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jeyben [28]
2 years ago
12

An ideal gas in a closed container initially has a volume V and Temperature T the final tempera is 5/4T and the final pressure i

s 2P what is the final gas
Chemistry
1 answer:
Anettt [7]2 years ago
5 0

Answer:

V_2 = \frac{5V}{8}

Explanation:

I am assuming you are saying what is the final volume of the gas

Known :

Initial volume (V1) = V

Initial temperature (T1) = T

Final temperature (T2) = 5/4 T

Initial pressure (P1) = P

Final pressure (P2) = 2P

<u>Wanted: Final volume (V2)</u>

<u />

<u />\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\\\frac{PV}{T} = \frac{(2P)V_2}{(5/4)T}\\\frac{V}{1} = \frac{(2)V_2}{5/4}\\5/4V = 2V_2\\\\V_2 = \frac{5V}{8}

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To know more about chemicals and how to utilise them to solve man's probl

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Ilya [14]
P = 203
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5 0
3 years ago
Give five examples of structures with this formula (c6h12). at least one should contain a ring, and at least one should contain
hammer [34]

Answer: -

Following are five examples of structures with the chemical formula C₆H₁₂

Compound A is Hexene.

Compound B is 2-Hexene.

Compound C is 3-Hexene.

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3 years ago
1. What are the possible genotypes of offspring when crossing BB, Bb parents?
marissa [1.9K]

Answer:

1, C: BB, Bb, Bb, BB

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Explanation:

1. If u do a punnet square for BB and Bb you will get: BB, Bb, Bb, Bb

B| B|

B| BB. BB

b| Bb Bb

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<u>B </u> <u>B</u>

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4 0
3 years ago
Read 2 more answers
The reaction 2NO(g)+O2(g)−→−2NO2(g) is second order in NO and first order in O2. When [NO]=0.040M, and [O2]=0.035M, the observed
Oksanka [162]

Answer:

(a) The rate of disappearance of O_{2} is: 4.65*10^{-5} M/s

(b) The value of rate constant is: 0.83036 M^{-2}s^{-1}

(c) The units of rate constant is:  M^{-2}s^{-1}

(d) The rate will increase by a factor of 3.24

Explanation:

The rate of a reaction can be expressed in terms of the concentrations of the reactants and products in accordance with the balanced equation.

For the given reaction:

2NO(g)+O_{2}->2NO_{2}

rate = -\frac{1}{2} \frac{d}{dt}[NO] = -\frac{d}{dt}[O_{2}] = \frac{1}{2}\frac{d}{dt}[NO_{2}] -----(1)

According to the question, the reaction is second order in NO and first order in  O_{2}.

Then we can say that, rate = k[NO]^{2}[O_{2}] -----(2)

where k is the rate constant.

The rate of disappearance of NO is given:

-\frac{d}{dt}[NO] = 9.3*10^{-5} M/s.

(a) From (1), we can get the rate of disappearance of O_{2}.

    Rate of disappearance of  O_{2} = -\frac{d}{dt}[O_{2}] = (0.5)*(9.3*10^{-5}) M/s = 4.65*10^{-5} M/s.

(b) The rate of the reaction can be obtained from (1).

    rate = -\frac{1}{2} \frac{d}{dt}[NO] = (0.5)*(9.3*10^{-5})

    rate = 4.65*10^{-5} M/s

   The value of rate constant can be obtained by using (2).

    rate constant = k = \frac{rate}{[NO]^{2}[O_{2}]}

    k = \frac{4.65*10^{-5}}{(0.040)^{2}(0.035)} = 0.83036 M^{-2}s^{-1}

(c) The units of the rate constant can be obtained from (2).

    k = \frac{rate}{[NO]^{2}[O_{2}]}

    Substituting the units of rate as M/s and concentrations as M, we get:

\frac{Ms^{-1} }{M^{3}} = M^{-2}s^{-1}

(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.

     rate\alpha [NO]^{2}

If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of (1.8)^{2} = 3.24

     

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