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3 years ago

5

When the ideal gas law is arranged as shown below, what property of the gas is being solved for (represented by the X)? x equals

PV over nR

1 answer:

Mkey [24]3 years ago

3 0

That's kind of a ponderous way to describe it, but your 'X' represents
the absolute temperature of the ideal gas.

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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

7 0

3 years ago

When an aqueous solution of magnesium nitrate is mixed with an aqueous solution of potassium carbonate, ____________.?

spin [16.1K]

<span>Ca(NO3)2 + Na2CO3 = CaCO3 + 2NaNO3
Yes a precipitate of Calcium Carbonate is formed since it is insoluble in water.
Mol Wt of Calcium Nitrate is 164. And that of Calcium Carbonate is 100.
One mole of Calcium Nitrate produces one mole of Calcium Carbonate.
i.e. 164 gms will produce 100gms of precipitate
So, 1.74gms of Calcium Carbonate will be obtained from 2.85gms Calcium Nitrate present in the original solution.</span>

3 0

3 years ago

What is energy being added to the water cycle?

Komok [63]

Answer:

yes

Explanation:

6 0

3 years ago

How many moles of CO gas are in 34.6 L?

Troyanec [42]

moles of CO gas : 1.545

<h3>Further explanation</h3>

Standard conditions for temperature and pressure are used as a reference in certain calculations or conditions

There are 2 conditions that are usually used as a reference : STP and RTP

Assuming the STP state :

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.

Then for 34.6 L of CO gas :

$\tt moles=\dfrac{34.6}{22.4}=1.545$

6 0

3 years ago

Why aren't descriptive investigations repeatable?

Bad White [126]

In descriptive investigations, we still haven't formed any hypothesis yet so we seek information by asking question.

It's not repeatable because repeating the questions over and over again without any clue about what we want to seek is completely waste of time.

Hope this helps xox :)

8 0

3 years ago

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