Answer: Th enthalpy of combustion for the given reaction is 594.244 kJ/mol
Explanation: Enthalpy of combustion is defined as the decomposition of a substance in the presence of oxygen gas.
W are given a chemical reaction:
To calculate the enthalpy change, we use the formula:
This is the amount of energy released when 0.1326 grams of sample was burned.
So, energy released when 1 gram of sample was burned is = 
Energy 1 mole of magnesium is being combusted, so to calculate the energy released when 1 mole of magnesium ( that is 24 g/mol of magnesium) is being combusted will be:
Answer: The limiting reactant is Na
Explanation:
Answer:
NH3
Explanation:
2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l)
So for two moles of NH3 we need one mole of CO2. So let's count moles for each reagent.
n(NH3)=m(NH3)/M(NH3)=135700/17,03=7968.29 mol
n(CO2)=m(CO2)/M(CO2)=211400/44.01=4803.45 mol
From equation we have to divide n(NH3) by 2 because we need two equivalent per one CO2. That will be 3984.145. So the limiting agent is NH3 because it's not enough of it to react with all CO2
Answer:
1.8g
Explanation:
Initial volume = 43.5ml
Final volume = 49.4ml
Mass = 10.88g
Density = ?
Volume = Final volume - initial volume
= 49.4 - 43.5
= 5.9ml
Density = Mass/volume
Density = 10.88/5.9
= 1.8g/ml
