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Vinvika [58]
3 years ago
14

Applying the rules of significant figures, which of the following would be the correct value for the density of a substance whic

h has a mass if 32.40 g and a volume of 2.0mL?
16.20 g/mL
16 g/mL
0.06 g/mL
5.00 g/mL​
Chemistry
1 answer:
lilavasa [31]3 years ago
6 0

16 g/ml

<h3>Further explanation</h3>

Given

mass = 32.40 g

volume = 2.0mL

Required

Density

Solution

Density is a quantity derived from the mass and volume  

Density is the ratio of mass per unit volume  

Density formula:  

\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}

ρ = density  

m = mass  

v = volume  

Input the value :

ρ = 32.40 g / 2.0 ml

ρ = 16.2 g/ml

Rules for division: The least number of significant figures in the problem determines the number of significant figures in the answer.

32.40 = 4 sig.fig

2.0 = 2 sig fig

The answer must be 2 sig fig

So the answer = 16 g/ml

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9. If 28.56 g of K2O is produced when 25.00 g K is reactedaccording to the following equation, what is the percent yieldof the r
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b. 95%

Explanation

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Mass of K₂O produced (actual yield) = 28.56 g

Mass of K that reacted = 25.00 g

Equation: 4K(s) + O₂(g) → 2K₂0(s)

What to find:

The percent yield of K₂O.

Step-by-step solution:

The first step is to calculate the theoretical yield of K₂O produced.

From the balanced equation, 4 mol K produced 2 mol K₂O

Molar mass of K₂O = 94.20 g/mol)

Molar mass of K = 39.10 g/mol)

This means 4 mol x 39.10 g/mol = 156.40 g K produced 2 mol x 94.20 g/mol = 188.40 g K₂O

So 25.00 g K will produce:

\frac{25.00\text{ }g\text{ }K\times188.40\text{ }g\text{ }K₂O}{156.40\text{ }g\text{ }K}=30.1151\text{ }g\text{ }K₂O

Actual yield of K₂O = 28.56 g

Theoretical yield of k₂O = 30.1151 g

The percent yield for the reaction can now be calculated using the formula below:

\begin{gathered} Percent\text{ }yield=\frac{Actual\text{ }yield}{Theoretical\text{ }yield}\times100\% \\  \\ Percent\text{ }yield=\frac{28.56\text{ }g}{30.1151\text{ }g}\times100\% \\  \\ Percent\text{ }yield=0.9484\times100\% \\  \\ Percent\text{ }yield=94.84\%\approx95\% \end{gathered}

Therefore, the percent yield for the reaction is 95%.

3 0
1 year ago
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