2.2311 moles of gas are there in a 50. 0 l container at 22. 0 °c and 825 torrs.
<h3>What is an ideal gas?</h3>
An Ideal gas is a hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas laws exactly.
Assuming the gas is ideal, we can solve this problem by using the following equation:
PV = nRT
Where:
P = 825 torr ⇒ 825 / 760 = 1.08 atm
V = 50 L
n = ?
R = 0.082 atm·L·mol⁻¹·K⁻¹
T = 22 °C ⇒ 22 + 273.16 = 295.16 K
We input the data:
1.08 atm x 50 L = n x 0.082 atm·L·mol⁻¹·K⁻¹ x 295.16 K
And solve for n:
24.20312
n = 2.2311 mol
Hence, 2.2311 moles of gas are there in a 50. 0 L container at 22. 0 °c and 825 torrs.
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For question 1:
A. Carbon, hydrogen, oxygen
Calcium, carbon, oxygen
Carbon, hydrogen
Carbon, hydrogen, oxygen
Silicon, oxygen
B. Carbon: 12, hydrogen: 22, oxygen: 11
Calcium: 1, Carbon: 1, oxygen: 3
Carbon:1, Hydrogen: 4
Carbon: 3, hydrogen: 8, Oxygen: 1
Silicon: 1 oxygen: 2
C. Table sugar: 45
Marble: 5
Natural gas: 5
Rubbing alcohol: 12
Glass: 3
For question 2: N2O
Chemical reaction: SO₄²⁻ + Ba²⁺ → BaSO₄.
m(sample) = 1,543 g.
m(BaSO₄) = 0,2243 g.
n(BaSO₄) = m(BaSO₄) ÷ M(BaSO₄).
n(BaSO₄) = 0,2243 g ÷ 233,4 g/mol.
n(BaSO₄) = 0,00096 mol.
n(BaSO₄) = n(SO₄²⁻).
ω(SO₄²⁻) = m(SO₄²⁻) ÷ m(sample).
ω(SO₄²⁻) = 0,00096 mol · 96 g/mol ÷ 1,543 g.
ω(SO₄²⁻) = 0,059 = 5,9%.
The reactant is Mercury (II) Oxide while the products are Mercury and Oxygen separately.
This is because the reactants are typically always on the left side of the yields symbol. In this decomposition reaction, it would still be the same as at the end of the reaction, there were to products produced: Mercury and Oxygen.
Products tend to always be on the right side of the yields symbol, they're what comes out of a reaction no matter what type.
Hope this helps!
Unrestricted populations of organisms experience Exponential growth.
As a population reaches is carrying capacity there is an increase in competition for All of the above.
Hope I helped.
