Calculate the percentage error in the molar mass of our natural gas which is 16.9g/mol
1 answer:
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Answer : The enthalpy of the reaction = -1839.6 KJ
Solution : Given,
= -520.0 KJ/mole
= -1699.8 KJ/mole
The balanced chemical reaction is,
Formula used :
We know that the standard enthalpy of formation of the element is equal to Zero.
Therefore, the enthalpy of formation of (Mn) and (Al) is equal to zero.
Now, put all the values in above formula, we get
= (-3399.6) + (1560)
= -1839.6 KJ
Answer :
The concentration of before any titrant added to our starting material is 0.200 M.
The pH based on this ion concentration is 0.698
Explanation :
First we have to calculate the concentration of before any titrant is added to our starting material.
As we are given:
Concentration of HBr = 0.200 M
As we know that the HBr is a strong acid that dissociates complete to give hydrogen ion and bromide ion
.
As, 1 M of HBr dissociates to give 1 M of
So, 0.200 M of HBr dissociates to give 0.200 M of
Thus, the concentration of before any titrant added to our starting material is 0.200 M.
Now we have to calculate the pH based on this ion concentration.
pH : It is defined as the negative logarithm of hydrogen ion concentration.
Thus, the pH based on this ion concentration is 0.698
The percentage composition of this compound : 40%Ca, 12%C and 48%O
<h3>Further explanation</h3>Given
20.0 g of calcium,
6.0 g of carbon,
and 24.0 g of oxygen.
Required
The percentage composition
Solution
Total mass of compound :
=mass calcium + mass carbon + mass oxygen
=20 g + 6 g + 24 g
=50 g
Percentage composition :
- Ca-calcium
- C-carbon
- O-oxygen