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Tems11 [23]
3 years ago
6

HELP!!! OPTIONS: gas, liquid, solid

Chemistry
1 answer:
erica [24]3 years ago
3 0

Answer:

Melting: solid liquid, Freezing: liquid to solid, Evaporation: liquid to gas.

Explanation:

mark me the brainliest

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Atoms 'like' to have 8 valence electrones. sodium has one electron 'in excess' and chlorine 'lacks' in one. so they just exchange :) both are happy. then the ionic bond is formed because chlorine is now charged negative (additional electron) and sodium is charged positive (gave up one electrone) - the opposite charges cause electrostatic pull. many atoms, many bonds and thats how salt is built.
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In the first step of the Ostwald process for the synthesis of nitric acid, ammonia is converted to nitric oxide by the high-temp
Paul [167]

Solution: The given balanced equation is:

4NH_3+5O_2\rightarrow 4NO+6H_2O

If we have a hypothetical equation:

A+2B\rightarrow 3C+5D

Then the rate could be written as:

rate=-\frac{\Delta [A]}{\Delta t}=-\frac{1}{2}\frac{\Delta [B]}{\Delta t}=\frac{1}{3}\frac{\Delta [C]}{\Delta t}=\frac{1}{5}\frac{\Delta [D]}{\Delta t}

From above expression one thing could easily be noticed that the coefficients of all are inverted. Also, there is negative sign in front of reactants and positive sign in front of the products. Negative sign stands for rate of consumption where as positive sign stands for rate of formation.

Like the above example, we can write the rate for the given equation and it would be looking as:

rate=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}=-\frac{1}{5}\frac{\Delta [O_2]}{\Delta t}=\frac{1}{4}\frac{\Delta [NO]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_2O]}{\Delta t}

Now we can easily answer all the parts of the question.

(a) From above expression, the rate of consumption of O_2 related to rate of consumption of NH_3 as:

-\frac{1}{5}\frac{\Delta [O_2]}{\Delta t}=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}

multiply both sides by -5

\frac{\Delta [O_2]}{\Delta t}=\frac{5}{4}\frac{\Delta [NH_3]}{\Delta t}

So, rate of consumption of oxygen is \frac{5}{4} the rate of consumption of ammonia.

(b) The relationship between rate of formation of NO to the rate of consumption of ammonia will be written as:

\frac{1}{4}\frac{\Delta [NO]}{\Delta t}=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}

Multiply both sides by 4

\frac{\Delta [NO]}{\Delta t}=-\frac{\Delta [NH_3]}{\Delta t}

So, rate of formation of NO equals to the rate of consumption of ammonia.

Now, the rate of formation of H_2O to the rate of consumption of ammonia would be:

\frac{1}{6}\frac{\Delta [H_2O]}{\Delta t}=-\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t}

Multiply both sides by 6

\frac{\Delta [H_2O]}{\Delta t}=-\frac{6}{4}\frac{\Delta [NH_3]}{\Delta t}

So, the rate of formation of H_2O is \frac{6}{4} times that is 1.5 times to the rate of consumption of ammonia.

5 0
3 years ago
Ammonia is a?<br> Base<br> Acid<br> Ketone<br> Ester
BaLLatris [955]

Answer:

it is a base

Explanation:

8 0
3 years ago
How many electrons are necessary to produce a charge of "-0.80" C
docker41 [41]

Answer:

5.0x10^18

Explanation:

6 0
3 years ago
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