Explanation:
Let us assume that the concentration of [
and 
is equal to x. Then expression for 
for the given reaction is as follows.
![K_{w} = [OH^{-}][H^{+}]](https://tex.z-dn.net/?f=K_%7Bw%7D%20%3D%20%5BOH%5E%7B-%7D%5D%5BH%5E%7B%2B%7D%5D)
Now, we will take square root on both the sides as follows.
![[H^{+}] = 8.2 \times 10^{-8}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%208.2%20%5Ctimes%2010%5E%7B-8%7D)
M
Thus, we can conclude that the 
concentration in neutral water at this temperature is 
M.
Answer:
1.14 atm and 1.139 mol
Explanation:
The <em>total pressure</em> of the container is equal to the <u>sum of the partial pressure of the three gasses</u>:
- P = Poxygen + Pnitrogen + Pcarbon dioxide
- 2.50 atm = 0.52 + 0.84 + Pcarbon dioxide
Now we <u>solve for the pressure of carbon dioxide</u>:
- Pcarbon dioxide = 1.14 atm
To c<u>alculate the number of CO₂ moles </u>we use <em>PV=nRT</em>:
- R = 0.082 atm·L·mol⁻¹·K⁻¹
- T = 32 °C ⇒ 32 + 273.16 = 305.16 K
1.14 atm * 25.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 305.16 K
Answer:
64.52 mg.
Explanation:
The following data were obtained from the question:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Final amount (N) =.?
Next, we shall determine the rate constant (K).
This is illustrated below:
Half life (t½) = 1590 years
Rate/decay constant (K) =?
K = 0.693 / t½
K = 0.693/1590
K = 4.36×10¯⁴ / year.
Finally, we shall determine the amount that will remain after 1000 years as follow:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Rate constant = 4.36×10¯⁴ / year.
Final amount (N) =.?
Log (N₀/N) = kt/2.3
Log (100/N) = 4.36×10¯⁴ × 1000/2.3
Log (100/N) = 0.436/2.3
Log (100/N) = 0.1896
Take the antilog
100/N = antilog (0.1896)
100/N = 1.55
Cross multiply
N x 1.55 = 100
Divide both side by 1.55
N = 100/1.55
N = 64.52 mg
Therefore, the amount that remained after 1000 years is 64.52 mg
Answer:
The different angles the moon has from sunlight to the earth
Explanation:
