Please help!

Chemistry

1 answer:

kozerog [31]1 year ago

7 0

Answer:

they are metals so they are:-

-malleable

-ductile

because metallic bonding is the same in all direction throughout the solid

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Which of the following characterizes a reaction at equilibrium?

Alekssandra [29.7K]

I’m pretty sure the correct answer is D.

4 0

2 years ago

Iron(III) sulfide reacts with hydrochloric acid to produce hydrogen sulfide and iron(II) chloride. Wright the formula equation f

NeX [460]

Answer:

Fe2S3 + 4HCl —> 2H2S + 2FeCl2 + S

Explanation:

From the question, we were told that:

Iron(III) sulfide reacts with hydrochloric acid to produce hydrogen sulfide and iron(II) chloride.

Please note that elemental sulphur is produced along with this reaction. The equation for the reaction is given below:

Fe2S3 + HCl —> H2S + FeCl2 + S

Now, we can balance the equation as shown below:

There are 3 atoms of S on the left side of the equation and 2 atoms on the right. It can be balance by putting 2 in front of H2S as shown below:

Fe2S3 + HCl —> 2H2S + FeCl2 + S

Now, we have 4 atoms of H on the right side and 1atom on the left side. It can be balance by putting 4 in front of HCl as shown below:

Fe2S3 + 4HCl —> 2H2S + FeCl2 + S

Also, there are 4 atoms of Cl on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of FeCl2 as shown below:

Fe2S3 + 4HCl —> 2H2S + 2FeCl2 + S

Now, we can see that the equation is balanced

5 0

3 years ago

15.0 moles of gas are in a 8.00 L tank at 22.3 ∘C∘C . Calculate the difference in pressure between methane and an ideal gas unde

leva [86]

Answer:

$\Delta P=4.10atm$

Explanation:

Hello!

In this case, since the ideal gas equation is used under the assumption of no interaction between molecules and perfectly sphere-shaped molecules but the van der Waals equation actually includes those effects, we can compute each pressure as shown below, considering the temperature in kelvins (22.3+273.15=295.45K):

$P^{ideal}=\frac{nRT}{V}=\frac{15.0mol*0.08206\frac{atm*L}{mol*K}*295.45K}{8.00L}=45.5atm$

Next, since the VdW equation requires the molar volume, we proceed as shown below:

$v=\frac{8.00L}{15.0mol}=0.533\frac{L}{mol}$

Now, we use its definition:

$P^{VdW}=\frac{RT}{v-b} -\frac{a}{v^2}$

Thus, by plugging in we obtain:

$P^{VdW}=\frac{0.08206\frac{atm*L}{mol*K}*295.45K}{0.533mol/L-0.0430L/mol} -\frac{2.300L^2*atm/mol^2}{(0.533L/mol)^2}\\\\P^{VdW}=49.44atm-8.09atm\\\\P^{VdW}=41.4atm$