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2 years ago

4

Chemistry

1 answer:

matrenka [14]2 years ago

5 0

Answer:

Yes.

Explanation:

Yes, this difference of readings will definitely affect the results of the experiment as well as the E values because the readings taken by both students are different from one another. There is a fault in one of the thermometer because both shows different readings of temperature of the same solution. This will affect the overall experiment and due to this error, we are unable to tell that which one reading is correct so the answer is uncertain or unsure.

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Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1

Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

Explanation:

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of $SbCl_5$ is increasing .So, the equilibrium will shift in the right direction.

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

Concentration of $SbCl_5$ = 0.195 M

Concentration of $SbCl_3$ = $6.98\times 10^{-2} M$

Concentration of $Cl_2$ = $6.98\times 10^{-2} M$

On adding more $[SbCl_5$ to 0.370 M at equilibrium :

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

Initially

0.370 M $6.98\times 10^{-2}M$

At equilibrium:

(0.370-x)M $(6.98\times 10^{-2}+x)M$

The equilibrium constant of the reaction = $K_c$

$K_c=2.50\times 10^{-2}$

The equilibrium expression is given as:

$K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}$

$2.50\times 10^{-2}=\frac{(6.98\times 10^{-2}+x)M\times (6.98\times 10^{-2}+x)M}{(0.370-x) M}$

On solving for x:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

3 0

3 years ago

You wash dishes for a chemistry laboratory to make extra money for laundry. You earn 12 dollars/hour, and each shift lasts 75 mi

nadezda [96]

Answer:

A)You must work 2 shifts to wash 10 loads of laundry

B)You can wash 15 loads pf laundry if you work for 3 shifts

C) each shift will pay for 3.75 loads of laundry if the cost per load rises to 16 quarters

Explanation:

Given:

amount earned = 12 dollars/hour

Duration for one shift = 75 minutes

Amount needed for Laundry = 12 quarters/load.

A. How many shifts must you work if you wish to wash 10 loads of laundry

To wash 10 loads, you will be requiring

=> $10 \times 12 \text{quarters}$

=> 120 quarters

1 quarter = 0.25 dollars---------------------------------(1)

120 quarters

= $120 \times 0.25 \text{ dollars}$

= 30 dollars

You need 30 dollars for washing 10 loads

In one hour i,e 60 minutes you earn= 12 dollars

In one minute the amount earned will be

$= \frac {12}{60}$

= 0.2 dollar-------------------------------------------------------(2)

Now to earn 30 dollars you have to work for

= $30 \times 0.2$ minutes

= 150 minutes

Thus the number of shifts will be

= $\frac{150}{75}$

= 2 shifts

B. How many loads of laundry can you wash if you work 3 shifts

We Know that 1 shift = 75 minutes

So 3 shifts = $75 \times 3$ = 225 minutes

From (2),

In 1 minute = 0.2 dollars is earned

So in 225 minutes = 225 \ times 0.2 = 45 dollar is earned

45 dollar = $45 \times 4$ = 180 quarters

You can wash 1 load with 12 quarters

With 180 quarters you can wash = $\frac{180}{12}$ = 15 loads

C. How many loads of laundry will each shift pay for if the cost per load rises to 16 quarters

Cost per load = 16 quarters

From (1)

16 quarters = $0.25 \times 16$ = 4 dollars per load

From (2) In 1 shift the amount earned will be

=> $75 \times 0.2$

=15 dollars

With 4 dollars you can wash 1 load

So with 15 dollars = $\frac{15}{4}$ = 3.75 loads

4 0

3 years ago

I need help with this bill nye worksheet. someone help. It's due tomorrow cant mess my grade up. I just picked a random subject

viva [34]

send a better picture please and thank you

5 0

2 years ago

A solution contains 90 mL of methanol, 18 mL of propanol, and 2 mL of diethyl ether. Which is the solvent in this solution?

otez555 [7]

Answer:

Methanol

Explanation:

5 0

3 years ago

You have a series circuit with a current of 6 amps and three resistors on it, with resistances of 10 ohms, 5 ohms, and 6 ohms, r

Mamont248 [21]

For this problem, you first find the net resistance of the circuit which can be done by adding up the resistors since this circuit is a series. 10Ω+5Ω+6Ω=21Ω

Since we are given the current of 6A, we can use Ohm’s Law of V=IR.

V=6A*21Ω

V=126 volts

3 0

3 years ago