Answer:
= 1 X 10⁻⁻¹²M
Explanation:
At 25°C & 1atm [H⁺][OH⁻] = 1 x 10⁻¹⁴ => [H⁺] = 1 X 10⁻¹⁴/[OH⁻] = 1 X 10⁻¹⁴/1 X 10⁻²
= 1 X 10⁻⁻¹²M
Answer:
0.025g
Explanation:
Firstly, we need to write a balanced chemical equation.
Ca + 2H2O —-> Ca(OH)2 + H2
From the chemical equation, we can see that one mole of calcium metal yielded one mole of hydrogen gas.
Now, we need to know the actual number of moles of hydrogen gas given off.
The number of moles is the mass divided by the atomic mass. The atomic mass of calcium is 40g/mol. The number of moles is thus 0.5/40 = 0.0125 mole
Since the mole ratio is 1:1, the number of moles of hydrogen gas produced too is 0.0125 moles.
We now Proceed to get the mass of the hydrogen gas produced. This is the number of moles of hydrogen multiplied by the molar mass of the diatomic hydrogen gas. The molar mass of the diatomic hydrogen gas is 2g/mol.
The amount given off is thus 2 * 0.0125 = 0.025g
The concentration of the solution is 4.25 M
Explanation
molarity=moles/volume in liters
moles = mass/molar mass
molar mass of HF = 19 + 1 = 20 g/mol
moles is therefore = 17.0 g/ 20 g/mol = 0.85 moles
volume in liters = 2 x10^2ml/1000 = 0.2 liters
therefore molarity = 0.85/0.2 = 4.25 M
Answer:
5.97 mol
Explanation:
To find the number of moles when given the mass of a substance, we divide the mass of the sample by its molar mass.
so, we get,
nN203 = 454 g / 76.01 g /mol
= 5.97 mol
Answer:
The answer to your question is 47.44 g of Oxygen
Explanation:
Data
mass of Ammonia = 14.4 g
mass of Oxygen = ?
Balanced chemical reaction
4NH₃ + 7O₂ ⇒ 4NO₂ + 6H₂O
Process
1.- Calculate the molar mass of Ammonia
NH₃ = 4[(1 x 14) + (3 x 1)] = 4[14 + 3] = 4[17] = 68 g
2.- Calculate the molar mass of Oxygen
O₂ = 7[16 x 2] = 7[32] = 224 g
3.- Use proportions to calculate the mass of Oxygen
68g of NH₃ --------------------- 224 g of O₂
14.4 g of NH₃ ----------------- x
x = (14.4 x 224) / 68
x = 3225.6/ 68
x = 47.44 g
