Answer:
Magnesium
0.003mole
Explanation:
The problem here entails we find the metal in the carbonate.
For group 2 member, let the metal = X;
The carbonate is XCO₃;
If we sum the atomic mass of the elements in the metal carbonate, we should arrive at 84g/mol
Atomic mass of C = 12g/mol
O = 16g/mol
Atomic mass of X + 12 + 3(16) = 84
Atomic mass of X = 84 - 60 = 24g/mol
The element with atomic mass of 24g is Magnesium
B.
Number of moles in 0.3g of CaCO₃:
Molar mass of CaCO₃ = 40 + 12 + 3(16) = 100g/mol
Number of moles = 
Number of moles = 
= 0.003mole
6 miles of H2O is equal to 12 Hydrogen molecules and 6 oxygen molecules. Equaling 18 in total.
There are 34 g of oxygen in the container.
We can use the<em> Ideal Gas Law</em> to solve this problem.
But 
, so
and
STP is 0 °C and 1 bar, so
Answer:
11.2 M → [HCl]
Explanation:
Solution density = Solution mass / Solution volume
35.38 % by mass, is the same to say 35.38 g of solute in 100 g of solution.
Let's determine the moles of our solute, HCl
35.38 g . 1 mol/36.45 g = 0.970 moles
Let's replace the data in solution density formula
1.161 g/mL = 100 g / Solution volume
Solution volume = 100 g / 1.161 g/mL → 86.1 mL
Let's convert the volume to L → 86.1 mL . 1L / 1000 mL = 0.0861 L
Molarity (M) → mol/L = 0.970 mol / 0.0861 L → 11.2 M
