<h3><u>Answer;</u></h3>
<u>= 128.772 g of water</u>
2C6H14 + 13 O2 → 6CO2 + 14 H2O
<h3><u>Explanation</u>;</h3>
1 mole of Hexane contains 86 g
Therefore;
87.91 grams of hexane will contain
= 87.91 g/86 g
= 1.022 moles
The balanced reaction for the combustion of Hexane is given by
2C6H14 + 13 O2 → 6CO2 + 14 H2O
Therefore; the mole ratio of C6H14 :H2O is
= 2 : 14
= 1: 7
therefore moles of water from 1.022 moles of Hexane will be;
= 1.022 ×7
= 7.154 moles
Mass of water will be;
= 7.154 moles× 18
<u>= 128.772 g of water</u>
Answer:
Compound formula : HCl
Element H
# atoms:1
Elements Cl
#atoms : 1
gram formula weight (g) = 36.46
Answer:
1. analyze the samples while they are frozen
stuff can react differently when it's colder
hope I could help good luck
Answer:
6.2846
Explanation:
Given that:-
Concentrations at equilibrium :-
![[CH_4]=0.126\ M](https://tex.z-dn.net/?f=%5BCH_4%5D%3D0.126%5C%20M)
![[H_2O]= 0.242\ M](https://tex.z-dn.net/?f=%5BH_2O%5D%3D%200.242%5C%20M)
![[CO]= 0.126\ M](https://tex.z-dn.net/?f=%5BCO%5D%3D%200.126%5C%20M)
![[H_2]= 1.15\ M](https://tex.z-dn.net/?f=%5BH_2%5D%3D%201.15%5C%20M)
The equilibrium reaction is:-
The expression for equilibrium constant is:
![K_{c}=\frac {\left [ H_2 \right ]^3\left [ CO \right ]}{\left [ CH_4 \right ]\left [ H_2O \right ]}](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%20%7B%5Cleft%20%5B%20H_2%20%5Cright%20%5D%5E3%5Cleft%20%5B%20CO%20%5Cright%20%5D%7D%7B%5Cleft%20%5B%20CH_4%20%5Cright%20%5D%5Cleft%20%5B%20H_2O%20%5Cright%20%5D%7D)
Applying the values as:-
<u>The equilibrium constant for the reaction is:- 6.2846</u>
