350.0-mL of 0.50 M hydrogen sulfate solution is reacted with 15.0 grams of sodium hydroxide. What volume of water will be produc
1 answer:
The volume of water that will be produced from the reaction will be 6.3 mL
<h3>Stoichiometric calculation</h3>From the equation of the reaction:
The mole ratio of hydrogen sulfate to sodium hydroxide is 1:2.
Mole of hydrogen sulfate = 0.50 x 350/1000 = 0.175 moles
Mole of 15 grams sodium hydroxide = 15/40 = 0.375 moles
Thus, hydrogen sulfide is the limiting reagent.
Mole ratio of hydrogen sulfide to water = 1:2.
Equivalent mole of water = 0.175 x 2 = 0.35 moles
Mass of 0.35 moles of water = 0.35 x 18 = 6.3 grams.
1 gram of water = 1 ml.
Thus, 6.3 grams of water will be equivalent to 6.3 mL
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Answer:
Part A is just T2 = 58.3 K
Part B ∆U = 10967.6 x C You can work out C
Part C
Part D
Part E
Part F
Explanation:
P = n (RT/V)
V = (nR/P) T
P1V1 = P2V2
P1/T1 = P2/T2
V1/T1 = V2/T2
P = Pressure(atm)
n = Moles
T = Temperature(K)
V = Volume(L)
R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.
bar = 0.986923 atm
N = 14g/mol
N2 Molar Mass 28g
n = 3.5 mol N2
T1 = 350K
P1 = 1.5 bar = 1.4803845 atm
P2 = 0.25 bar = 0.24673075 atm
Heat Capacity at Constant Volume
Q = nCVΔT
Polyatomic gas: CV = 3R
P = n (RT/V)
0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))
V = (nR/P) T
V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K
V = (0.28721/1.4803845) x 350
V = 0.194 x 350
V = 67.9036 L
So V1 = 67.9036 L
P1V1 = P2V2
1.4803845 atm x 67.9036 L = 0.24673075 x V2
100.52343693 = 0.24673075 x V2
V2 = P1V1/P2
V2 = 100.52343693/0.24673075
V2 = 407.4216 L
P1/T1 = P2/T2
1.4803845 atm / 350 K = 0.24673075 atm / T2
0.00422967 = 0.24673075 /T2
T2 = 0.24673075/0.00422967
T2 = 58.3 K
∆U= nC ∆T
Polyatomic gas: C = 3R
∆U= nC ∆T
∆U= 28g x C x (350K - 58.3K)
∆U = 28C x 291.7
∆U = 10967.6 x C
I have provided the steps and solution within the attachment. The pH of the solution would be 12.30, this indicates that the solution is basic, as a higher value of pH indicates presence of more hydroxide ions and less of hydrogen ions in the solution.
