subtracting 4.003 u4.003\text{ u} from 4.028 u4.028\text{ u} yields a mass defect of 0.025 u0.025\text{ u}. Multiplying this mass defect by 931.5 MeV/u931.5\text{ MeV/u} gives the total energy released as 23.2875 MeV23.2875\text{ MeV}.

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1
Calculate the mass defect (\Delta m)
\Delta m = 4.028\text{ u} - 4.003\text{ u} = 0.025\text{ u}
Mass defect is the loss of mass during nuclear fusion that gets converted into energy.
2
Convert mass defect into energy released
E = 0.025\text{ u} \times 931.5\text{ MeV/u} = 23.2875\text{ MeV}
According to mass-energy equivalence, each atomic mass unit (u) of missing mass yields 931.5 MeV of energy.

Anahtar Kavram

Mass Defect and Energy Release in Nuclear Reactions
Soru 516Soru

A radioactive sample has a half-life of 12 minutes12\text{ minutes}. If its initial activity is 96 Bq96\text{ Bq}, what is the remaining activity of the sample, in Bq\text{Bq}, after an elapsed time of 36 minutes36\text{ minutes}?

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Cevap: 12

Cevap

The remaining activity of the radioactive sample after 36 minutes36\text{ minutes} is 12 Bq12\text{ Bq}.
In 36 minutes36\text{ minutes}, exactly 33 half-lives elapse (36/12=336 / 12 = 3). The remaining activity reduces to (1/2)3=1/8(1/2)^3 = 1/8 of the initial value, giving 96 Bq/8=12 Bq96\text{ Bq} / 8 = 12\text{ Bq}.

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1
Calculate the number of half-lives that have elapsed
n=3n = 3
Divide the total elapsed time (36 minutes36\text{ minutes}) by the half-life duration (12 minutes12\text{ minutes}).
2
Calculate the remaining activity of the isotope
A=12 BqA = 12\text{ Bq}
After n=3n = 3 half-lives, the remaining activity fraction is (12)3=18\left(\frac{1}{2}\right)^3 = \frac{1}{8}. Multiplying initial activity 96 Bq96\text{ Bq} by 18\frac{1}{8} yields 12 Bq12\text{ Bq}.

Anahtar Kavram

Radioactive Decay Law and Half-life
Soru 517Soru

A weak monobasic acid, HA\text{HA}, has an acid dissociation constant (KaK_a) of 4.5×105 mol dm34.5 \times 10^{-5}\text{ mol dm}^{-3} at 25C25^\circ\text{C}. Calculate the percentage ionization of a 0.05 mol dm30.05\text{ mol dm}^{-3} solution of this acid.

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Cevap: 3

Cevap

The percentage ionization of the weak monobasic acid solution is 3.0%.
According to Ostwald's dilution law for weak monobasic acids, Ka=α2CK_a = \alpha^2 C. Rearranging to solve for the degree of ionization gives α=Ka/C=(4.5×105)/0.05=9.0×104=0.03\alpha = \sqrt{K_a / C} = \sqrt{(4.5 \times 10^{-5}) / 0.05} = \sqrt{9.0 \times 10^{-4}} = 0.03. Expressed as a percentage, 0.03×100%=3.0%0.03 \times 100\% = 3.0\%.

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1
Relate acid dissociation constant (KaK_a), initial molar concentration (CC), and degree of ionization (α\alpha)
Ka=α2CK_a = \alpha^2 C
For a weak monobasic acid undergoing partial ionization (HAH++A\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-), Ostwald's dilution law simplifies to Ka=α2C1αα2CK_a = \frac{\alpha^2 C}{1 - \alpha} \approx \alpha^2 C because α1\alpha \ll 1.
2
Substitute the given values into the simplified expression and calculate α\alpha
\alpha = \sqrt{\frac{4.5 \times 10^{-5}}{0.05}} = \sqrt{9.0 \times 10^{-4}} = 0.03
Rearranging the equation yields α=Ka/C\alpha = \sqrt{K_a / C}. Dividing the acid dissociation constant by the molar concentration gives 9.0×1049.0 \times 10^{-4}, and taking the square root yields 0.030.03.
3
Convert the fractional degree of ionization to percentage ionization
3.0%
Multiplying the fractional degree of ionization by 100% gives the percentage of acid molecules ionized in solution.

Anahtar Kavram

Ostwald's Dilution Law and Degree of Ionization of Weak Acids
Tahmini Süre:1m 30s
Soru 518Soru

At a specific temperature and pressure, 2.0 moles2.0\text{ moles} of a real gas occupy a volume of 18.0 dm318.0\text{ dm}^3. Under the exact same conditions, 2.0 moles2.0\text{ moles} of an ideal gas occupy a volume of 22.5 dm322.5\text{ dm}^3. What is the compressibility factor (ZZ) of the real gas under these conditions?

