The partial atmospheric pressure (atm) of hydrogen in the mixture is 0.59 atm.
<h3>How do we calculate the partial pressure of gas?</h3>
Partial pressure of particular gas will be calculated as:
p = nP, where
- P = total pressure = 748 mmHg
- n is the mole fraction which can be calculated as:
- n = moles of gas / total moles of gas
Moles will be calculated as:
- n = W/M, where
- W = given mass
- M = molar mass
Moles of Hydrogen gas = 2.02g / 2.014g/mol = 1 mole
Moles of Chlorine gas = 35.90g / 70.9g/mol = 0.5 mole
Mole fraction of hydrogen = 1 / (1+0.5) = 0.6
Partial pressure of hydrogen = (0.6)(748) = 448.8 mmHg = 0.59 atm
Hence, required partial atmospheric pressure of hydrogen is 0.59 atm.
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Answer:
2,909 M
Explanation:
molair mass is of.ethylene is 26,04 g/mol
first you need to calculate how much mL 3 kg is. You can do this by using the density of ethylene: 1,1 g/mL.
3000 g x 1.1 = 3300 mL = 3,3 L
Next you need to calculate the amount of moles:
250 g / 26,04 g/mol = 9,60 mol
Now you can calculate the molarity:
9,6/3.3 = 2,909 M
I don't know the answer for the second question. I'm sorry.
Answer:
The reaction would shift toward the reactants
When the reaction reach equilibrium the partial pressure of NH3 will be greater than 1atm
Explanation:
For the reaction:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
Where K is defined as:
As initial pressures of all 3 gases is 1.0atm, reaction quotient, Q, is:
As Q > K, <em>the reaction will produce more NH₃ until Q = K consuming N₂ and H₂.</em>
Thus, there are true:
<h3>The reaction would shift toward the reactants</h3><h3>When the reaction reach equilibrium the partial pressure of NH3 will be greater than 1atm</h3>
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D. Common characteristics
