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2 years ago

If atoms are so tiny how do we know they exist

1 answer:

5 0

There are three ways that scientists have proved that these sub-atomic particles exist. They are direct observation, indirect observation or inferred presence and predictions from theory or conjecture. Scientists in the 1800's were able to infer a lot about the sub-atomic world from chemistry.

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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

A chemist needs to find the concentration of a solution of barium hydroxide.

Brums [2.3K]

Method 1: gravimetry

advantages: Impurities in the sample can be identified

disadvantages: The process is long, because it goes through several stages

Method 2: titration

advantages: the process is fast, because the titrate and titrant react immediately

disadvantages: Sometimes the determination of the end point of the titration is carried out too fast or too slowly so that the calculations carried out are inaccurate

3 0

1 year ago

Which phrase is the best summery of the model shown?

aliina [53]

D.) a transformation of light energy into chemical energy

3 0

3 years ago

Which solution will form a precipitate when mixed with a solution of aqueous na2co3? which solution will form a precipitate when

trapecia [35]

Answer:

$CuCl_{2}(aq.)$ will form a precipitate of insoluble $CuCO_{3}$ when aqueous $Na_{2}CO_{3}$ is added.

Explanation:

According to solubility rule-

all carbonates are insoluble except group IA compounds and $NH_{4}^{+}$

all salts of sodium are soluble

When $Na_{2}CO_{3}$ is added to given solutions, a double displacement reaction takes place in each solution to form a sodium salt and a carbonate salt.

So, in accordance with solubility rule, addition of $Na_{2}CO_{3}$ into $CuCl_{2}(aq.)$ will result precipitation of insoluble $CuCO_{3}$

Reaction: $Na_{2}CO_{3}(aq.)+CuCl_{2}(aq.)\rightarrow 2NaCl(aq.)+CuCO_{3}(aq.)$

4 0

3 years ago

A toy car is pushed 10 meters in 5 seconds. What is the speed of the toy car in meters per second?

dedylja [7]

The answer is C because the toy car goes 2 meters per each second which in the end 2x5=10.

4 0

2 years ago