Answer:
25 grams of Mg(OH)2 will be produced by 14.424 gram of Mg3N2
Explanation:
The balanced equation is
Mg3N2 + 6H2O -> 3Mg(OH)2 + 2NH3
Molecular weight of magnesium nitride = 100.9494 g/mol
Molecular weight of magnesium hydroxide = 58.3197 g/mol
one mole of Mg3N2 produces three moles of 3Mg(OH)2
100.9494 g/mol of Mg3N2 produces 3* 58.3197 g/mol of Mg(OH)2
1 gram of Mg3N2 produces
grams of Mg(OH)2
Or 1.733 grams of Mg(OH)2 will be produced by 1 gram of Mg3N2
25 grams of Mg(OH)2 will be produced by 14.424 gram of Mg3N2
Answer:
D. (16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%
Explanation:
Step 1: Detemine the mass of O in SO₂
There are 2 atoms of O in 1 molecule of SO₂. Then,
m(O) = 2 × 16.0 g = 16.0 g + 16.0 g = 32.0 g
Step 2: Determine the mass of SO₂
m(SO₂) = 1 × mS + 2 × mO = 1 × 32.1 g + 2 × 16.0 g = 32.1 g + 16.0 g + 16.0 g = 64.1 g
Step 3: Detemine the mass percent of oxygen in SO₂
We will use the following expression.
m(O)/m(SO₂) × 100%
(16.0 g + 16.0 g) × 100% / (32.1 g + 16.0 g + 16.0 g) = 49.9%
Answer:
Explanation:
Hello,
In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):
Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:
Best regards.
Answer:
Molality = 1.13 m
Explanation:
Molality is defined as the moles of the solute present in 1 kilogram of the solvent.
Given that:
Mass of 
= 26.5 g
Molar mass of = 32.04 g/mol
The formula for the calculation of moles is shown below:
Thus,
Mass of water = 735 g = 0.735 kg ( 1 g = 0.001 kg )
So, molality is:
<u>Molality = 1.13 m</u>
