Answer:
ΔH = 125.94kJ
Explanation:
It is possible to make algebraic sum of reactions to obtain ΔH of reactions (Hess's law). In the problem:
1. 2W(s) + 3O2(g) → 2WO3(s) ΔH = -1685.4 kJ
2. 2H2(g) + O2(g) → 2H2O(g) ΔH = -477.84 kJ
-1/2 (1):
WO3(s) → W(s) + 3/2O2(g) ΔH = 842.7kJ
3/2 (2):
3H2(g) + 3/2O2(g) → 3H2O(g) ΔH = -716.76kJ
The sum of last both reactions:
WO3(s) + 3H2(g) → W(s) + 3H2O(g)
ΔH = 842.7kJ -716.76kJ
<h3>ΔH = 125.94kJ </h3>
Answer:
0.11 mol
Explanation:
<em>This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar): CH₃CO₂H. An analytical chemist has determined by measurements that there are 0.054 moles of oxygen in a sample of acetic acid. How many moles of hydrogen are in the sample?</em>
Step 1: Given data
- Formula of acetic acid: CH₃CO₂H
- Moles of oxygen in the sample of acetic acid: 0.054 moles
Step 2: Establish the appropriate molar ratio
According to the chemical formula of acetic acid, the molar ratio of H to O is 4:2.
Step 3: Calculate the moles of atoms of hydrogen
We will use the theoretical molar ratio for acetic acid.
0.054 mol O × (4 mol H/2 mol O) = 0.11 mol H
Answer:
449730.879 cal/g
Explanation:
Given data:
Mass of sample = 4.9 g
Change in temperature = 2.08 °C (275.23 k)
Heat capacity of calorimeter = 33.50 KJ . K⁻¹
Solution:
C(candy) = Q/m
Q = C (calorimeter) × ΔT
C(candy) = C (calorimeter) × ΔT / m
C(candy) = 33.50 KJ . K⁻¹ × 275.23 K / 4.90 g
C(candy) = 9220.205 KJ / 4.90 g
C(candy) = 1881.674 KJ / g
It is known that,
1 KJ /g = 239.006 cal/g
1881.674 × 239.006 = 449730.879 cal/g
The number of grams : 17.082 g
<h3>Further explanation</h3>
Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution
Where
M = Molarity
n = Number of moles of solute
V = Volume of solution
453.9 mL of 0.237 M calcium acetate
MW Ca(C₂H₃OO)₂ : 158,17 g/mol
You just have to pass your final with a grade above 75
