The answer is 267.93 g
Molar mass of CaBr2 is the sum of atomic masses of Ca and Br:
Mr(CaBr2) = Ar(Ca) + 2Ar(Br)
Ar(Ca) = 40 g/mol
Ar(Br) = 79.9 g/mol
Mr(CaBr2) = 40 + 2 * 79.9 = 199.8 g/mol
The percentage of Br in CaBr2 is:
2Ar(Br) / Mr(CaBr2) * 100 = 2 * 79.9 / 199.8 * 100 = 79.98%
Now make a proportion:
x g in 79.98%
335 g in 100%
x : 79.98% = 335 g : 100%
x = 79.98% * 335 g : 100%
x = 267.93 g
Answer:
194.6 mL of SO₂
Explanation:
The reaction that takes place is:
P₄S₃ + 6O₂(g) → P₄O₁₀ + 3SO₂(g)
<u>To solve this problem we need to use PV=nRT</u>, so first let's convert the given units:
- 23.8 °C → 23.8 + 273.15 = 296.95 K
- 747 torr → 747/760 = 0.983 atm
We need to calculate V, so in order to do that we calculate n, using the mass of the reactant (P₄S₃):
0.576 g P₄S₃ *
= 7.85 * 10⁻³ mol SO₂ = n
PV=nRT
0.983 atm * V = 7.85 * 10⁻³ mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 296.95 K
V = 0.1946 L
- Finally we convert L into mL:
0.1946 * 1000 = 194.6 mL
Answer:
A. Boiling point = 59 °C, Melting point = -7.2°C, triple point = -7.3°C
Explanation:
The answer is option E. All of the above