<h3>
Answer:</h3>
The pressure increases by 10% of the original pressure
Thus the new pressure is 1.1 times the original pressure.
<h3>
Explanation:</h3>
We are given;
- Initial temperature as 30°C, but K = °C + 273.15
- Thus, Initial temperature, T1 =303.15 K
- Final temperature, T2 is 333.15 K
We are required to state what happens to the pressure;
- We are going to base our arguments to Pressure law;
- According to pressure law, the pressure of a gas and its temperature are directly proportional at a constant volume
- That is; P α T
- Therefore, at varying pressure and temperature
Assuming the initial pressure, P1 is P
Rearranging the formula;
[tex]P2=\frac{P1T2}{T1}[/tex]
= 1.10 P
The new pressure becomes 1.10P
This means the pressure has increased by 10%
We can conclude that, the new pressure will be 1.1 times the original pressure.
Answer:
K = 2.7x10⁻⁵ at 25ºC
Explanation:
A way to write Arrhenius equation is:
ln K = - Ea/R × (1/T) + lnA
If you graph ln K as Y and 1/T as X (Absolute temperature in K), the equation you will obtain is:
Y = -13815X +35.817
R² = 0.9927
(Taking the last k point as 0.0386) (ln 0.0386), <em>0.1386 has no sense</em>)
Your slope is -13815
-13815K = - Ea/R
-13815K×8.314J/molK = 114858J/mol = Ea
And your intercept =
lnA = 35.817
A = 3.59x10¹⁵
Now, you want to know rate constant at 25ºC = 298.15K. Replacing in the equation (Where Y is ln (activation energy) and X is 1/T):
Y = -13815X +35.817
Y = -13815(1/298.15K) +35.817
Y = -10.5187
lnK = -10.5187
<h3>K = 2.7x10⁻⁵ at 25ºC</h3>
Answer:
Red
Explanation:
Red color is evidenced when thymol blue indicator is in a solution having a pH of 11. A pH of 11 means that the solution is basic or alkaline. Therefore, the indicator turns red, indicating that the solution is alkaline.
When thymol blue indicator is in a acidic solution, the indicator remains blue.
Answer:
1) The value of Kc:
C. remains the same.
2) The value of Qc:
A. is greater than Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium.
4) The concentration of N2 will:
B. decrease.
Explanation:
Hello,
In this case, by means of the Le Chatelier's principle which is based on the shift a chemical reaction could have under some modifications, we have:
1) The value of Kc:
C. remains the same, since it just depend the reaction's thermodynamics as it is computed via:
2) The value of Qc:
A. is greater than Kc, since the reaction quotient is:
![Qc=\frac{[N_2][H_2]^3}{[NH_3]^2}](https://tex.z-dn.net/?f=Qc%3D%5Cfrac%7B%5BN_2%5D%5BH_2%5D%5E3%7D%7B%5BNH_3%5D%5E2%7D)
Thus, the lower the concentration of ammonia, the higher Qc, making Qc>Kc.
3) The reaction must:
B. run in the reverse direction to restablish equilibrium, since ammonia was withdrawn and should be regenerated to reach the equilibrium.
4) The concentration of N2 will:
B. decrease, since less reactant is forming the products.
Best regards.
Answer:
They reduce the bond angle to be slightly lower than the tetrahedral bond angle, approximately 104.45 degrees.
Explanation:
The unshared pair of electrons or lone pair electrons in order to have the minimum repulsion possible with each other pushes the other bonding pairs closer together making the bond angle smaller or bent.
The bond angle is slightly lower than the tetrahedral bond angle of 108 degrees, leaving the water molecule with a bent molecular geometry.
