Answer:
A sample of a gas (5.0 mol) at 1.0 atm is expanded at constant temperature from 10 L to 15 L. The final pressure is 0.67 atm.
Step by Step Explanation?
Boyle's law states that in constant temperature the variation volume of gas is inversely proportional to the applied pressure.
The formula is,
P₁ x V₁ = P₂ × V₂
Where,
P₁ is initial pressure = 1 atm
P2 is final pressure = ? (Not Known)
V₁ is initial volume = 10 L
V₂ is final volume = 15 L
Now put the values in the formula,
\begin{gathered}\rm 1\times 10 = P_2\times 15\\\\\rm P_2 = \frac{10}{15\\} \\\\\rm P_2 = 0.67\end{gathered]
Therefore, the answer is 0.67 atm.
Answer:
9.6 moles O2
Explanation:
I'll assume it is 345 grams, not gratis, of water. Hydrogen's molar mass is 1.01, not 101.
The molar mass of water is 18.0 grams/mole.
Therefore: (345g)/(18.0 g/mole) = 19.17 or 19.2 moles water (3 sig figs).
The balanced equation states that: 2H20 ⇒ 2H2 +02
It promises that we'll get 1 mole of oxygen for every 2 moles of H2O, a molar ratio of 1/2.
get (1 mole O2/2 moles H2O)*(19.2 moles H2O) or 9.6 moles O2
Answer:
kp= 3.1 x 10^(-2)
Explanation:
To solve this problem we have to write down the reaction and use the ICE table for pressures:
2SO2 + O2 ⇄ 2SO3
Initial 3.4 atm 1.3 atm 0 atm
Change -2x - x + 2x
Equilibrium 3.4 - 2x 1.3 -x 0.52 atm
In order to know the x value:
2x = 0.52
x=(0.52)/2= 0.26
2SO2 + O2 ⇄ 2SO3
Equilibrium 3.4 - 0.52 1.3 - 0.26 0.52 atm
Equilibrium 2.88 atm 1.04 atm 0.52 atm
with the partial pressure in the equilibrium, we can obtain Kp.

Answer:
67.5%
Explanation:
Step 1: Write the balanced equation for the electrolysis of water
2 H₂O ⇒ 2 H₂ + O₂
Step 2: Calculate the theoretical yield of O₂ from 17.0 g of H₂O
According to the balanced equation, the mass ratio of H₂O to O₂ is 36.04:32.00.
17.0 g H₂O × 32.00 g O₂/36.04 g H₂O = 15.1 g O₂
Step 3: Calculate the percent yield of O₂
Given the experimental yield of O₂ is 10.2 g, we can calculate its percent yield using the following expression.
%yield = (exp yield / theoret yield) × 100%
%yield = (10.2 g / 15.1 g) × 100% = 67.5%