Explanation:
It is known that 1 SCF produces approximately 1000 Btu of thermal energy.
As it is not mentioned for how many hours the gas is used in this process. Therefore, we assume that the total number of hours natural gas used in this process are as follows.
= 8760 hours
Now, we will calculate the annual cost of natural gas used in the process as follows.
= 555384000 SCF
Hence, annual cost of natural gas used in this process = loss of thermal energy
This will be equal to,
= 555,384,000,000 BTU
Thus, we can conclude that the annual cost of natural gas used in the process is 555,384,000,000 BTU.
Hey there!:
Temperature in kelvin : 45 + 273 => 318 K
Number of moles H2:
p * V = n * R * T
4.46 * 0.579 = n * 0.082 * 318
2.58234 = n * 26.076
n = 2.58234 / 26.076
n = 0.09903 moles of H2O
Given the reaction:
XeF6(s) + 3 H2(g) → Xe(g) + 6 HF(g)
Molar mass XeF6 = 245.3 g/mol
0.09903 mol H2O * ( 1 mol XeF6 / 3 mol H2 ) * 245.3
= 8.09 g of XeF6
Hope that helps
Answer:
- <u><em>The volume of CO₂(g) produced at STP when 0.05 moles of C₂H₄(g) was burnt in O₂(g) is 2.24dm</em></u><em><u>³</u></em>
Explanation:
The question is incomplete.
This is the complete question:
<em>Consider the reaction by the following equation:</em>
<em />
<em> C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g)</em>
<em />
<em>The volume of CO₂(g) produced at s.t.p when 0.05 moles of C₂H₄(g) was burnt in O₂(g) is ________</em>
<em />
<em>[Molar Volume of gas = 22.4dm³]</em>
<em />
<em> A. 1.12dm³</em>
<em> B. 2.24dm³</em>
<em> C. 3.72dm³</em>
<em> D. 4.48dm³</em>
<h2>Solution</h2>
<u />
<u>1. Write the mole ratio between CO₂(g) and C₂H₄(g)</u>
- 1 mol C₂H₄(g) : 2 mol CO₂(g)
<u>2. Multiply the 0.05 moles of C₂H₄(g) by the mole ratio</u>
- 0.05 mol C₂H₄ × 2 mol CO₂ / 1 mol C₂H₄(g) = 0.10 mol CO₂
<u>3. Convert moles of CO₂ to volume at STP using the molar volume at STP</u>
- 0.10 mol CO₂ × 22.4 dm³ / mol = 2.24 dm³
The increase in the boiling point of a solvent is a colligative property.
That means that the increase in the boling point will be related to the number of particles (molecules or ions) present in the solution.
The higher the number of particles (molecules or ions) the higher the increase in the boiling point.
All the aqueous solutions presented are electrolytes, i.e. the solutes are ionic compounds.
Then, you have to compare the number of ions that you have in each solution.
A) 1.0 M KCl ---> 1.0 M K+ + 1.0 MCl- = 2 moles of particles / liter
B) 1.0 M CaCl2 --> 1.0M Ca(2+) + 1.0M * 2 Cl (-) = 3 moles of particle / liter
C) 2.0M KCl ---> 2.0 M K+ + 2.0 M Cl- = 4 moles of particle / liter
D) 2.0 M CaCl2 ----> 2.0 M Ca (2+) + 2.0M * 2 Cl (-) = 6 moles of particle / liter.
Then, the solution 2.0M CaCl2(aq) has the highest increase in the boiling point.
Answer: option D) 2.0 M Ca Cl2(aq)
Answer:
162 kJ
Explanation:
The reaction given by the problem is:
- NBr₃(g) + 3 H₂O(g) → 3 HOBr(g) + NH₃(g) ∆H = +81 kJ
If we turn it around, we have:
- 3 HOBr(g) + NH₃(g) → NBr₃(g) + 3 H₂O(g) ∆H = -81 kJ
If we think now of HOBr and NH₃ as our reactants, then now we need to find out <u>which one will be the </u><em><u>limiting reactant</u></em> when we have 9 moles of HOBr and 2 moles of NH₃:
- When we have 1 mol NH₃, we need 3 mol HOBr. So when we have 2 moles NH₃, we need 6 moles HOBr. We have more than 6 moles HOBr so that's our <em>reactant in excess</em>, thus NH₃ is our limiting reactant.
-81 kJ is our energy change when there's one mol of NH₃ reacting, so we <u>multiply that value by two when there's two moles of NH₃ reacting</u>. The answer is 81*2 = 162 kJ.