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timofeeve [1]
3 years ago
11

Write a chemical equation that describes the dissolving of solid magnesium nitrate, Mg(NO3)2, in water.

Chemistry
2 answers:
serious [3.7K]3 years ago
7 0
<span>Given below is the chemical equation that describes the dissolving of solid magnesium nitrate, Mg(NO3)2, in water Mg(NO3)2(aq) → Mg2+(aq) + 2NO3-(aq)</span>
Leviafan [203]3 years ago
5 0

Answer:

Your answer is Mg(NO3)2, in water Mg(NO3)2(aq) → Mg2+(aq) + 2NO3-(aq)

Explanation:

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What is the importance of a chemical symbol?
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Chemical symbols are used to standardize the 'language of chemistry' and to identify elements and atoms in a chemical formula easily. Chemical symbols consist of one or two letters, most often derived from the name of the element.

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Effect of azeotrope system​
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3 years ago
Covalent compounds are generally not very hard. Justify the statement.
Eduardwww [97]

Covalent compounds are generally not very hard because they are formed by two or more nonmetallic atoms.

<h3>COVALENT COMPOUNDS:</h3>

Covalent compounds are compounds whose constituent elements are joined together by covalent bonds.

Covalent bonding occurs when two or more nonmetallic atoms of an element share valence electrons. This means that covalent compounds will not be physically hard since they constitute non-metals.

Examples of covalent compounds are:

  1. H2 - hydrogen
  2. H2O - water
  3. HCl - hydrogen chloride
  4. CH4 - methane

Learn more about covalent compounds at: brainly.com/question/21505413

4 0
2 years ago
The pyrolysis of ethane proceeds with an activation energy of about 300 kJ/mol. How much faster is the decomposition at 625°C th
Alona [7]

Answer:

The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_2 = rate of reaction at T_2

K_1 = rate of reaction at T_1

Ea = activation energy of the reaction

R = gas constant = 8.314 J/K mol

E_a=300 kJ/mol=300,000 J/mol

T_2=625^oC=898.15 K,T_1=525^oC=798.15 K

\log (\frac{K_2}{K_1})=\frac{300,000 J/mol}{2.303\times 8.314 J/K mol}[\frac{1}{798.15 K}-\frac{1}{898.15 K}]

\log (\frac{K_2}{K_1})=2.185666

K_2=153.344\times K_1

The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.

4 0
3 years ago
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