Answer:
Explanation:
Firstly, write the expression for the equilibrium constant of this reaction:
Secondly, we may relate the change in Gibbs free energy to the equilibrium constant using the equation below:
From here, rearrange the equation to solve for K:
Now we know from the initial equation that:
Let's express the ratio of ADP to ATP:
Substitute the expression for K:
Now we may use the values given to solve:
Answer:
metal can hold heat keep it near fire
Molar mass:
O2 = 16 x 2 = 32.0 g/mol Mg = 24 g/mol
<span>2 Mg(s) + O2(g) --->2 MgO(s)
</span>
2 x 24.0 g Mg -------------> 32 g O2
5.00 g Mg -----------------> ( mass of O2)
mass of O2 = 5.00 x 32 / 2 x 24.0
mass of O2 = 160 / 48
= 3.33 g of O2
hope this helps!
Answer:
C5H12 + 8O2 --> 5CO2 + 6H2O
Explanation:
Complete question
Write a balanced chemical equation to represent the combustion of CH3CH2CH2CH2CH3
Solution
The given compound is pentane
C5H12
The empirical equation representing combustion of pentane is
C5H12 + O2 --> CO2 + H2O
We will first balance the carbon atoms
C5H12 + O2 --> 5CO2 + H2O
Now we will balance the Hydrogen molecule
C5H12 + O2 --> 5CO2 + 6H2O
Now we will balance the oxygen molecule
C5H12 + 8O2 --> 5CO2 + 6H2O