This is equivalent to having a standard enthalpy change of reaction equal to 10.611 kJ
<u>Explanation</u>:
The standard enthalpy change of reaction, Δ
H
∘
, is given to you in kilojoules per mole, which means that it corresponds to the formation of one mole of carbon dioxide.
C
(s] + O
2(g]
→
CO
2(g]
Remember, a negative enthalpy change of reaction tells you that heat is being given off, i.e. the reaction is exothermic.
First to convert grams of carbon into moles,
use carbon's molar mass(12.011 g).
Moles of C = mass in gram / molar mass
= 0.327 g / 12.011 g
Moles of C = 0.027 moles
Now, in order to determine how much heat is released by burning of 0.027 moles of carbon to form carbon-dioxide.
= 0.027 moles C 
393 kJ
Heat released = 10.611 kJ.
So, when 0.027 moles of carbon react with enough oxygen gas, the reaction will give off 10.611 kJ of heat.
This is equivalent to having a standard enthalpy change of reaction equal to 10.611 kJ
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Answer:
0 M is the silver ion concentration in a solution prepared mixing both the solutions.
Explanation:
Moles of silver nitrate = n
Volume of the solution = 425 mL = 0.425 L (1 mL = 0.001 L)
Molarity of the silver nitrate solution = 0.397 M
Moles of sodium phosphate = n'
Volume of the sodium phosphate solution = 427 mL = 0.427 L (1 mL = 0.001 L)
Molarity of the sodium phosphate solution = 0.459 M
According to reaction, 3 moles of silver nitrate reacts with 1 mole of sodium phosphate, then 0.1687 moles of silver nitrate will recat with :
of sodium phosphate
This means that only 0.05623 moles of sodium phosphate will react with all the 0.1687 moles of silver nitrate , making silver nitrate limiting reagent and sodium phosphate as an excessive reagent.
So, zero moles of silver nitrate will be left in the solution after mixing of the both solutions and hence zero moles of silver ions will left in the resulting solution.
0 M is the silver ion concentration in a solution prepared mixing both the solutions.
Answer: pH of HCl =5, HNO3 = 1,
NaOH = 9, KOH = 12
Explanation:
pH = -log [H+ ]
1. 1.0 x 10^-5 M HCl
pH = - log (1.0 x 10^-5)
= 5 - log 1 = 5
2. 0.1 M HNO3
pH = - log (1.0 x 10 ^ -1)
pH = 1 - log 1 = 1
3. 1.0 x 10^-5 NaOH
pOH = - log (1.0 x 10^-5)
pOH = 5 - log 1 = 5
pH + pOH = 14
Therefore , pH = 14 - 5 = 9
4. 0.01 M KOH
pOH = - log ( 1.0 x 10^ -2)
= 2 - log 1 = 2
pH + pOH = 14
Therefore, pH = 14 - 2 = 12
