Answer:
0.11mole
Explanation:
Let us assume that the condition is at standard temperature and pressure(STP);
Given parameters:
Volume of water = 2.45L
Unknown:
Number of moles found in this volume of water = ?
Solution;
At STP;
Number of moles = 
Input the parameters and solve;
Number of moles of water = 
= 0.11mole
The number of moles of water found is 0.11mole
Answer:
The final pressure is 90.1 atm.
Explanation:
Assuming constant temperature, we can solve this problem by using <em>Boyle's Law</em>, which states:
Where in this case:
We <u>input the given data</u>:
- 159 atm * 463 L = P₂ * 817 L
And <u>solve for P₂</u>:
The final pressure is 90.1 atm.
Answer:
ΔH°rxn = - 433.1 KJ/mol
Explanation:
- CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)
⇒ ΔH°rxn = 4ΔH°HCl(g) + ΔH°CCl4(g) - 4ΔH°Cl2(g) - ΔH°CH4(g)
∴ ΔH°Cl2(g) = 0 KJ/mol.....pure element in its reference state
∴ ΔH°CCl4(g) = - 138.7 KJ/mol
∴ ΔH°HCl(g) = - 92.3 KJ/mol
∴ ΔH°CH4(g) = - 74.8 KJ/mol
⇒ ΔH°rxn = 4(- 92.3 KJ/mol) + (- 138.7 KJ/mol) - 4(0 KJ/mol) - (- 74.8 KJ/mol)
⇒ ΔH°rxn = - 369.2 KJ/mol - 138.7 KJ/mol - 0 KJ/mol + 74.8 KJ/mol
⇒ ΔH°rxn = - 433.1 KJ/mol
First, you need to convert kg to g.
So, 1 kg =1000g.
3.5 x 1000 = 3500g Ca(OH)2
We need to know the molar mass of Ca(OH)2.
Ca= 40.08 g
O=2(15.999)
H=2(1.0079)
Add them all together and you get 74.0938 g.
Put it in the formula from mass to moles.
# of moles = grams Ca(OH)2 x 1 mol Ca(OH)2
--------------------
molar mass Ca(OH)2
3500 g Ca(OH)2 x 1 mol Ca(OH)2
---------------------
74.0938 g Ca(OH)2
So divide 1/74.0938 and multiply by 3500.
You will get about 47.24 moles Ca(OH)2.
Hope this helps! :)
Answer:
what are the orbitals present in the fifth principal energy level
Explanation:
