1) V(CH₄) = 0,376 L.T(CH₄) = 304 K.p(CH₄) = 1,5 atm 101325 Pa/atm = 151987,5 Pa = 151,9875 kPa.R = 8,314 J/K·mol.Use ideal gas law: p·V = n·R·T.n(CH₄) = p · V ÷ R · T.n(CH₄) = 151,9875 kPa · 0,376 L ÷ 8,314 J/K· mol · 304 K.n(CH₄) = 0,0226 mol.V(CH₄) = n(CH₄) · Vm.V(CH₄) = 0,0226 mol · 22,4 dm³/mol.V(CH₄) = 0,506 dm³ = 0,506 L.
2) V(SO₂) = 5,2 L.p(SO₂) = 45,2 atm = 45,2 atm · 101,325 kPa/atm = 4579,89 kPa.T(SO₂) = 293 K.R = 8,314 J/K·mol.Use ideal gas law: p·V = n·R·T.n(SO₂) = p · V ÷ R · T.n(SO₂) = 4579,89 kPa · 5,2 L ÷ 8,314 J/K· mol · 293 K.n(CH₄) = 9,77 mol.There is not enogh SO₂, 225 mol - 9,77 mol = 215,23 mol is needed.
3) p(He) = 3,50 atm · 101,325 kPa/atm = 354,63 kPa.V(He) = 4,00 L.n(He) = 0,410 mol.R = 8,314 J/K·mol.Use ideal gas law: p·V = n·R·T.T = p · V ÷ R · n.T(He) = 354,63 kPa · 4,00 L ÷ 8,314 J/K· mol · 0,410 mol.T(He) = 416,14 K.n - amount of substance.
4) p(Ar) = 1,00 atm · 101,325 kPa/atm = 101,325 kPa.V(Ar) = 3,4 L.T(Ar) = 263 K.R = 8,314 J/K·mol.Use ideal gas law: p·V = n·R·T.n(Ar) = p · V ÷ R · T.n(Ar) = 101,325 kPa · 3,4 L ÷ 8,314 J/K· mol · 263 K.n(Ar) = 0,157 mol.n(Ar) = 0,157 mol + 2,5 mol = 2,657 mol.p(Ar) = 2,657 mol · 8,314 J/K· mol · 263 K ÷ 3,4 L.p(Ar) = 1708,74 kPa.
Answer:
I would say that the answers are an increase in gases in the atmosphere that absorb heat an increase in surfaces that radiate energy into the atmosphere, and a decrease in the amount of water on Earth’s surface
Explanation:
Answer:b
Explanation:
The Kelvin degree is the same size as the Celsius degree; hence the two reference temperatures for Celsius, the freezing point of water (0°C), and the boiling point of water (100°C), correspond to 273.15°K and 373.15°K, respectively.
Answer:
Percentage abundance of B 10 is = 20 %
Percentage abundance of B 11 is = 80 %
Explanation:
The formula for the calculation of the average atomic mass is:
Given that:
Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.
For first isotope, B 10:
% = x %
Mass = 10.0129 u
For second isotope, B 11:
% = 100 - x
Mass = 11.0093 u
Given, Average Mass = 10.81 u
Thus,
Solving for x, we get that:
x = 20 %
Thus percentage abundance of B 10 is = 20 %
Percentage abundance of B 11 is = 100 - 20 % = 80 %
