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3 years ago

1) The term which indicates that the element occurs in pairs is ____. Select the letter of the correct answer.

2 answers:

7 0

1. Diatomic
2. Products
3. Reactants
4. Coefficient
5. Subscript
6. True
7. False
8. First you must write the reactants. Ensure the valencies of the reactants are correct. Draw an arrow. Write the products. Ensure the valencies of the products are correct and apply subscripts as necessary. Balance the equation by ensuring both sides have equal amounts of each elements - use only coefficients to complete this NOT subscripts.

Vaselesa [24]3 years ago

5 0

The answer is a 2 is b

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Describe the position of metal ,non metals and metollieds in perodic tables

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3 years ago

Consider the following reaction at a high temperature. Br2(g) ⇆ 2Br(g) When 1.35 moles of Br2 are put in a 0.780−L flask, 3.60 p

UNO [17]

Answer : The equilibrium constant $K_c$ for the reaction is, 0.1133

Explanation :

First we have to calculate the concentration of $Br_2$ .

$\text{Concentration of }Br_2=\frac{\text{Moles of }Br_2}{\text{Volume of solution}}$

$\text{Concentration of }Br_2=\frac{1.35moles}{0.780L}=1.731M$

Now we have to calculate the dissociated concentration of $Br_2$ .

The balanced equilibrium reaction is,

$Br_2(g)\rightleftharpoons 2Br(aq)$

Initial conc. 1.731 M 0

At eqm. conc. (1.731-x) (2x) M

As we are given,

The percent of dissociation of $Br_2$ = $\alpha$ = 1.2 %

So, the dissociate concentration of $Br_2$ = $C\alpha=1.731M\times \frac{1.2}{100}=0.2077M$

The value of x = 0.2077 M

Now we have to calculate the concentration of $Br_2\text{ and }Br$ at equilibrium.

Concentration of $Br_2$ = 1.731 - x = 1.731 - 0.2077 = 1.5233 M

Concentration of $Br$ = 2x = 2 × 0.2077 = 0.4154 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be :

$K_c=\frac{[Br]^2}{[Br_2]}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.4154)^2}{1.5233}=0.1133$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.1133

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3 years ago

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lara31 [8.8K]

Answer:

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Explanation:

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3 years ago

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tekilochka [14]

Answer:

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Explanation:

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4 years ago

A sample of neon occupies a volume of 461 mL at 1 atm and 273 K. What will be the volume of the neon when the pressure is reduce

Softa [21]

Answer:

0.501 L

Explanation:

To solve this problem we will use Boyle,s Law. According to this law "The volume of given amount of gas is inversely proportional to applied pressure at constant temperature".

V∝ 1/P

V= K/P

VP=K

Here the K is proportionality constant.

so,

P1V1 = P2V2

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V= volume

Given data:

P1= 1 atm

V1= 461 mL

P2= 0.92 atm

V2= ? (L)

To solve this problem we have to convert the mL into L first.

1 L = 1000 mL

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Now we will put the values in the equation,

P1V1 = P2V2

V2= P1V1/ P2

V2= 1 atm × 0.461 L / 0.92 atm

V2= 0.501 L

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4 years ago