Answer:
20.53 g.
Explanation:
- Knowing that every 15.0 g of pure calcium fluoride contains 7.70 g of calcium.
<u>Using cross multiplication:</u>
15.0 g of pure calcium fluoride contains → 7.70 g of calcium.
40.0 g of pure calcium fluoride contains → ??? g of calcium.
<em>∴ The mass of Ca in 40.0 g of calcium fluoride</em> = (7.70 g)(40.0 g)/(15.0 g) = <em>20.53 g.</em>
Exothermic.
To remember this, ‘exo’ means outside, and ‘thermic’ means heat. It gives the outside (exo), heat (thermic).
Answer:
The answer to your question is 2.32 atm
Explanation:
Data
P = ?
n = 0.214
V = 2.53 L
T = 61°C
R = 0.082 atm L/mol°K
Formula
PV = nTR
solve for P
P = nRT/V
Process
1.- Calculate the temperature in K
°K = °C + 273
°K = 61 + 273
= 334
2.- Substitution
P = (0.214 x 0.082 x 334) / 2.53
3.- Simplification
P = 5.86/2.53
4.- Result
P = 2.32 atm
Answer: 12 L fluorine gas at STP can be collected from the decomposition of 90.7 g of 
Explanation:
The balanced decomposition reaction is shown as

moles of 
According to stoichiometry:
2 moles of
gives = 3 moles of flourine gas
Thus 0.36 moles of
gives =
of flourine gas
Using ideal gas equation :

P = pressure of gas = 1 atm ( at STP)
V = Volume of gas = ?
n = moles of gas = 0.54
R = gas constant = 0.0821 L atm/Kmol
T = temperature = 273 K ( at STP)
Putting the values we get :


Thus 12 L fluorine gas at STP can be collected from the decomposition of 90.7 g of 