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Radda [10]
3 years ago
13

15.50 g of NH4Cl reacts with an excess of AgNO3. In the reaction 35.50 g AgCl is produced. what is the theoretical yield of AgCl

calculated in grams?
Chemistry
1 answer:
Readme [11.4K]3 years ago
8 0

Answer:

The answer to your question is 41.6 g of AgCl

Explanation:

Data

mass of NH₄Cl = 15.5 g

mass of AgNO₃ = excess

mass of AgCl = 35.5 g

theoretical yield = ?

Process

1.- Write the balanced chemical reaction.

              NH₄Cl  +  AgNO₃   ⇒   AgCl  +  NH₄NO₃

2.- Calculate the molar mass of NH₄Cl and AgCl

NH₄Cl = 14 + 4 + 35.5 = 53.5 g

AgCl = 108 + 35.5 = 143.5 g

3.- Calculate the theoretical yield

                 53.5 g of NH₄Cl -------------------- 143.5 g of AgCl

                  15.5 g of NH₄Cl  -------------------    x

                         x = (15.5 x 143.5) / 53.5

                         x = 2224.25 / 53.5

                         x = 41.6 g of AgCl

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Subtract 5.01x10^7-30x10^-9
professor190 [17]
<span>
</span><span>7/9 x 45/1 = 7/1 x 5/1
by dividing both sides by 9
7/1 x 5/1 = 35/1 = 35</span>
8 0
4 years ago
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A rock is believed to have been formed 1.25 billion years ago, as calculated by using potassium-40 dating. If the half-life of p
Leya [2.2K]

Answer:

The answer to the question is

50 % of the original amount of potassium 40 will be left after one half life or 1.25 billion years

Explanation:

To solve the question we note that the half life is the time for half of the quantity of  substance that undergoes radioactive decay to  disintegrate, thus

we have

half life of potassium 40 K₄₀ = 1.25 billion years

To support the believe tht the rock was formed 1.25 billion years ago we have

N_{(t)} =N_{(0)} (\frac{1}{2}) ^{\frac{t}{t_{\frac{1}{2} } } }

After 1.25 billion years we have

N_{(t)} =N_{(0)} (\frac{1}{2}) ^{\frac{1.25billion}{1.25 billion}  } } }  = N_{(t)} =N_{(0)} (\frac{1}{2}) ^{1 } } } =0.5 of N_{(0)} will be left or 50 % of the original amount of potassium 40 will be left

4 0
4 years ago
At 1000°C, cyclobutane (C4H8) decomposes in a first-order reaction, with the very high rate constant of 79 1/s, to two molecules
leva [86]

Explanation:

It is known that for first order reaction, the equation is as follows.

           t = \frac{2.303}{K} log \frac{[C_{4}H_{8}]_{o}}{[C_{4}H_{8}]_{t}}

    t = ?,        K = rate constant = 79 1/s

Initial conc. of C_{4}H_{8} = 1.68

Decompose amount of C_{4}H_{8} = 52% of 1.68

                                                   = \frac{52}{100} \times 1.68

                                                   = 0.8736

                                                   = 0.87

Now, [C_{4}H_{8}]_{t} = (1.68 - 0.87)

                         = 0.81

Therefore, calculate the value of t as follows.

               t = \frac{2.303}{K} log \frac{[C_{4}H_{8}]_{o}}{[C_{4}H_{8}]_{t}}

                  = \frac{2.303}{79 s^{-1}} log \frac{1.68}{0.81}

                  = \frac{2.303}{79 s^{-1}} \times 0.316

                  = 9.212 \times 10^{-3} s

                  = 0.00921 s

Thus, we can conclude that 0.00921 s will be taken for 52% of the cyclobutane to decompose.

7 0
3 years ago
What is the name of two areas where rivers leave rich, fertile soil?
Tom [10]

Answer:

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5 0
3 years ago
PLEASE HELP!!!
poizon [28]

The net ionic equation of the reaction is:

  • Pb²⁺ (aq) +2 I⁻ (aq) → PbI2(s)

<h3>What are net ionic equations?</h3>

Net ionic equations are equations where ions which remain in solution known as spectator ions are not shown in the equation. Only ions involved in formation of product are shown.

In the given equation, sodium and nitrate ions are spectator ions.

The net ionic equation of the given reaction is as follows:

  • Pb²⁺ (aq) +2 I⁻ (aq) → PbI2(s)

In conclusion, spectator ions are not shown in net ionic equation.

Learn more about net ionic equations at: brainly.com/question/19705645
#SPJ1

6 0
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