150/30 = 5
HF1 20/2 = 10
HF2 10/2 = 5
HF3 5/2 = 2.5
HF4 2.5/2 = 1.25
HF5 1.25/2 = 0.625
Answer: 0.63g
Answer:
ΔH of the reaction is -802.3kJ.
Explanation:
Using Hess's law, you can know ΔH of reaction by the sum of ΔH's of half-reactions.
Using the reactions:
<em>(1) </em>Cgraphite(s)+ 2H₂(g) → CH₄(g) ΔH₁ = −74.80kJ
<em>(2) </em>Cgraphite(s)+ O₂(g) → CO₂(g) ΔH₂ = −393.5k
J
<em>(3) </em>H₂(g) + 1/2 O₂(g) → H₂O(g) ΔH₃ = −241.80kJ
The sum of (2) - (1) produce:
CH₄(g) + O₂(g) → CO₂(g) + 2H₂(g) ΔH' = -393.5kJ - (-74.80kJ) = -318.7kJ
And the sum of this reaction with 2×(3) produce:
CH₄(g) + 2 O₂(g) → CO₂(g) + 2H₂O(g) And ΔH = -318.7kJ + 2×(-241.80kJ) =
<em>-802.3kJ</em>
<em>m Na₂CO₃: 23g×2 + 12g + 16g×3 = 106 g/mol</em>
------------------------------
1 mol ------- 106g
X ------------ 10,6g
X = 10,6/106
<u>X = 0,1 mol Na₂CO₃</u>
Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:
Answer:
Thorium-234 option a .......