Answer: Pentane C5H12
Explanation:
The boiling point of a substance is simply defined as the temperature whereby a liquid's vapor pressure is equal to the pressure that is surrounding the liquid and hence, the liquid will changes into vapor.
The likely molecular formula for this compound is Pentane i.e C5H12 due to the fact that its boiling point is between Butane with formula C4H10 and Hexane with formula C6H14 boiling points.
Answer: 625 kj/mol
Explanation:
As shown below this expression gives the activation energy of the reverse reaction:
EA reverse reaction = EA forward reaction + | enthalpy change |
1) The activation energy, EA is the difference between the potential energies of the reactants and the transition state:
EA = energy of the transition state - energy of the reactants.
2) The activation energy of the forward reaction given is:
EA = energy of the transition state - energy of [ NO2(g) + CO(g) ] = 75 kj/mol
3) The negative enthalpy change - 250 kj / mol for the forward reaction means that the products are below in the potential energy diagram, and that the potential energy of the products, [NO(g) + CO2(g) ] is equal to 375 kj / mol - 250 kj / mol = 125 kj/mol
4) For the reverse reaction the reactants are [NO(g) + CO2(g)], and the transition state is the same than that for the forward reaction.
5) The difference of energy between the transition state and the potential energy of [NO(g) + CO2(g) ] will be the absolute value of the change of enthalpy plus the activation energy for the forward reaction:
EA reverse reaction = EA forward reaction + | enthalpy change |
EA reverse reaction = 375 kj / mol + |-250 kj/mol | = 375 kj/mol + 250 kj/mol = 625 kj/mol.
And that is the answer, 625 kj/mol
Thank you for posting your question here at brainly. Below is the answer:
At 25 C and 1 atm pressure (the standard state): P is a solid, O2 is a gas, and Cl2 is a gas.
<span>P(s) + O2(g) + Cl2(g) ==> POCl3(l) </span>
<span>To make 1 mole of POCl3, we need to start with 1 mole of P, 1/2 mole of O2, and 3/2 mole of Cl2. </span>
<span>P(s) + 1/2O2(g) + 3/2Cl2(g) ==> POCl3(l) </span>
<span>NOTE: Some people write P as P4(s), in which case you would need 1/4 mole P4.</span>
Answer:
a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.
b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.
c)If concentration of ![[H^+]](https://tex.z-dn.net/?f=%5BH%5E%2B%5D)
is changed to 0.0001 M than rate will be increased by the factor of 0.01.
d) If concentration when [sucrose] and both are changed to 0.1 M than rate will be increased by the factor of 1.
Explanation:
Sucrose + 
fructose+ glucose
The rate law of the reaction is given as:
![R=k[H^+][sucrose]](https://tex.z-dn.net/?f=R%3Dk%5BH%5E%2B%5D%5Bsucrose%5D)
![[H^+]=0.01M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.01M)
[sucrose]= 1.0 M
![R=k[0.01M][1.0 M]](https://tex.z-dn.net/?f=R%3Dk%5B0.01M%5D%5B1.0%20M%5D)
..[1]
a)
The rate of the reaction when [Sucrose] is changed to 2.5 M = R'
![R'=[0.01 M][2.5 M]](https://tex.z-dn.net/?f=R%27%3D%5B0.01%20M%5D%5B2.5%20M%5D)
..[2]
[2] ÷ [1]
![\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}](https://tex.z-dn.net/?f=%5Cfrac%7BR%27%7D%7BR%7D%3D%5Cfrac%7B%5B0.01%20M%5D%5B2.5%20M%5D%7D%7Bk%5B0.01M%5D%5B1.0%20M%5D%7D)
If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.
b)
The rate of the reaction when [Sucrose] is changed to 0.5 M = R'
![R'=[0.01 M][0.5 M]](https://tex.z-dn.net/?f=R%27%3D%5B0.01%20M%5D%5B0.5%20M%5D)
..[2]
[2] ÷ [1]
![\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}](https://tex.z-dn.net/?f=%5Cfrac%7BR%27%7D%7BR%7D%3D%5Cfrac%7B%5B0.01%20M%5D%5B0.5%20M%5D%7D%7Bk%5B0.01M%5D%5B1.0%20M%5D%7D)
If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.
c)
The rate of the reaction when is changed to 0.001 M = R'
![R'=[0.0001 M][1.0 M]](https://tex.z-dn.net/?f=R%27%3D%5B0.0001%20M%5D%5B1.0%20M%5D)
..[2]
[2] ÷ [1]
![\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}](https://tex.z-dn.net/?f=%5Cfrac%7BR%27%7D%7BR%7D%3D%5Cfrac%7B%5B0.0001%20M%5D%5B1.0M%5D%7D%7Bk%5B0.01M%5D%5B1.0%20M%5D%7D)
If concentration of is changed to 0.0001 M than rate will be increased by the factor of 0.01.
d)
The rate of the reaction when [sucrose] and both are changed to 0.1 M = R'
![R'=[0.1M][0.1M]](https://tex.z-dn.net/?f=R%27%3D%5B0.1M%5D%5B0.1M%5D)
..[2]
[2] ÷ [1]
![\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}](https://tex.z-dn.net/?f=%5Cfrac%7BR%27%7D%7BR%7D%3D%5Cfrac%7B%5B0.1M%5D%5B0.1M%5D%7D%7Bk%5B0.01M%5D%5B1.0%20M%5D%7D)
If concentration when [sucrose] and both are changed to 0.1 M than rate will be increased by the factor of 1.
Answer:
The concentration after 20 mins is 0.832 M
Explanation:
Zero order rate law is given by;
R = K [A₀]⁰
A zero order reaction, rate is independent of the initial concentration
R = K
Where;
R is the rate of reaction
K is the rate constant = 0.0416 M/min
Since R = K,
Then, R = 0.0416 M/min
After 20 min, the concentration will be;
A = Rt
A = (0.0416 M/min)(20 min)
A = 0.832 M
Therefore, the concentration after 20 mins is 0.832 M
