Explanation:
The reaction equation will be as follows.
Calculate the amount of 
dissolved as follows.
It is given that 
= 0.032 M/atm and 
= 
atm.
Hence, ![[CO_{2}]](https://tex.z-dn.net/?f=%5BCO_%7B2%7D%5D)
will be calculated as follows.
= 
= 
= 
or, = 
It is given that 
As, ![K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5E%7B2%7D%7D%7B%5BCO_%7B2%7D%5D%7D)
![4.46 \times 10^{-7} = \frac{[H^{+}]^{2}}{0.608 \times 10^{-5}}](https://tex.z-dn.net/?f=4.46%20%5Ctimes%2010%5E%7B-7%7D%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5E%7B2%7D%7D%7B0.608%20%5Ctimes%2010%5E%7B-5%7D%7D)
![[H^{+}]^{2}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%5E%7B2%7D)
= 
![[H^{+}]](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D)
= 
Since, we know that pH = ![-log [H^{+}]](https://tex.z-dn.net/?f=-log%20%5BH%5E%7B%2B%7D%5D)
So, pH = 
= 5.7
Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.
Answer: 66.2 g
Explanation:
1) The ratio of Al in the molecule is 1 mol to 1 mol .
2) The mass of 1 mol of molecules of Al (CH2H3O2)3 is the molar mass of the compound.
3) You calculate the molar mass of the compound using the atomic masses of each atom, in this way:
Al: 27 g/mol
C: 2 * 3 * 12 g/mol = 72 g/mol
H: 3 * 3 * 1 g/mol = 9 g/mol
O: 2 * 3 * 16 g/mol = 96 g/mol
Molar mass = 27 g/mol + 72 g/mol + 9 g/mol + 96 g/mol = 204 g/mol
4) Set a proportion:
27 g/mol x
-------------------- = ----------
204 g/mol 500 g
5) Solve for x:
x = 500 g * 27 g/mol / 204 g/mol = 66.2 g
Answer:
The combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Explanation:
Given;
CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = -890 kJ/mol
From the combustion reaction above, it can be observed that;
1 mole of methane (CH₄) released 890 kilojoules of energy.
Now, we convert 59.7 grams of methane to moles
CH₄ = 12 + (1x4) = 16 g/mol
59.7 g of CH₄ 
1 mole of methane (CH₄) released 890 kilojoules of energy
3.73125 moles of methane (CH₄) will release ?
= 3.73125 moles x -890 kJ/mol
= -3320.81 kJ
Therefore, the combustion of 59.7 grams of methane releases 3320.81 kilojoules of energy
Answer:
52 da
Step-by-step explanation:
Whenever a question asks you, "How long to reach a certain concentration?" or something similar, you must use the appropriate integrated rate law expression.
The i<em>ntegrated rate law for a first-order reaction </em>is
ln([A₀]/[A] ) = kt
Data:
[A]₀ = 750 mg
[A] = 68 mg
t_ ½ = 15 da
Step 1. Calculate the value of the rate constant.
t_½ = ln2/k Multiply each side by k
kt_½ = ln2 Divide each side by t_½
k = ln2/t_½
= ln2/15
= 0.0462 da⁻¹
Step 2. Calculate the time
ln(750/68) = 0.0462t
ln11.0 = 0.0462t
2.40 = 0.0462t Divide each side by 0.0462
t = 52 da
Answer:
A)Trial 1 because the average rate of reaction is lower.
Explanation:
I accidentally gave myself low rating my bad
