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kupik [55]
3 years ago
5

Which of the reactions below are catalyzed? Check all that apply.

Chemistry
2 answers:
Verdich [7]3 years ago
7 0

Answer:

B. Salt, NaCl, is produced by the process of evaporation of seawater or brine. If the surface area of the water is increased, the same volume of water evaporates faster.

C. The Haber process combines hydrogen and nitrogen to make ammonia. The two gases are passed through a reactor under pressure and at high temperatures. If iron is added to the reactor, the yield of ammonia increases.

Explanation:

Evaporation of water is responsible for the production of sodium chloride also known as table salt. Sodium and chlorine are present in water. When more evaporation of water occurs, sodium and chlorine come close together forming sodium chloride. Haber process is responsible for the production of ammonia which is used as fertilizer. For speed up the process, catalyst is used such as iron in order to complete the reaction in less time. Iron binds hydrogen and nitrogen with each other.

Alexus [3.1K]3 years ago
3 0

Answer:a c d edj

Explanation:

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It is a completely true statement that a <span>base increases the OH- ion concentration of water. The correct option among the two options that are given in the question is the first option. I hope that this is the answer that you were looking for and the answer has actually come to your desired help.</span>
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Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe
leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as <u>molarity</u>, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

c = \frac{n}{V}

n = moles (unit mol)

V = volume (unit L)

<u>Find the molar mass (M) of potassium hydroxide.</u>

M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}

M_{KOH} = 56.106 \frac{g}{mol}

<u>Calculate the moles of potassium hydroxide.</u>

n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}

n_{KOH} = 0.25941(9)mol

Carry one insignificant figure (shown in brackets).

<u>Convert the volume of water to litres.</u>

V = \frac{500.0mL}{1}*\frac{1L}{1000mL}

V = 0.5000L

Here, carrying an insignificant figure doesn't change the value.

<u>Calculate the concentration.</u>

c = \frac{n}{V}

c = \frac{0.25941(9)mol}{0.5000 L}              

c = 0.5188(3) \frac{mol}{L}         <= Keep an insignificant figure for rounding

c = 0.5188 \frac{mol}{L}              <= Rounded up

c = 0.5188M               <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

7 0
3 years ago
In one sample of a compound of copper and oxygen, 3.12g of the compound contains 2.50g of copper and the remainder is oxygen. Ca
blsea [12.9K]

Answer:

% composition O = 19.9%

% composition Cu = 80.1%

Explanation:

Given data:

Total mass of compound = 3.12 g

Mass of copper = 2.50 g

Mass of oxygen = 3.12 - 2.50 = 0.62 g

% composition = ?

Solution:

Formula:

<em>% composition = ( mass of element/ total mass)×100</em>

% composition Cu = (2.50 g / 3.12 g)×100

% composition Cu = 0.80 ×100

% composition Cu = 80.1%

For oxygen:

<em>% composition = ( mass of element/ total mass)×100</em>

% composition O = (0.62 g / 3.12 g)×100

% composition O = 0.199 ×100

% composition O = 19.9%

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Gain .....
It loses electron when a positive charge is formed..ex Na+

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Will the different flavors of khool aid effect the temperature of the boiling point
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No it does not effect the temperature of boiling point
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