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Orlov [11]
3 years ago
8

If a mixture of gases contained 78% nitrogen at a pressure of and 22% carbon dioxide at , what is the total pressure of the syst

em? 1,329 atm 17.5 cm hg 639 torr 1.75 atm none of these
Chemistry
1 answer:
Naddika [18.5K]3 years ago
4 0
The pressures given are the partical pressures of the two gases.

The law of Dalton or of the partial pressures states that the total pressure of a mixture of gases is equal to the sum of the pressures of all the gases.

So, in this case:

Total pressure = pressure of nitrogen + pressure of carbon dioxyde.

Of course both terms must be in the same units.

i have found that the pressures for this problem are:

Pressure of nitrogen = 984 torr

Pressure of carbon dioxyde = 345 torr

Total pressure = 984 torr + 345 torr = 1329 torr.

Now convert to atm: 1 atm = 760 torr

=> 1329 torr * 1 atm / 760 torr = 1.74868 atm ≈ 1.75 atm

Answer: 1.75 atm

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Aspartame (c14h18n2o5) is a solid used as an artificial sweetener. its combustion produces carbon dioxide gas, liquid water, and
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Aspartame (C₁₄H₁₈N₂O₅) is a solid used as an artificial sweetener. its combustion produces carbon dioxide gas, liquid water, and nitrogen gas

C₁₄H₁₈N₂O₅ + 16O₂-----> 14CO₂ + 9H₂O + N₂.

As it can be seen from the equation, that the coefficient of nitrogen gas in the balanced equation for the reaction is 1.

So the answer here is 1 only that is coefficient of nitrogen gas in the balanced equation for the reaction is 1.

5 0
3 years ago
In a machine shop, two cams are produced, one of aluminum and one of iron. Both cams have the same mass. Which cam is larger? (a
Lelu [443]
The best and most correct answer among the choices provided by your question is the second choice or letter B.

The iron cam is larger than the aluminum cam even if with the same size.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
5 0
3 years ago
Read 2 more answers
What is the average kinetic energy of 1 mole of a gas that is at 100 degrees Celsius?
Paha777 [63]
The average kinetic energy of an ideal gas is calculated as 

KE_avg = 3/2 kT

where T is the temperature in Kelvin and k=R/N_A; R is the universal gas constant and N_A is the number of moles.

Thus, upon substitution we get

KE_avg = 3/2(8.314/1)(100+273)
KE_avg = 3/2(8.314)(373)
KE_avg = 4651.683

The average kinetic energy of 1 mole of a gas at 100 degree Celsius is 4651.683 J.


3 0
3 years ago
Potassium + water → potassium hydroxide<br> +<br> hydrogen<br> formula
kvasek [131]

Answer:

If you help I'll help you deal?

3 0
2 years ago
Determine the rate law and the value of k for the following reaction using the data provided.
blsea [12.9K]

<u>Answer:</u> The rate law expression is \text{Rate}=k[NO]^2[O_2]^1 and value of 'k' is 1.727\times 10^3M^{-2}s^{-1}

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

2NO(g)+O_2(g)\rightarrow 2NO_2(g)

Rate law expression for the reaction:

\text{Rate}=k[NO]^a[O_2]^b

where,

a = order with respect to nitrogen monoxide

b = order with respect to oxygen

  • <u>Expression for rate law for first observation:</u>

8.55\times 10^{-3}=k(0.030)^a(0.0055)^b       ....(1)

  • <u>Expression for rate law for second observation:</u>

1.71\times 10^{-2}=k(0.030)^a(0.0110)^b       ....(2)

  • <u>Expression for rate law for third observation:</u>

3.42\times 10^{-2}=k(0.060)^a(0.0055)^b      ....(3)

Dividing 1 from 2, we get:

\frac{1.71\times 10^{-2}}{8.55\times 10^{-3}}=\frac{(0.030)^a(0.0110)^b}{(0.030)^a(0.0055)^b}\\\\2=2^b\\b=1

Dividing 1 from 3, we get:

\frac{3.42\times 10^{-2}}{8.55\times 10^{-3}}=\frac{(0.060)^a(0.0055)^b}{(0.030)^a(0.0055)^b}\\\\2^2=2^a\\a=2

Thus, the rate law becomes:

\text{Rate}=k[NO]^2[O_2]^1

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

8.55\times 10^{-3}=k[0.030]^2[0.0055]^1\\\\k=1.727\times 10^3M^{-2}s^{-1}

Hence, the rate law expression is \text{Rate}=k[NO]^2[O_2]^1 and value of 'k' is 1.727\times 10^3M^{-2}s^{-1}

4 0
3 years ago
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