Answer:
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Answer:
The answer to your question is 25.9 g of KCl
Explanation:
Data
Grams of KCl = ?
Volume = 0.75 l
Molarity = 1 M
Formula
Solve for number of moles
Substitution
Number of moles = 1 x 0.75
Simplification
Number of moles = 0.75 moles
Molecular mass KCl = 39 + 35.5 = 34.5
Use proportions to find the grams of KCl
34.5 g of KCl ---------------- 1 mol
x ---------------- 0.75 moles
x = (0.75 x 34.5) / 1
x = 25.9 g of KCl
Answer:
2Na2O2+2H2O⟶O2+4NaOH
2×78g 32g
156g of Na2O2 produces 32g of O2,
12g of Na2O2 produces =15632×12=10.66g.
Density of O2 at NTP=1.428g/mL.
DensityMass=Vol.
1.42810.66=7.46mL
Vol. of O2 at NTP is 7.46mL.
Explanation:
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Answer:
8.33 hours
Explanation:
In order to solve this problem, we must apply Graham's law of diffusion in gases. Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of its vapour density. For two gases we can write;
R1/R2=√d2/d1
Where;
R1= rate of diffusion of hydrogen
R2= rate diffusion of unknown gas
d1= vapour density of hydrogen
d2= vapour density of the unknown gas
Volume of hydrogen gas = 360cm^3
Time taken for hydrogen gas to diffuse= 1 hour =3600 secs
R1 = 360 cm^3/3600 secs = 0.1 cm^3 s-1
Vapour density of unknown gas = 25
Vapour density of hydrogen = 1
Substituting values,
0.1/R2 = √25/1
0.1/R2 = 5/1
5R2 = 0.1 × 1
R2 = 0.1/5
R2= 0.02 cm^3s-1
Volume of unknown gas = 600cm^3
Time taken for unknown gas to diffuse= volume of unknown gas/ rate of diffusion of unknown gas
Time taken for unknown gas to diffuse= 600/0.02
Time= 30,000 seconds or 8.33 hours
