Answer:
0.189 g.
Explanation:
- This problem is an application on <em>Henry's law.</em>
- Henry's law states that the solubility of a gas in a liquid is directly proportional to its partial pressure of the gas above the liquid.
- Solubility of the gas ∝ partial pressure
- If we have different solubility at different pressures, we can express Henry's law as:
<em>S₁/P₁ = S₂/P₂,</em>
S₁ = 0.0106/0.792 = 0.0134 g/L and P₁ = 0.321 atm
S₂ = ??? g/L and P₂ = 5.73 atm
- So, The solubility of the gas at 5.73 atm (S₂) = S₁.P₂/P₁ = (0.0134 g/L x 5.73 atm) / (0.321 atm) = 0.239 g/L.
<em>The quantity in (g) = S₂ x V = (0.239 g/L)(0.792 L) = 0.189 g.</em>
<em></em>
Answer:
The amount in grams of hydrogen gas produced is 0.551 grams
Explanation:
The parameters given are;
Number of atoms of potassium, aₙ = 3.289 × 10²³ atoms
Chemical equation for the reaction is given as follows;
2K + 2H₂O 
KOH + H₂
Avogadro's number, 
, regarding the number of molecules or atom per mole is given s follows;
= 6.02 × 10²³ atoms/mole
Therefore;
The number of moles of potassium present = 3.289 × 10²³/(6.02 × 10²³) = 0.546 moles
2 moles of potassium produces one mole of hydrogen gas, therefore;
1 moles of potassium produces 1/2 mole of hydrogen gas, and 0.546 moles of potassium will produce 0.546/2 moles of hydrogen which is 0.273 moles of hydrogen gas
The molar mass of hydrogen gas = 2.016 grams
Therefore, 0.273 moles will have a mass of 0.273×2.016 = 0.551 grams.
The amount in grams of hydrogen gas produced = 0.551 grams.
Molarity of Ag+ is less than the molar solubility thus ppt will not occur.
Balanced reaction-:
<h3>2AgNO3(aq)+K2CrO4(aq)→Ag2CrO4(s)+2KNO3(aq)</h3>
Moles of AgNO3=mass(g)molar mass (g/mol) =2.7×10−5g / 169.86 gmol
=1.589⋅10^−7 mol
Molarity of Ag+=moles of solute(L)=1.589⋅10−7 mol0.015 L=1.059⋅10−5M
Ksp of Ag2CrO4
=[Ag+]2[CrO42−]
1.2⋅10−12=[2s]2[s]
4s3=1.2⋅10−12
s=6.69⋅10−5 M
Molarity of Ag+ is less than the molar solubility thus ppt will not occur.
<h3>What is the molarity calculation formula?</h3>
The volume of solvent required to dissolve the provided solute is multiplied by the ratio of the moles of the solute whose molarity has to be computed. (M=frac{n}{V}) The molality of the solution that needs to be computed in this case is M. n is the solute's molecular weight in moles.
Learn more about Molarity:
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