Answer:
D. freezing point depression
Explanation:
Answer:
The mass percentage of bromine in the original compound is 81,12%
Explanation:
<u>Step 1: Calculate moles AgBr</u>
moles AgBr = mass AgBr / molar mass AgBr
= 0.8878 g / 187.77 g/mol
= 0.00472812 moles AgBr
⇒
Since 1 mol AgBr contains 1 mol Br-
Then the amount of moles Br- in the original sample must also have been 0.00472812 moles
<u>Step 2:</u> Calculating mass Br-
mass Br- = molar mass Br x moles Br-
= 79.904 g/mol x 0.00472812 mol
= 0.377796 g Br-
⇒
There were 0.377796 g Br- in the original sample
<u>Step 3:</u> Calculating mass percentage Br-
⇒mass percentage = actual mass Br- / total mass x 100%
% mass Br = 0.377796 g / 0.4657 g x 100 %
= 81.12%
Answer:
Partial pressure O₂ = 1.78 atm
Explanation:
We can apply the mole fraction to solve the question:
Moles of gas / Total moles = Partial pressure of gas / Total pressure
Total moles = 1 H₂ + 2.5 He + 2O₂ = 4.5 moles
2 mol O₂ / 4.5 mol = Partial pressure O₂ / 4 atm
(2 mol O₂ / 4.5 mol ) . 4 atm = Partial pressure O₂ → 1.78 atm
If you're talking about atoms then its a molecule
