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zzz [600]
3 years ago
9

Need help in science earths moving surface

Chemistry
1 answer:
mel-nik [20]3 years ago
3 0

Answer:

All answers stated below:

Explanation:

1. Plate Tectonics

2. lithosphere

3. asthenosphere

4. divergent boundary

5. (not sure)

6. convergent

7. transform

8. convection

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What happens to the partial pressure of oxygen in a sample of air if the temperature is increased?
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Air is mainly composed of N2  (78%), O2 (21%) and other trace gases. Now, the total pressure of air is the sum of the partial pressures of the constituent gases. The partial pressure of each gas, for example say O2, can be expressed as:

p(O2) = mole fraction of O2 * P(total, air)  ----(1)

Thus, the partial pressure is directly proportional to the total pressure. If we consider a sealed container then, as the temperature of air increases so will its pressure. Based on equation (1) an increase in the pressure of air should also increase the partial pressure of oxygen.

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4 years ago
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In a titration, 25.0 mL of KHP is titrated to the equivalence point with NaOH. The final solution volume is 45.5 mL. What was th
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3 years ago
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During studies of the reaction below,
Leni [432]

<u>Answer:</u> The percent yield of the nitrogen gas is 11.53 %.

<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 NO:</u>

Given mass of NO = 11.5 g

Molar mass of NO = 30 g/mol

Putting values in equation 1, we get:

\text{Moles of NO}=\frac{11.5g}{30g/mol}=0.383mol

  • <u>For N_2O_4 :</u>

Given mass of N_2O_4 = 102.1 g

Molar mass of N_2O_4 = 92 g/mol

Putting values in equation 1, we get:

\text{Moles of }N_2O_4=\frac{102.1g}{92g/mol}=1.11mol

For the given chemical reactions:

2N_2H_4(l)+N_2O_4(l)\rightarrow 3N_2(g)+4H_2O(g)      ......(2)

N_2H_4(l)+2N_2O_4(l)\rightarrow 6NO(g)+2H_2O(g)       .......(3)

  • <u>Calculating the experimental yield of nitrogen gas:</u>

By Stoichiometry of the reaction 3:

6 moles of NO is produced from 2 moles of N_2O_4

So, 0.383 moles of NO will be produced from = \frac{2}{6}\times 0.383=0.128mol of N_2O_4

By Stoichiometry of the reaction 2:

1 mole of N_2O_4 produces 3 moles of nitrogen gas

So, 0.128 moles of N_2O_4 will produce = \frac{3}{1}\times 0.128=0.384mol of nitrogen gas

Now, calculating the experimental yield of nitrogen gas by using equation 1, we get:

Moles of nitrogen gas = 0.384 moles

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

0.384mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(0.384mol\times 28g/mol)=10.75g

  • <u>Calculating the theoretical yield of nitrogen gas:</u>

By Stoichiometry of the reaction 2:

1 mole of N_2O_4 produces 3 moles of nitrogen gas

So, 1.11 moles of N_2O_4 will produce = \frac{3}{1}\times 1.11=3.33mol of nitrogen gas

Now, calculating the theoretical yield of nitrogen gas by using equation 1, we get:

Moles of nitrogen gas = 3.33 moles

Molar mass of nitrogen gas = 28 g/mol

Putting values in equation 1, we get:

3.33mol=\frac{\text{Mass of nitrogen gas}}{28g/mol}\\\\\text{Mass of nitrogen gas}=(3.33mol\times 28g/mol)=93.24g

  • To calculate the percentage yield of nitrogen gas, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of nitrogen gas = 10.75 g

Theoretical yield of nitrogen gas = 93.24 g

Putting values in above equation, we get:

\%\text{ yield of nitrogen gas}=\frac{10.75g}{93.24g}\times 100\\\\\% \text{yield of nitrogen gas}=11.53\%

Hence, the percent yield of the nitrogen gas is 11.53 %.

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3 years ago
How can the rate constant be determined from the rate law?
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Answer:

You can determine x and y in the rate law by looking at experimental data and noticing how the change in concentration of a reactant is related to the reaction rate.

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