Answer:
by using ideal gas law
Explanation:
ideal gas law:
PV=nRT
where:
P is pressure measured in Pascal (pa)
V is volume measured in letters (L)
n is number of moles
R is ideal gas constant
T is temperature measured in Kelvin (K)
by applying the given:
P(initial) V(initial)=nRT(initial)
P(final) V(final)=nRT(final)
nR is constant in both equations since same gas
then,
P(initial) V(initial) / T(initial) = P(final) V(final) / T(final)
then by crossing multiply both equations
V (final)= { (P(initial) V(initial) / T(initial)) T(final) } /P (final)
P(initial)=P(final)= 1 atm = 101325 pa
V(initial)= 6 L
T(initial) = 28°c = 28+273 kelvin
T(final) = 39°c = 39+273 kelvin
by substitution
V(final) = 6.21926 L
Answer:
E° = 0.65 V
Explanation:
Let's consider the following reductions and their respective standard reduction potentials.
Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V
Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:
Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V
The overall cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.80 V - 0.15 V = 0.65 V
Answer:
Explanation:
Hello,
At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:
Once we've got the moles we compute the final volume via:
Thus, the molarity of the sodium atoms turn out into:
Now, we perform the same procedure but now for the bromide ions:
Finally, its molarity results:
Best regards.
Answer: 0.0944 gram of H2
Explanation:
Raising the T from 25 C (298 K) to 700 C (973 K) increases the pressure of each gas by:
2.0 atm x (973 K / 298 K) = 6.53 atm
Where
Kc = Kp because the moles of product equals the moles of reactants.
At equilibriuim, the amounts are
P(H2) = 6.53 - x
P(CO2) = 6.53 - x
P(H2O) = x
P(CO) = x
Kc = Kp = .534 = (x)(x) / [(6.53 - x)(6.53 - x)]
Take the square root of each side
(.534)^0.5 = x / (6.53 - x)
x = 0.731 (6.53 - x)
x = 4.77 - 0.731x
1.731x = 4.77
x = 4.77 / 1.731 = 2.76 atm
P(H2) at equilibriuim = 6.53 - 2.76 = 3.77 atm
P(CO2) at equilibrium = 6.53 - 2.76 = 3.77 atm
PV = nRT
n = PV/RT = [(3.77 atm)(1.00 L)] / [(0.08206 L atm/K mol)(973 K)] = 0.0472 mol H2
0.0472 mol H2 x (2.00 g / 1.00 mol) = 0.0944 g
