Partial pressure of gas A is 1.31 atm and that of gas B is 0.44 atm.
The partial pressure of a gas in a mixture can be calculated as
Pi = Xi x P
Where Pi is the partial pressure; Xi is mole fraction and P is the total pressure of the mixture.
Therefore we have Pa = Xa x P and Pb = Xb x P
Let us find Xa and Xb
Χa = mol a/ total moles = 2.50/(2.50+0.85) = 2.50/3.35 = 0.746
Xb = mol b/total moles = 0.85/(2.50+0.85) = 0.85/3.35 = 0.254
Total pressure P is given as 1.75 atm
Pa = Xa x P = 0.746 x 1.75 = 1.31atm
Partial pressure of gas A is 1.31 atm
Pb = Xb x P = 0.254 x 1.75 = 0.44atm
Partial pressure of gas B is 0.44 atm.
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In the reaction Sn(s) + 2H+(aq) → Sn2+ (aq) + H2(g)
from this reaction, we get that Sn loses from 0 to 2 electrons so it's oxidized So it is the reducing agent.
and H gains from 0 to 1 electrons so, it's reduced so ∴ it is the oxidizing agent
Answer is: mass of unused sulfur is 5.87 grams.
Balanced chemical reaction: C + 2S → CS₂.
m(C) = 12.0 g; mass of carbon.
m(S) = 70.0 g; mass of sulfur.
n(C) = m(C) ÷ M(C).
n(C) = 12 g ÷ 12 g/mol.
n(C) = 1 mol; amount of substance.
n(S) = m(S) ÷ M(S).
n(S) = 70 g ÷ 32.065 g/mol.
n(S) = 2.183 mol.
From chemical reaction: n(C) : n₁(S) = 1 : 2.
n₁(S) = 1 mol · 2 = 2 mol.
Δn(S) = n(S) - n₁(S).
Δn(S) = 2.183 mol - 2 mol.
Δn(S) = 0.183 mol; amount of unused sulfur.
Δm(S) = 0.183 mol · 32.065 g/mol.
Δm(S) = 5.87 g.
(H+)(OH-) = Kw
kw= 1 x10^-14
OH-= 1 x10 ^-11
(H+)= KW / OH-
concentration of H+ = (1x10^-14) /.(1 x 10 ^-11) = 1 x10 ^-3
Ph= -log (H+)
PH=-log ( 1 x 10 ^-3) = 3 therefore the solution is acidic since the PH less than 7
Actually mass and weight are two different things but most people did not understand the difference between them. And they are used synonymously on the earth. Mass is the measure of the amount of matter and weight is the measure of how the force of gravity acts on the mass that is Weight = mass x force of gravity. And they are also directly proportional to each other that is also the reason both terms are used synonymously.
