I would go with C because that seems like the best answer choice
The answer would be 118.68 g.
Explanation for this is:4 moles of NH3 give 4 moles of NO2
so 1mole of NH3 will give 1 mole of NO2
43.9 grams of NH3 contains 2.58 moles
so 2.58 moles will be produced of NO2
which is 118.7 grams this the amount of oxygen that is used.
Answer:
a0 = 2
a1= 9
a2= 6
a3= 8
Explanation:
The equation for the reaction is;
C3H7OH + O2 → CO2 + H2O
To balance the chemical equation we introduce coefficients;
Therefore the balanced chemical equation will be;
2C3H7OH + 9O2 → 6CO2 + 8H2O
Chemical equations are balanced to ensure the law of conservation of mass is obeyed, such that the mass of the reactants is equivalent to that of the products.
15 grams of NH3 can be dissolved
<h3>Further explanation</h3>
Given
50 grams of water at 50°C
Required
mass of NH3
Solution
Solubility is the maximum amount of a substance that can dissolve in some solvents. Factors that affect solubility
- 1. Temperature:
- 2. Surface area:
- 3. Solvent type:
- 4. Stirring process:
We can use solubility chart (attached) to determine the solubility of NH3 at 50°C
From the graph, we can see that the solubility of NH3 in 100 g of water at 50 C is 30 g
So that the solubility in 50 grams of water is:
= 50/100 x 30
= 15 grams
Answer:
Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em> by adding 1 mol of
to 1 kg of water.
Explanation:
1) Moles of NaCl ,
Mass of water = m= 1 kg = 1000 g
Moles of water = 
Vapor pressure of the solution = 
Vapor pressure of the pure solvent that is water = 
Mole fraction of solute(NaCl)= 



The vapor pressure for the NaCl solution at 17.19 Torr.
2) Moles of sucrose ,
Mass of water = m = 1 kg = 1000 g
Moles of water = 
Vapor pressure of the solution = 
Vapor pressure of the pure solvent that is water = 
Mole fraction of solute ( glucose)= 



The vapor pressure for the glucose solution at 17.19 Torr.
p = p' = 17.19 Torr
Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent by adding 1 mol of
to 1 kg of water.