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Zorluk: OrtaScientific Notation

A city water purification system processes 8.5×1058.5 \times 10^5 liters of water per day. A specialized chemical treatment requires 4.0×10−44.0 \times 10^{-4} grams of a purifying agent per liter of water. What is the total mass, in grams, of the purifying agent needed each day? Express the answer in proper scientific notation.

  1. A
    3.4×1003.4 \times 10^0
  2. B
    3.4×10−193.4 \times 10^{-19}
  3. 3.4×1023.4 \times 10^2Cevap
  4. D
    3.4×10103.4 \times 10^{10}

Cevap

The correct answer is 3.4×1023.4 \times 10^2.
The total mass is calculated by multiplying the daily volume by the required agent per liter: (8.5×105)×(4.0×10−4)(8.5 \times 10^5) \times (4.0 \times 10^{-4}). Multiplying the coefficients gives 3434, and adding the exponents yields 10110^1. The intermediate result, 34×10134 \times 10^1, normalizes to 3.4×1023.4 \times 10^2 by shifting the decimal one place to the left and adding 11 to the exponent.

Adım Adım Çözüm

1
Set up the multiplication expression to find the total mass.
(8.5×105)×(4.0×10−4)(8.5 \times 10^5) \times (4.0 \times 10^{-4})
The total mass is the product of the total daily volume of water and the mass of the purifying agent required per liter.
2
Group the coefficients and the powers of ten.
(8.5×4.0)×(105×10−4)(8.5 \times 4.0) \times (10^5 \times 10^{-4})
The commutative and associative properties of multiplication allow grouping similar numerical components.
3
Multiply the coefficients and add the exponents.
34×10134 \times 10^1
8.5×4.0=348.5 \times 4.0 = 34, and following the product rule of exponents, 105×10−4=105+(−4)=10110^5 \times 10^{-4} = 10^{5 + (-4)} = 10^1.
4
Normalize the result into proper scientific notation.
3.4×1023.4 \times 10^2
The coefficient 3434 must be converted to 3.43.4 (a number between 1 and 10). Because the decimal point is moved one place to the left, the exponent is increased by 1.

Anahtar Kavram

Multiplying Numbers in Scientific Notation and Normalization
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Konular

Scientific Notation

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