All categories

2 years ago

What is this temperature in degree celsius of 77.2k

Chemistry

1 answer:

nalin [4]2 years ago

7 0

Degree celsius plus 273.15 equals degree kelvin so 77.2 minus 273.15 equals -195.95

You might be interested in

Module 2 dba for chemistry questions? flvs

NeX [460]

Can you translate to english?

4 0

3 years ago

Which substance gets broken down in a homogeneous mixture? O colloid solution solute Osolvent

astra-53 [7]

Answer:

Solute

Explanation:

i just did the quiz

3 0

3 years ago

1. A container holds four times as many moles of CO2 as O2. If the total pressure in the container is 20. atm,

Hitman42 [59]

Answer:

Partial pressure of CO2 = 16 atm

Explanation:

Total number of moles of gases = 4+1 = 5 moles

Mole fraction of CO2 = 4/5

Partial pressure of CO2 = mole fraction of CO2 × total pressure

Partial pressure of CO2 = (4/5) × 20

Partial pressure of CO2 = 16 atm

7 0

2 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

7 0

2 years ago

What is the difrense between soft water and hard water?

Nutka1998 [239]

Hard water... is water that contains an appreciable quantity of dissolved minerals (like calcium and magnesium). Soft water... is treated water in which the only ion is sodium. As rainwater falls, it is naturally soft.

6 0

2 years ago