First write the balanced equation of this reaction:
2H2 + O2 —> 2H2O
mol of H2= 0.60 gH2/2.02 gH2 = 0.297 mol
There are 2 mol of H2 for every 2 mol of H2O so the number of mol of H2 is equal to the number of mol of H2O.
g of H2O = 0.297 mol H2O • 18.02 gH2O = 5.35 g H2O
Do the same thing for O2:
mol of O2 = 4.8 gO2/32.0 gO2 = 0.15 mol of O2
There is 1 mol of O2 for every 2 mol of H2O so multiply 0.15 • 2 to get the number of mol of H2O
g of H2O = 0.30 mol H2O • 18.02 gH2O = 5.41 g H2O
The correct answer is 5.35 g H2O (or 5.4 g if checking significant figures) because O2, in this case, is the limiting reactant of this reaction.
<u>Answer:</u> No crystals of potassium sulfate will be seen at 0°C for the given amount.
<u>Explanation:</u>
We are given:
Mass of potassium nitrate = 47.6 g
Mass of potassium sulfate = 8.4 g
Mass of water = 130. g
Solubility of potassium sulfate in water at 0°C = 7.4 g/100 g
This means that 7.4 grams of potassium sulfate is soluble in 100 grams of water
Applying unitary method:
In 100 grams of water, the amount of potassium sulfate dissolved is 7.4 grams
So, in 130 grams of water, the amount of potassium sulfate dissolved will be 
As, the soluble amount is greater than the given amount of potassium sulfate
This means that, all of potassium sulfate will be dissolved.
Hence, no crystals of potassium sulfate will be seen at 0°C for the given amount.
Answer:
He put segregationists in charge of federal agencies...
Explanation:
...
47% yield.
First, let's determine how many moles of ethane was used and how many moles of CO2 produced. Start with the respective atomic weights.
Atomic weight carbon = 12.0107
Atomic weight hydrogen = 1.00794
Atomic weight oxygen = 15.999
Molar mass C2H6 = 2 * 12.0107 + 6 * 1.00794 = 30.06904 g/mol
Molar mass CO2 = 12.0107 + 2 * 15.999 = 44.0087 g/mol
Moles C2H6 = 8 g / 30.06904 g/mol = 0.266054387 mol
Moles CO2 = 11 g / 44.0087 g/mol = 0.249950578 mol
Looking at the balanced equation, for every 2 moles of C2H6 consumed, 4 moles of CO2 should be produced. So at 100% yield, we should have 0.266054387 / 2 * 4 = 0.532108774 moles of CO2. But we only have 0.249950578 moles, or 0.249950578 / 0.532108774 = 0.46973587 =
46.973587% of what was expected.
Rounding to 2 significant figures gives 47% yield.
