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: The correct option is, 2.
Explanation:
Nuclear fission: It is defined as the reactions in which a larger unstable nuclei breaks into two or more smaller stable nuclei.
Nuclear fusion: It is defined as the reaction where two smaller nuclei combine together to form a larger nuclei.
Option 1 & option 2 shows fusion reaction and option 3 & option 4 shows fission reaction.
In option 1, helium itself is forming but in option 2, element is heavier than forming which is, neon.
Hence, the correct option is 2.
Molarity = (Mass/ molar mass) x (1/ volume of solution in Litres)
Mass = Molarity x molar mass x volume of solution in Litres
Molarity of Tris = 100 mM = 0.1 M
volume of Tris sol. = 100 mL = 0.1 L
molar mass of Tris = 121.1 g/mol
Hence,
mass of Tris = Molarity of Tris x molar mass ofTris x volume of Tris solution
= 0.1 M x 121.1 g/mol x 0.1 L
= 1.211 g
mass of Tris = 1.211 g