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Cevap: 0.8

Cevap

0.8
The compressibility factor ZZ measures the deviation of a real gas from ideal gas behavior and is calculated as Z=VrealVidealZ = \frac{V_{\text{real}}}{V_{\text{ideal}}}. Substituting the given values yields Z=18.022.5=0.80Z = \frac{18.0}{22.5} = 0.80. A value of Z<1Z < 1 indicates that intermolecular attractive forces dominate, causing the real gas to occupy less volume than predicted by the ideal gas equation.

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1
Identify the formula for compressibility factor
Z=VrealVidealZ = \frac{V_{\text{real}}}{V_{\text{ideal}}}
The compressibility factor ZZ quantifies deviation from ideality as the ratio of molar volume of a real gas to that of an ideal gas at the same temperature and pressure.
2
Substitute the given values into the equation
Z=18.0 dm322.5 dm3Z = \frac{18.0\text{ dm}^3}{22.5\text{ dm}^3}
Both volumes are measured under identical pressure, temperature, and mole count.
3
Calculate the numerical value
Z=0.80Z = 0.80
Dividing 18.0 by 22.5 gives 0.80, indicating Z<1Z < 1 due to predominant intermolecular attraction.

Anahtar Kavram

Compressibility factor (Z = V_real / V_ideal) measuring real gas deviation
Soru 519Soru

During a fiscal year, the Ministry of Transportation of a state government recorded the following financial disbursements:

- Construction of a modern bus terminal: 85,000,000₦85,000,000
- Routine servicing and minor repairs of fleet vehicles: 4,200,000₦4,200,000
- Payment of monthly salaries and allowances to ministry staff: 38,500,000₦38,500,000
- Major structural rehabilitation and capacity expansion of existing railway bridges: 52,000,000₦52,000,000
- Purchase of new traffic monitoring equipment: 19,800,000₦19,800,000
- Refueling of operational vehicles and office stationery supplies: 3,500,000₦3,500,000

What is the total recurrent expenditure (in ) of the Ministry for the fiscal year?

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Cevap: 46200000

Cevap

The total recurrent expenditure of the Ministry for the fiscal year is ₦46,200,000.
Recurrent expenditures are ongoing operational expenses essential for daily administration, personnel payments, and regular asset maintenance. Adding routine servicing (4,200,000₦4,200,000), salaries (38,500,000₦38,500,000), and refueling/stationery (3,500,000₦3,500,000) gives 46,200,000₦46,200,000. Capital expenditures such as building new terminals, acquiring equipment, and major bridge expansions are excluded because they yield long-term physical assets.

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1
Classify each public sector expenditure item as either recurrent expenditure or capital expenditure.
Recurrent expenditures: Routine servicing (4,200,000₦4,200,000), staff salaries (38,500,000₦38,500,000), and refueling/stationery (3,500,000₦3,500,000). Capital expenditures: Bus terminal construction (85,000,000₦85,000,000), major bridge expansion (52,000,000₦52,000,000), and equipment acquisition (19,800,000₦19,800,000).
Recurrent expenditures represent operational costs consumed within the financial year, whereas capital expenditures create non-current assets or increase their long-term capacity/lifespan.
2
Sum the classified recurrent expenditure amounts to obtain the total.
Total Recurrent Expenditure = 4,200,000+38,500,000+3,500,000=46,200,000₦4,200,000 + ₦38,500,000 + ₦3,500,000 = ₦46,200,000.
Aggregating all operational and recurring personnel expenses provides the true total of public recurrent spending.

Anahtar Kavram

Public sector recurrent expenditure comprises regular operational running costs, staff salaries, administrative expenses, and routine maintenance of existing public infrastructure, whereas capital expenditure involves creating or expanding long-term physical assets.
Soru 520Soru

An agro-processing enterprise expands its production facility in the long run by increasing all of its input factors by 30%30\%. Consequently, its monthly output of processed cassava flour increases from 800 bags800\text{ bags} to 1,104 bags1,104\text{ bags}. What is the percentage increase in total output resulting from this scale expansion?

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Cevap: 38

Cevap

The percentage increase in total output is 38%.
The absolute increase in output is 1,104800=304 bags1,104 - 800 = 304\text{ bags}. Expressing this increase as a percentage of the base output gives 304800×100=38%\frac{304}{800} \times 100 = 38\%. Because the output increase (38%38\%) exceeds the input increase (30%30\%), the firm experiences increasing returns to scale.

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1
Find the absolute change in total output.
Change in output = 1,104800=304 bags1,104 - 800 = 304\text{ bags}.
To calculate a percentage change, the absolute difference between final and initial quantities must first be established.
2
Divide the change in output by the initial output and express as a percentage.
Percentage increase = (304800)×100=38%\left(\frac{304}{800}\right) \times 100 = 38\%.
Percentage change measures the relative expansion of output relative to the original baseline quantity.

Anahtar Kavram

Calculating Percentage Change in Output for Returns to Scale Analysis
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