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3 years ago

PLEASEEEE HELPPPPP ASAP!!!!!

Chemistry

2 answers:

Wewaii [24]3 years ago

8 0

Answer: Adding a reactant will shift the reaction towards the products.

Adding a product will shift the reaction towards the reactants.

Removing a reactant will shift the reaction towards the reactants.

Removing a product will shift the reaction towards the products.

Explanation:

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.

Adding a reactant will increase the concentration of reactants and hence the reaction will try to undo the increase by shifting the reaction towards the products.

Adding a product will increase the concentration of products and hence the reaction will try to undo the increase by shifting the reaction towards the reactants.

Removing a reactant will decrease the concentration of reactants and hence the reaction will try to undo the decrease by shifting the reaction towards the reactants.

Removing a product will decrease the concentration of products and hence the reaction will try to undo the decrease by shifting the reaction towards the products.

Alex3 years ago

5 0

Answer:

Adding a reactant - shift towards product

Adding a product - shift towards reactant

Removing a reactant - shift towards reactant

Removing a product- shift towards product

Explanation:

a) Reactant added – When a reactant is added to a chemical reaction which has not yet reached the equilibrium, then the added reactant is partially consumed and the reaction shifts towards the product. However, there is no effect on the K value

b) Adding a product – When a product is added, then it is partially consumed and the reaction shifts towards the reactant. Again, there is no change in K value.

c) Removing a reactant – The reaction will shift towards the reactant

d) Removing a product- The reaction will shift towards the product

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Answer : The current passing between the electrodes is, $1.056\times 10^{-2}A$

Explanation :

First we have to calculate the charge of sodium ion.

$q=ne$

where,

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Now put all the given values in the above formula, we get:

$q=(2.68\times 10^{16})\times (1.6\times 10^{-19}C)=4.288\times 10^{-3}C$

Now we have to calculate the charge of chlorine ion.

$q'=ne$

where,

q' = charge of chlorine ion

n = number of chlorine ion = $3.92\times 10^{16}$

e = charge on electron = $1.6\times 10^{-19}C$

Now put all the given values in the above formula, we get:

$q'=(3.92\times 10^{16})\times (1.6\times 10^{-19}C)=6.272\times 10^{-3}C$

Now we have to calculate the current passing between the electrodes.

$I=\frac{q}{t}+\frac{q'}{t}$

$I=\frac{4.288\times 10^{-3}}{1.00}+\frac{6.272\times 10^{-3}}{1.00}$

$I=1.056\times 10^{-2}A$

Thus, the current passing between the electrodes is, $1.056\times 10^{-2}A$

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A substance registers a temperature change from 20 to 40 to what incremental temperature change does this correspond

ss7ja [257]

Given:

Initial temperature, T1 = 20 C

Final temperature, T2 = 40 C

To determine:

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Explanation:

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