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Step2247 [10]
3 years ago
12

The density of benzene is 0.879g/mL. The volume (with 1 decimal) of 131.9g sample of benzene is how many mL?

Chemistry
1 answer:
timurjin [86]3 years ago
5 0

Answer:

150.1 mL

Explanation:

Step 1: Given data

  • Density of benzene (ρ): 0.879 g/mL
  • Mass of the sample of benzene (m): 131.9 g
  • Volume of the sample of benzene (V): ?

Step 2: Calculate the volume of the sample of benzene

Density is an intrinsic property. It is equal to the quotient between the mass and the volume of the sample of benzene.

ρ = m/V

V = m/ρ

V = 131.9 g/(0.879 g/mL)

V = 150.1 mL

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Read 2 more answers
Given the following reaction: 2N2O5=2N2O4+O2, if the rate of oxygen production is 0.15M/min, determine:
astraxan [27]

Answer:

a. r_{N_2O_5}=-0.075M/min

b. r_{N_2O_4}=0.075M/min

Explanation:

Hello.

In this case, according to the balanced chemical reaction, we can write the law of rate proportions:

\frac{r_{N_2O_5}}{-2} =\frac{r_{N_2O_4}}{2} =\frac{r_{O_2}}{1}

Thus, we proceed as follows:

a. Since the rate of oxygen production is 0.15 M/min, we can make the following setup:

\frac{r_{N_2O_5}}{-2}  =\frac{r_{O_2}}{1}\\\\r_{N_2O_5}=\frac{r_{O_2}}{-2}  =\frac{0.15M/min}{-2}\\\\ r_{N_2O_5}=-0.075M/min

b. Since the rate of oxygen production is 0.15 M/min, we can make the following setup:

\frac{r_{N_2O_4}}{2}  =\frac{r_{O_2}}{1}\\\\r_{N_2O_4}=\frac{r_{O_2}}{2}  =\frac{0.15M/min}{2}\\\\ r_{N_2O_4}=0.075M/min

Best regards!

8 0
3 years ago
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