Answer:
C. One mole of water was produced from this reaction.
Explanation:
- From the balanced equation:
<em>2H₂ + O₂ → 2H₂O,</em>
It is clear that 2 mol of H₂ react with 1 mol of O₂ to produce 2 mol of H₂O.
- So, if one mole of hydrogen was used in this reaction:
1/2 mol of O₂ was used in the reaction, and
1 mol of water was produced from this reaction.
- So, the correct statement is:
<em>C. One mole of water was produced from this reaction.
</em>
Answer:
V₂ = 946.72 mL
Explanation:
Given data;
Initial pressure = 0.926 atm
Initial volume = 457 mL
Temperature = constant = 29.5°C
Final pressure = 0.447 atm
Final volume = ?
Solution:
The given problem will be solved through the Boyle's law,
Mathematical expression:
P₁V₁ = P₂V₂
P₁ = Initial pressure
V₁ = initial volume
P₂ = final pressure
V₂ = final volume
by putting values,
P₁V₁ = P₂V₂
0.926 atm × 457 mL = 0.447 atm × V₂
V₂ = 423.18 atm. mL/ 0.447 atm
V₂ = 946.72 mL
Answer:
True, in as far as greater magnitude = greater power.
Answer:
The enthalpy change during the reaction is -199. kJ/mol.
Explanation:
Mass of solution = m
Volume of solution = 100.0 mL
Density of solution = d = 1.00 g/mL
First we have to calculate the heat gained by the solution in coffee-cup calorimeter.
where,
m = mass of solution = 100 g
q = heat gained = ?
c = specific heat = 
= final temperature = 
= initial temperature = 
Now put all the given values in the above formula, we get:
Now we have to calculate the enthalpy change during the reaction.
where,
= enthalpy change = ?
q = heat gained = 2.242 kJ
n = number of moles fructose = 
Therefore, the enthalpy change during the reaction is -199. kJ/mol.
