In chemistry, yes...yes it is.
I think the answer is covalent bonds.
Answer:
wood burning and cookies baking
Explanation:
took the test <333
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>
<u>Answer:</u> The equilibrium constant for the above reaction is 1.31
<u>Explanation:</u>
We are given:
Initial moles of nitrogen gas = 0.3411 moles
Initial moles of hydrogen gas = 1.661 moles
Equilibrium moles of nitrogen gas = 0.2001 moles
For the given chemical reaction:

<u>Initial:</u> 0.3411 1.661
<u>At eqllm:</u> 0.3411-x 1.661-3x 2x
Evaluating for 'x', we get:

Volume of the container = 2.50 L
The expression of
for the above equation follows:
![K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5Ctimes%20%5BH_2%5D%5E3%7D)
We are given:
![[NH_3]=\frac{2\times 0.141}{2.50}=0.1128M](https://tex.z-dn.net/?f=%5BNH_3%5D%3D%5Cfrac%7B2%5Ctimes%200.141%7D%7B2.50%7D%3D0.1128M)
![[N_2]=\frac{0.2001}{2.5}=0.08004M](https://tex.z-dn.net/?f=%5BN_2%5D%3D%5Cfrac%7B0.2001%7D%7B2.5%7D%3D0.08004M)
![[H_2]=\frac{1.661-(3\times 0.141)}{2.5}=0.4952M](https://tex.z-dn.net/?f=%5BH_2%5D%3D%5Cfrac%7B1.661-%283%5Ctimes%200.141%29%7D%7B2.5%7D%3D0.4952M)
Putting values in above expression, we get:

Hence, the equilibrium constant for the above reaction is 1.31