The formula used for determining gas pressure, volume and temperature interaction would be PV=nRT.
<span>• What is the temperature in Kelvins?
</span>You already right at this part. Kelvin temperature formula from celsius should be:
K= C+273.15=
<span>K= 27 +273.15 = 300.15
It is important to remember that the formula in this question is using Kelvin unit at temperature, not Celcius or Fahrenheit.
</span>
<span>• Assuming that everything else remains constant, what will happen to the pressure if the temperature decreases to -15 ºC?
</span>In this case, the temperature is decreased from 27C into -15C and you asked the change in the pressure.
Using PV=nRT formula, you can derive that the temperature will be directly related to pressure. If the temperature decreased, the pressure will be decreased too.
<span> If you increase the number of moles to 6 moles, increase temperature to 400K and reduce the volume to 25 L, what will the new pressure be?
</span>PV=nRT
P= nRT/V
P= 6 moles* <span>0.0821 L*atm/(mol*K) * 400K/25L= 7.8816 atm</span>
Answer:
[O₃]= 8.84x10⁻⁷M
Explanation:
<u>The photodissociation of ozone by UV light is given by:</u>
O₃ + hν → O₂ + O (1)
<u>The first-order reaction of the equation (1) is:</u>
(2)
<em>where k: is the rate constant and Δ[O₃]/Δt: is the variation in the ozone concentration with time, and the negative sign is by the decrease in the reactant concentration </em>
<u>We can get the following expression of the </u><u>first-order integrated law</u><u> of the reaction (1), by resolving the equation (2):</u>
(3)
<em>where [O₃](t): is the ozone concentration in the elapsed time and [O₃]₀: is the initial ozone concentration</em>
We can calculate the initial ozone concentration using equation (3):
So, the ozone concentration after 10 days is 8.84x10⁻⁷M.
I hope it helps you!
I think it means the reaction doesn't have any intermediate phases - the order doesn't change, and it moves at the same rate, it doesn't speed up or slow down at any time