Hey there!
I believe the answer is Combination (or Synthesis) Reaction.
Answer:
100.8 °C
Explanation:
The Clausius-clapeyron equation is:
-Δ
Where 'ΔHvap' is the enthalpy of vaporization; 'R' is the molar gas constant (8.314 j/mol); 'T1' is the temperature at the pressure 'P1' and 'T2' is the temperature at the pressure 'P2'
Isolating for T2 gives:

(sorry for 'deltaHvap' I can not input symbols into equations)
thus T2=100.8 °C
Answer: The mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams
Explanation:
Depression in freezing point is given by:
= Depression in freezing point
i= vant hoff factor = 1 (for non electrolyte)
= freezing point constant =
m= molality =


Let Mass of solute (KBr) = x g
Thus the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams
Answer:
[O₂(g)] = 0.0037M
Explanation:
2SO₂(g) + O₂(g) => 2SO₃(g)
Conc: [SO₂(g)] [O₂(g)] [SO₃(g)] and [SO₂(g)] = [SO₃(g)]
Kc = [SO₃(g)]²/[O₂(g)][SO₂(g)]² => Kc = 1/[O₂(g)] = 270 if [SO₂(g)] = [SO₃(g)]
∴ [O₂(g)] = (1/270)M = 0.0037M
Answer:
1.25 moles
Explanation:
First, we need to balance the equation. Essentially, this means making sure we have the same number of each atom on each side.
On the left side, we currently have:
- 1 Co atom
- 2 F aomts
On the right side, we have:
- 1 Co atom
- 3 F atoms
To balance it, add a 2 to Co on the left, 3 to F2 on the left, and 2 to CoF3 on the right:
→ 
Now, we have 1.25 moles of Co, and since the ratio between Co and CoF3 is 1:1, we also have 1.25 moles of CoF3.
Thus, the answer is 1.25 moles.