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Lera25 [3.4K]
2 years ago
14

When I’m the imposter in among us....

Chemistry
2 answers:
pav-90 [236]2 years ago
8 0
THANKS FOR THE POINT also FRRRR LOL
bagirrra123 [75]2 years ago
4 0

Answer:

when the imposter is sus

Explanation:

ඞ Don ඞ Don ඞ don don don don DON ඞ don don donඞ

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What is the mass of carbon monoxide (CO) gas if the gas occupies a volume of 4.1 L at 2.0 atm of pressure and -73oC?
Stels [109]

Answer:

1.7

Explanation:

4 0
2 years ago
A student adds 0. 0030 mol of hcl to 100 ml of a 0. 10 m solution of a r2nh, a weak base. The ph of the solution is found to be
egoroff_w [7]

Since an acidic salt solution is produced when a strong acid neutralizes a weak base, the pH of the salt solution formed when HCl is added to R2NH will be less than 7.

<h3>What is a neutralization reaction?</h3>

A neutralization reaction is the react ion between an acid and a base to form salt and water only.

Neutralization reactions can either produce a neutral solution, an acidic solution or an alkaline solution at equivalence point.

When a strong acid is added to a weak base, the pH of the salt solution formed will be less than 7.

Therefore, the pH of the salt solution formed when HCl is added to R2NH will be less than 7.

Learn more about pH at: brainly.com/question/940314

5 0
1 year ago
The energy available to primary consumers in an energy pyramid is 800 kilocalories. Approximately how much energy is used by the
ki77a [65]
The answer is B, Let me know if you get it right!
5 0
3 years ago
Read 2 more answers
In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char
lisabon 2012 [21]

Answer : The partial pressure of N_2,H_2\text{ and }NH_3 at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution :  Given,

Initial pressure of N_2 = 1.42 bar

Initial pressure of H_2 = 2.87 bar

K_p = 0.036

The given equilibrium reaction is,

                              N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)

Initially                   1.42      2.87             0

At equilibrium    (1.42-x)  (2.87-3x)     2x

The expression of K_p will be,

K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}

Now put all the values of partial pressure, we get

0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}

By solving the term x, we get

x=0.287\text{ and }3.889

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of NH_3 at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of N_2 at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of H_2 at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of N_2 + Partial pressure of H_2 + Partial pressure of NH_3

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0
3 years ago
An element has the mass number 12 and atomic number 6. The number of neutron in it is?
slamgirl [31]

Answer:

12-6=6

the answer is A)6

8 0
3 years ago
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