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Alika [10]
2 years ago
14

An aqueous solution was prepared containing 1.0mol of AgNO3 and 1.0mol of FeSO4 in 1.00 dm^3. of water. When equilibrium was est

ablished, there was 0.44mol of Ag+(aq) in the mixture. Ag + (aq) + Fe2 + (aq) <==> Ag(s) + Fe3+ (aq) . What is the numerical value of Kc? . A 0.35 B 0.62 C 1.62 D 2.89
Chemistry
1 answer:
Brums [2.3K]2 years ago
8 0
Kc = [Fe3+]/ ([Ag+] [Fe2+]) 
<span>Initial concentrations [Ag+]= [Fe2+]= 1.0 mol/ dm^3 [Fe3+] =0.0 mol/dm^3 </span>
<span>change: Ag+: (1.0 - .44) mol/dm^3 = 0.56 mol/ dm^3 = change of Fe2+. Change of Fe3+ is 0.56 mol/dm^3. </span>
<span>Equilibrium concentrations [Ag+] = [Fe2+] = 0.44 mol/dm^3 [Fe3+] = 0.56 mol/dm^3 </span>

<span>Kc = 0.56/0.44^2 dm^3/mol = 2.89 dm^3 /mol</span>
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First-order reaction that results in the destruction of a pollutant has a rate constant of 0.l/day. (a) how many days will it ta
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Rate equation for first order reaction is as follows:

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}

Here, k is rate constant of the reaction, t is time of the reaction, A_{0} is initial concentration and A_{t} is concentration at time t.

The rate constant of the reaction is 0.1 day^{-1}.

(a) Let the initial concentration be 100, If 90% of the chemical is destroyed, the chemical present at time t will be 100-90=10, on putting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{10}=23.03 days

Thus, time required to destroy 90% of the chemical is 23.03 days.

(b) Let the initial concentration be 100, If 99% of the chemical is destroyed, the chemical present at time t will be 100-99=1, on putting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{1}=46.06 days

Thus, time required to destroy 99% of the chemical is 46.06 days.

(c)  Let the initial concentration be 100, If 99.9% of the chemical is destroyed, the chemical present at time t will be 100-99.9=0.1, on putting the values,

t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{0.1}=69.09 days

Thus, time required to destroy 99.9% of the chemical is 69.09 days.

5 0
2 years ago
Which one is balanced or unbalanced force 1. Change object motion 2. Push or pull 3. Have direction 4. Do not change objects mot
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Answer:

Explanation:

1. Change object motion - unbalanced force

2. Push or pull - Balanced forces do not cause a change in motion. When balanced forces act on an object at rest, the object will not move. If you push against a wall, the wall pushes back with an equal but opposite force. Neither you nor the wall will move. Forces that cause a change in the motion of an object are unbalanced forces. So if object move it is unbalanced.

3. Have direction - unbalanced force

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2 years ago
A standard backpack is approximately 30cm x 30cm x 40cm. Suppose you find a hoard of pure gold while treasure hunting in the wil
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Explanation:

The dimensions of a standard backpack is 30cm x 30cm x 40cm

The mass of an average student is 70 kg

We know that, the density of gold is 19.3 g/cm³.

Let m be the mass of the backpack. So,

\text{density}=\dfrac{\text{mass}}{\text{volume}}\\\\m=d\times V\\\\m=19.3\ g/cm^3\times (30\times 30\times 40)\ cm^3\\\\m=694800\ g\\\\\text{or}\\\\m=694.8\ kg\approx 700\ kg

An average student has a mass of 70 kg. If we compare the mass of student and mass of backpack, we find that the backpack is 10 times of the mass of the student.

8 0
2 years ago
A medical class is hearing a lecture on vision. The doctor asks them what happens when the eye produces too much fluid, causing
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Answer:

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2 years ago
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Answer:

\boxed{ \text{8.40 g}} 

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After one half-life, half (50 %) of the original amount will remain.  

After a second half-life, half of that amount (25 %) will remain, and so on.  

We can construct a table as follows:  

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<u>half-lives</u>      <u>    t/yr      </u>       <u>Remaining</u>  

      0               0                          1

      1                1.3  billion              ½

      2              2.6                          ¼

      3              3.9                          ⅛

We see that after 2 half-lives, ¼ of the original mass remains.

Conversely, if two half-lives have passed, the original mass must have been four times the mass we have now.

Original  mass = 4 × 2.10 g = \boxed{ \text{8.40 g}}

7 0
2 years ago
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