Answer:
D. (16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%
Explanation:
Step 1: Detemine the mass of O in SO₂
There are 2 atoms of O in 1 molecule of SO₂. Then,
m(O) = 2 × 16.0 g = 16.0 g + 16.0 g = 32.0 g
Step 2: Determine the mass of SO₂
m(SO₂) = 1 × mS + 2 × mO = 1 × 32.1 g + 2 × 16.0 g = 32.1 g + 16.0 g + 16.0 g = 64.1 g
Step 3: Detemine the mass percent of oxygen in SO₂
We will use the following expression.
m(O)/m(SO₂) × 100%
(16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%
Answer:
15.0 L
Explanation:
To find the volume, you need to use the Ideal Gas Law:
PV = nRT
In this equation,
-----> P = pressure (mmHg)
-----> V = volume (L)
-----> n = moles
-----> R = Ideal Gas constant (62.36 L*mmHg/mol*K)
-----> T = temperature (K)
To calculate the volume, you need to (1) convert grams C₄H₁₀ to moles (via the molar mass), then (2) convert the temperature from Celsius to Kelvin, and then (3) calculate the volume (via the Ideal Gas Law).
Molar Mass (C₄H₁₀): 4(12.011 g/mol) + 10(1.008 g/mol)
Molar Mass (C₄H₁₀): 58.124 g/mol
32 grams C₄H₁₀ 1 moles
------------------------- x ----------------------- = 0.551 moles C₄H₁₀
58.124 grams
P = 728 mmHg R = 62.36 L*mmHg/mol*K
V = ? L T = 45.0 °C + 273.15 = 318.15 K
n = 0.551 moles
PV = nRT
(728 mmHg)V = (0.551 moles)(62.36 L*mmHg/mol*K)(318.15 K)
(728 mmHg)V = 10922.7632
V = 15.0 L
Answer:
+125.4 KJmol-1
Explanation:
∆H C4H10(g) = -2877.6kJ/mol
∆H C(s)=-393.5kJ/mol
∆H H2(g) = -285.8
∆H reaction= ∆Hproducts - ∆H reactants
∆H reaction= (-2877.6kJ/mol) - [4(-393.5kJ/mol) +5(-285.8)]
∆H reaction= +125.4 KJmol-1
B different skills as every subject varies from each other
