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olga_2 [115]
3 years ago
15

A piece of solid tin is submerged in silver nitrate solution a reaction occurs producing tin(IV) nitrate solution and solid silv

er
Write a word equation write a skeleton equation write a balanced chemical equation
Chemistry
1 answer:
Sauron [17]3 years ago
8 0

Answer:

Tin + silver trioxonitrate V -------->Tin IV nitrate + silver

Explanation:

The term word equation refers to an equation in which the reactants and products are written in words rather than chemical symbols.

Note than tin is above silver in the electrochemical series hence silver will be displaced by tin as follows;

Tin + silver trioxonitrate V -------->Tin IV nitrate + silver

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3 years ago
For the chemical equation SO 2 ( g ) + NO 2 ( g ) − ⇀ ↽ − SO 3 ( g ) + NO ( g ) the equilibrium constant at a certain temperatur
MakcuM [25]

Answer:

Moles of NO₂ = 0.158

Explanation:

                         SO 2 ( g ) + NO 2 ( g ) ⇄  SO 3 ( g ) + NO ( g )

              According to the law of mass equation

                                     

                                       K_{c} = \frac{[SO_{3} ][NO]}{[SO_{2}][NO_{2}  ]}

                              ⇒   3.10 = \frac{(1.00)(1.00)}{(2.30) [NO_{2} ]}    At equilibrium [SO₃] = [NO]

                              ⇒ [NO₂] = \frac{1}{6.3}

                              ⇒ [NO₂] = 0.158

So. number of moles of NO₂ at equilibrium added = 0.158

8 0
3 years ago
Which form of emission is commonly not written in nuclear equations because they do not affect charges, atomic numbers, or mass
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Answer: It is Gamma ray

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The rate constant for a certain reaction is measured at two different temperatures:
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Answer: The activation energy Ea for this reaction is 22689.8 J/mol

Explanation:

According to Arrhenius equation with change in temperature, the formula is as follows.

ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]

k_1 = rate constant at temperature T_1 = 2.3\times 10^8

k_2 = rate constant at temperature T_2 = 4.8\times 10^8

E_a= activation energy = ?

R= gas constant = 8.314 J/kmol

T_1 = temperature = 280.0^0C=(273+280)=553K

T_2 = temperature = 376.0^0C=(273+376)=649K

Putting in the values ::

ln \frac{4.8\times 10^8}{2.3\times 10^8} = \frac{-E_{a}}{8.314}[\frac{1}{649} - \frac{1}{553}]

E_a=22689.8J/mol

The activation energy Ea for this reaction is 22689.8 J/mol

3 0
3 years ago
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