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Zepler [3.9K]
3 years ago
10

Carry out the following calculation, paying special attention to the significant figures (where 4/3 is exact), rounding, and uni

ts. 
3.39x10^7 g/(4/3)(3.1416)(1.65x10^2 cm)^3=____g/cm^3
Chemistry
1 answer:
babunello [35]3 years ago
7 0

Answer:

Value = 1.80 g/cm³ (Approx)

Explanation:

Given:

\frac{3.39 \times 10^7g}{(\frac{4}{3} )(3.1416)(1.65 \times 10^2 cm)^3}

Computation:

\frac{3.39 \times 10^7g}{(\frac{4}{3} )(3.1416)(1.65 \times 10^2 cm)^3} \\\\\frac{3.39 \times 10^7g}{(\frac{4}{3} )(3.1416)(4.492125 \times 10^6 cm^3)} \\\\ \frac{3.39 \times 10^7g}{(\frac{4}{3} )(3.1416)(4.492125 \times 10^6 cm^3)}\\\\ \frac{3.39 \times 10^7g}{18.8166132\times 10^6 cm^3} \\\\ 1.80159945g/cm^3

Value = 1.80 g/cm³ (Approx)

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Explanation:

First things first, we need too write out the balanced equation between HBr and KOH.

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