The three patters are clumped, random, and uniform
Answer:
10
Explanation:
The unbalanced combustion reaction is shown below as:-
On the left hand side,
There are 3 hydrogen atoms and 1 nitrogen atom and 2 oxygen atoms
On the right hand side,
There are 1 nitrogen atom and 2 hydrogen atoms and 2 oxygen atoms
Thus,
Right side, must be multiplied by 6 to balance hydrogen.
Left side, must be multiplied by 4 to balance hydrogen.
Also, Right side, is multiplied by 4 so to balance nitrogen.
Left side, must be multiplied by 5 to balance the whole reaction.
Thus, the balanced reaction is:-
Sum of Coefficient of product - 4 + 6 = 10
Answer:
3 × 10⁴ kJ
Explanation:
Step 1: Write the balanced thermochemical equation
C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(g) ΔH = -2220 kJ
Step 2: Calculate the moles corresponding to 865.9 g of H₂O
The molar mass of H₂O is 18.02 g/mol.
865.9 g × 1 mol/18.02 g = 48.05 mol
Step 3: Calculate the heat produced when 48.05 moles of H₂O are produced
According to the thermochemical equation, 2220 kJ of heat are evolved when 4 moles of H₂O are produced.
48.05 mol × 2220 kJ/4 mol = 2.667 × 10⁴ kJ ≈ 3 × 10⁴ kJ
Answer:
- NH₃ is the limiting reactant.
- Theoretical yield = 120 kg
Explanation:
- 2NH₃ (aq) + CO₂ (aq) → CH₄N₂O (aq) + H₂O (l)
First we <u>convert the given masses of reactants to moles</u>, using their <em>respective molar masses</em>:
- 68.2 kg NH₃ ÷ 17 kg/kmol = 4.01 kmol NH₃
- 105 kg CO₂ ÷ 44 kg/kmol = 2.39 kmol CO₂
2.39 kmol of CO₂ would react completely with (2.39 * 2) 4.78 kmol of NH₃. There are not as many NH₃ kmoles so <u>NH₃ is the limiting reactant.</u>
We <u>calculate how much urea would form with a 100% yield</u>, using the <em>moles of limiting reactant</em>:
- 4.01 kmol NH₃ * = 2.00 kmol CH₄N₂O
We <u>convert that amount to kg</u>:
- 2.00 kmol CH₄N₂O * 60 kg/kmol = 120 kg CH₄N₂O
Finally we <u>calculate the percent yield</u>:
- 87.5 kg / 120 kg * 100% = 72.9 %
Answer:
P1V1=P2V2
Explanation:
because volume is inversely proportional to pressure provided that temperature remains constant