Answer:
Explanation:
Hello!
In this case, since the molecular formula of glycine is C₂H₅NO₂, we realize that the molar mass is 75.07 g/mol; thus, the moles in 130.0 g of glycine are:
Furthermore, we can notice 75.07 grams of glycine contains 14.01 grams of nitrogen; thus, the percent nitrogen turns out:
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At STP volume is 22.4 L
Molar mass NO₂ = 46.0 g/mol
1 mole ---------- 22.4 L
? mole ---------- 11.4 L
moles = 11.4 * 1 / 22.4
moles = 11.4 / 22.4
= 0.5089 moles of NO₂
Mass NO₂ :
moles NO₂ * molar mass
0.5089 * 46.0
= 23.4094 g of NO₂
hope this helps!
Answer:
V₂ = 1070 mL or 1.07 L
Solution:
Data Given;
P₁ = 1170 mmHg
V₁ = 915 mL
T₁ = 24 °C + 273 K = 297 K
P₂ = 842 mmHg
V₂ = ?
T₂ = - 23 °C + 273 K = 250 K
According to Ideal gas equation,
P₁ V₁ / T₁ = P₂ V₂ / T₂
Solving for V₂,
V₂ = P₁ V₁ T₂ / P₂ T₁
Putting Values,
V₂ = (1170 mmHg × 915 mL × 250 K) ÷ (842 mmHg × 297 K)
V₂ = 1070 mL or 1.07 L
Significant figures communicates the level of precision in measurements.
