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krok68 [10]
3 years ago
9

Billions of pounds of urea are produced annually for use as a fertilizer. Balance the skeletal equation for the synthesis of ure

a. Express your answer as a chemical equation. Identify all of the phases in your answer.
NH3(g)+CO2(g)==>CO(NH2)2(s)+H20(l)
Chemistry
1 answer:
klio [65]3 years ago
5 0
2NH₃(g) + CO₂(g) → CO(NH₂)₂(s) + H₂O(l)
is the balanced equation for the synthesis of urea.
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3 years ago
Which chemical equation demonstrates the law of conservation of mass!
zlopas [31]

Answer:

F. 2NO + 02 —> 2NO

H. 4NH3 + 502 —> 4NO + 6H20

Explanation:

The law of conservation of mass states that matter can neither be created nor destroyed during a chemical reaction but can be convert from one form to another.

2NO + 02 —> 2NO

From the above, the total number of N on the left balance the total number on the right i.e 2 atoms of N on both side of the equation.

The total number of O on the left balance the total number on the right i.e 2 atoms of O on both side of the equation. This is certified by the law of conservation of mass.

4NH3 + 502 —> 4NO + 6H20

From the above, the total number of N on the left balance the total number on the right i.e 4 atoms of N on both side of the equation.

The total number of O on the left balance the total number on the right i.e 10 atoms of O on both side of the equation.

The total number of H on the left balance the total number on the right i.e 12 atoms of O on both side of the equation.

This is certified by the law of conservation of mass.

The rest equation did not conform to the law of conservation of mass as the atoms on the left side did not balance those on the right side

5 0
3 years ago
A 20.00-mL sample of a weak base is titrated with 0.0568 M HCl. At the endpoint, it is found that 17.88 mL of titrant was used.
nata0808 [166]

Answer: 0.0508mL

Explanation: Using the basic formula that states: C acid * V acid = C base * V base. we have:0.568 * 17.88 = 20 * C base.

therefore concentration of the base is 1.0156/20 = 0.0508 mL

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It will explode...............
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