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3 years ago

Can someone help me plz!!! Calculate the mass, in grams, of 2 moles of CaCl2.

Chemistry

2 answers:

Annette [7]3 years ago

8 0

Ooh math is harddd?!

Luden [163]3 years ago

7 0

Answer:

222.2 grams of CaCl2 in 2 moles

Explanation:

Mole=given mass/gram mass formula

2moles=x/Gram Mass Formula

(Gram Mass Formula of CaCl2):

Ca=40.1

Cl= 35.5 x 2= 71

-----------------------

GFM=111.1

2 moles = x/111.1 g

x=222.2

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What is the change in enthalpy when 11. 0 g of liquid mercury is heated by 15°c?

Vlad1618 [11]

Enthalpy change is the difference between energy used and energy gained. The change in enthalpy of the liquid mercury is 0.0231 kJ.

<h3>What is the enthalpy change?</h3>

Enthalpy change is the difference between the energy used to break chemical bonds and the energy gained by the products formed in a chemical reaction.

The enthalpy change is given by,

$\rm \Delta H_{rxn} = \rm q_{rxn}$

and,

$\rm q = mc\Delta T$

Given,

Mass of the liquid mercury (m) = 11.0 gm

The specific heat of mercury (c) = 0.14 J per g per degree Celsius

Temperature change = 15 degrees Celsius

Enthalpy change is calculated as:

$\begin{aligned} \rm q &= \rm mc\Delta T\\\\&= 11 \times 0.14 \times 15\\\\&= 23.1 \;\rm J\end{aligned}$

Therefore, 0.0231 kJ is the change in enthalpy.

Learn more about enthalpy change here:

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6 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

3 years ago

Enter your answer in the provided box.

nexus9112 [7]

Answer:

V₂ = 50.93 L

Explanation:

Initial volume, $V_1=43.1\ L$

Initial temperature, $T_1=24^{\circ} C=24+273=297\ K$

Final temperature, $T_2=78^{\circ} C=78+273=351\ K$

We need to find the final volume of the gas. The relation between the volume and the temperature is given by :

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{43.1\times 351}{297}\\\\V_2=50.93\ L$

So, the final volume of the gas is 50.93 L.

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3 years ago

Relation between glancing and angle of deviation<br>

shtirl [24]

Answer:

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2 years ago

What compound name is As5O2

levacccp [35]

Answer:

Arsenic pentoxide

Explanation:

3 0

2 years ago