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Sever21 [200]
3 years ago
9

If 5.738 grams of AgNO3 is mixed with 4.115 grams of BaCl2 and allowed to react according to the balanced equation: BaCl2(aq) +

2 AgNO3(aq) → 2 AgCl(s) + Ba(NO3)2(aq) What is the limiting reagent? BaCl2AgNO3 How many grams of AgCl could be produced? grams AgCl What mass, in grams, of the excess reagent will remain? grams of excess reagent
Chemistry
1 answer:
IceJOKER [234]3 years ago
5 0

Answer:

Limiting reagent: AgNO3

grams AgCl : 2.44 g AgCl

grams of excess reagent remain: 0.62 g BaCl2

Explanation:

1. Change grams to mol:

<em>AgNO3:</em>

5.738g x (1mol/169.87g) = 0.034 mol AgNO3

<em>BaCl2:</em>

4.115g x (1 mol/208.23g) = 0.020 mol BaCl2

2. Limiting reagent:

<em>AgNO3:</em>

0.034 mol AgNO3 x (1 mol BaCL2/ 2mol AgNO3) = 0.017 mol BaCl2

<em>BaCl2:</em>

0.020 mol BaCl2 x (2 mol AgNO3/1 mol BaCl2) = 0.04 mol AgNO3

Limiting reagent: AgNO3

3. Grams of AgCl produced:

Using the limiting reagent:

0.017 mol AgNO3 x (2mol AgCl / 2 mol AgNO3) = 0.017 mol AgCl

4. Change mol to grams:

0.017 mol AgCl x ( 143.32 g AgCl /1mol AgCl) =2.44 g AgCl

5. Grams of the excess reagent:

0.034 mol AgNO3 x (1 mol BaCl2 / 2 mol AgNO3) = 0.017 mol BaCl2

0.020 mol BaCl2 - 0.017 mol BaCl2 = 0.003 mol BaCl2

0.003 mol BaCl2 x ( 208.23 g BaCl2 / 1 mol BaCl2) = 0.62 g BaCl2

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Lynna [10]

Answer:

Therefore the concentration of the reactant after 4.00 minutes will be 0.686M.

Explanation:

The unit of k is s⁻¹.

The order of the reaction = first order.

First order reaction: A first order reaction is  a reaction in which the rate of reaction depends only the value of the concentration of the reactant.

-\frac{d[A]}{dt} =kt

[A] = the concentration of the reactant at time t

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t= time

Here k= 4.70×10⁻³ s⁻¹

t= 4.00

[A₀] = initial concentration of reactant = 0.700 M

-\frac{d[A]}{dt} =kt

\Rightarrow -\frac{d[A]}{[A]}=kdt

Integrating both sides

\Rightarrow\int -\frac{d[A]}{[A]}=\int kdt

⇒ -ln[A] = kt +c

When t=0 , [A] =[A₀]

-ln[A₀]  = k.0 + c

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Therefore

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Putting the value of k, [A₀] and t

- ln[A] =4.70×10⁻³×4 -ln (0.70)

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5 0
3 years ago
Thorium-234 has a half-life of 24.1 days. How many grams of a 150 g sample would you have after 120.5 days?
chubhunter [2.5K]

Answer:

8.625 grams of a 150 g sample of Thorium-234  would be left after 120.5 days

Explanation:

The nuclear half life represents the time taken for the initial amount of sample  to reduce into half of its mass.

We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.

Initial amount of Thorium-234 available as per the question is 150 grams

So now  we start with 150 grams  of Thorium-234

150 \times \frac{1}{2}=24.1

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<span><span>
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<span><span>
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