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MrMuchimi
3 years ago
7

The weight of the buggy was 40N on Mars. When the buggy landed on Mars it rested on an area of 0.025 m2. Calculate the pressure

exerted by the buggy on the surface of Mars
Chemistry
1 answer:
elena-s [515]3 years ago
4 0

Answer:

The pressure exerted by the buggy on the surface of Mars is 1600 pascals.

Explanation:

The pressure is determined by the definition of stress, which is the force exerted by the buggy on the martian surface divided by the contact area of the latter:

\sigma = \frac{F}{A}

Where:

\sigma - Stress, measured in pascals.

F - Force, measured in newtons.

A - Area, measured in square meters.

The force is the weight of the buggy (40 N) and A = 0.025\,m^{2}, the stress is now calculated:

\sigma = \frac{40\,N}{0.025\,m^{2}}

\sigma = 1600\,Pa

The pressure exerted by the buggy on the surface of Mars is 1600 pascals.

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What is the vapor pressure of the solution if 35.0 g of water is dissolved in 100.0 g of ethyl alcohol at 25 ∘C? The vapor press
masya89 [10]

<u>Answer:</u> The vapor pressure of the solution is 43.55 mmHg

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For water:</u>

Given mass of water = 35.0 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

\text{Moles of water}=\frac{35.0g}{18g/mol}=1.944mol

  • <u>For ethyl alcohol:</u>

Given mass of ethyl alcohol = 100.0 g

Molar mass of ethyl alcohol = 46 g/mol

Putting values in equation 1, we get:

\text{Moles of ethyl alcohol}=\frac{100.0g}{46g/mol}=2.174mol

Total moles of solution = [1.944 = 2.174] moles = 4.118 moles

  • Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

<u>For water:</u>

\chi_{\text{water}}=\frac{n_{\text{water}}}{n_{\text{water}}+n_{\text{ethyl alcohol}}}

\chi_{water}=\frac{1.944}{4.118}=0.472

<u>For ethyl alcohol:</u>

\chi_{\text{ethyl alcohol}}=\frac{n_{\text{ethyl alcohol}}}{n_{\text{water}}+n_{\text{ethyl alcohol}}}

\chi_{\text{ethyl alcohol}}=\frac{2.174}{4.118}=0.528

Dalton's law of partial pressure states that the total pressure of the system is equal to the sum of partial pressure of each component present in it.

To calculate the vapor pressure of the solution, we use the law given by Dalton, which is:

P_T=\sum_{i=1}^n (p_i\times \chi_i)

Or,

P_T=[(p_{\text{water}}\times \chi_{\text{water}})+(p_{\text{ethyl alcohol}}\times \chi_{\text{ethyl alcohol}}

We are given:

Vapor pressure of water = 23.8 mmHg

Vapor pressure of ethyl alcohol = 61.2 mmHg

Putting values in above equation, we get:

p_T=[(23.8\times 0.472)+(61.2\times 0.528)]\\\\p_T=43.55mmHg

Hence, the vapor pressure of the solution is 43.55 mmHg

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