Answer:
= 913.84 mL
Explanation:
Using the combined gas laws
P1V1/T1 = P2V2/T2
At standard temperature and pressure. the pressure is 10 kPa, while the temperature is 273 K.
V1 = 80.0 mL
P1 = 109 kPa
T1 = -12.5 + 273 = 260.5 K
P2 = 10 kPa
V2 = ?
T2 = 273 K
Therefore;
V2 = P1V1T2/P2T1
= (109 kPa × 80 mL × 273 K)/(10 kPa× 260.5 K)
<u>= 913.84 mL</u>
NaOH+HCl-> NaCl+H2O
1 mole of NaOH
1 mole of HCl.
To calculate volume of NaOH
CaVa/CbVb= Na/Nb
Where Ca=2M
Cb=1M
Va=200cm³
Vb=xcm³
Substitute into the equation.
2×200/1×Vb=1/1
400/Vb=1/1
Cross multiply
Vb×1=400×1
Vb=400cm³
To calculate the mass of sodium chloride, NaCl from the neutralization rxn.
Mole of NaCl=1
Molar mass of NaCl= 23+35.5=58.5
Mass=xgrammes.
Mass of NaCl=Number of moles × Molar mass.
Substitute
Mass of NaCl= 1×58.5
=58.5g
This is what I could come up with.
Answer:
The amount of energy released from the combustion of 2 moles of methae is 1,605.08 kJ/mol
Explanation:
The chemical reaction of the combustion of methane is given as follows;
CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)
Hence, 1 mole of methane combines with 2 moles of oxygen gas to form 1 mole of carbon dioxide and 2 moles of water vapor
Where:
CH₄ (g): Hf = -74.6 kJ/mol
CO₂ (g): Hf = -393.5 kJ/mol
H₂O (g): Hf = -241.82 kJ/mol
Therefore, the combustion of 1 mole of methane releases;
-393.5 kJ/mol × 1 + 241.82 kJ/mol × 2 + 74.6 kJ/mol = -802.54 kJ/mol
Hence the combustion of 2 moles of methae will rellease;
2 × -802.54 kJ/mol or 1,605.08 kJ/mol.
Answer:
H₂SO₄ (aq) + 2LiOH (aq) ⇒ Li₂SO₄ (aq) + 2H₂O (l)
Explanation:
This is an acid-base reaction, so we know the products are going to a salt/ionic compound and water.
