Answer: 0.07868 mol H₂O
Explanation:
1) Chemical equation:
Cu₂O +H₂ → 2Cu + H₂O
2) mole ratios:
1 mol Cu₂O : 1 mol H₂ : 2 mol Cu : 1 mol H₂O
3) Convert 10.00 g of Cu to grams, using the atomic mass:
Atomic mass of Cu: 63.546 g/mol
number of moles = mass in grams / atomic mass = 10.00g / 63.546 g/mol
number of moles = 0.1574 mol
4) Use proportions
2mol Cu 0.1574 mol Cu
--------------- = ---------------------
1 mol H₂O x
⇒ x = 0.1574 mol Cu × 1 mol H₂O / 2mol Cu = 0.07868 mol H₂O
That is the answer
Answer:
A calculator has an endifiite shape because all of its atoms are touching each other
0.370 mol metal oxide = 55.45 g
<span>1 mol = 55.45/0.370 = 149.86 g </span>
<span>in 1 mol there are 3 mol O = 16 * 3 = 48 g of O </span>
<span>there is 48/149.86 * 100% O in the sample </span>
<span>the sample has 48/149.86 * 0.370 = 0.119 g O</span>
Answer:
V = 81.14 L
Explanation:
Given data:
Volume of gas = ?
Number of moles = 3.30 mol
Temperature of gas = 25°C
Pressure of gas = 0.995 atm
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Now we will convert the temperature.
25+273 = 298 K
now we will put the values in formula:
V = 3.30 mol 0.0821 atm.L/ mol.K 298 K / 0.995 atm
V = 80.74 L. atm / 0.995 atm
V = 81.14 L
Answer:
Approximately 75%.
Explanation:
Look up the relative atomic mass of Ca on a modern periodic table:
There are one mole of Ca atoms in each mole of CaCO₃ formula unit.
- The mass of one mole of CaCO₃ is the same as the molar mass of this compound:
. - The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element:
.
Calculate the mass ratio of Ca in a pure sample of CaCO₃:
.
Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be 
of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio 
:
.
In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:
.
