Any change in the volume or shape of earths crust is called??

The system co2(g) + h2(g) ⇀↽ h2o(g) + co(g) is at equilibrium at some temperature. at equilibrium a 4.00 l vessel contains 1.00

Marina CMI [18]

<u>Answer:</u> The moles of $CO_2$ added to the system is 7.13 moles

<u>Explanation:</u>

We are given:

Moles of $CO_2$ at equilibrium = 1.00 moles

Moles of $H_2$ at equilibrium = 1.00 moles

Moles of $H_2O$ at equilibrium = 2.40 moles

Moles of $CO$ at equilibrium = 2.40 moles

Volume of the container = 4.00 L

Concentration is written as:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

The given chemical equation follows:

$CO_2(g)+H_2(g)\rightleftharpoons H_2O(g)+CO(g)$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[CO][H_2O]}{[CO_2][H_2]}$

Putting values in above equation, we get:

$K_c=\frac{(\frac{2.40}{4.00})\times (\frac{2.40}{4.00})}{(\frac{1.00}{4.00})\times (\frac{1.00}{4.00})}\\\\K_c=5.76$

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of CO = 0.791 mol/L

Volume of solution = 4.00 L

Putting values in above equation, we get:

$0.791M=\frac{\text{Moles of CO}}{4.00L}\\\\\text{Moles of CO}=(0.791mol/\times 4.00L)=3.164mol$

Extra moles of CO = (3.164 - 2.40) = 0.764 moles

Let the moles of $CO_2$ needed be 'x' moles.

Now, equilibrium gets re-established:

$CO_2(g)+H_2(g)\rightleftharpoons H_2O(g)+CO(g)$

Initial: 1.00 1.00 2.40 2.40

At eqllm: (0.236+x) 0.236 3.164 3.164

Again, putting the values in the expression of $K_c$ , we get:

$5.76=\frac{(\frac{3.164}{4.00})\times (\frac{3.164}{4.00})}{(\frac{0.236+x}{4.00})\times (\frac{0.236}{4.00})}\\\\5.76=\frac{10.011}{0.056+0.236x}\\\\x=7.13$

Hence, the moles of $CO_2$ added to the system is 7.13 moles

4 0

3 years ago

What causes cold, deep ocean currents to form in the ocean currents near the poles?

lianna [129]

What causes deep, cold currents to form in the oceans near the poles? The sinking of dense, cold water paired with high salinity.

4 0

2 years ago

Gold forms a substitutional solid solution with silver. Calculate the number of gold atoms per cubic centimeter (in atoms/cm3) f

Leokris [45]

Answer:

Gold: 1.1 x 10²² atoms/cm³

Silver: 4.8 x 10²² atoms/cm³

Explanation:

100 g of the alloy will have 29 g of Au and 71 g of Ag.

19.32 g Au ____ 1 cm³

29 g Au ______ x

x = 1.5 cm³

10.49 g Ag ____ 1 cm³

71 g Ag _______ y

y = 6.8 cm³

The total volume of 100g of the alloy is x+y = 8.3 cm³.

Gold:

196.97 g Au____ 6.022 x 10²³ atoms Au

29 g Au _______ w

w = 8.9 x 10²² atoms Au

8.9 x 10²² atoms Au ____ 8.3 cm³

A ____ 1 cm³

A = 1.1 x 10²² atoms Au

Silver:

107.87 g Ag____ 6.022 x 10²³ atoms Ag

71 g Ag _______ w

w = 4.0 x 10²³ atoms Ag

4.0 x 10²³ atoms Ag ____ 8.3 cm³

B ____ 1 cm³

B = 4.8 x 10²² atoms Ag

6 0

3 years ago

Calculate the standard entropy of reaction at 298 K for the reaction Hg(liq) + Cl2(g) → HgCl2(s) The standard molar entropies of

never [62]

Answer:

−153.1 J / (K mol)

Explanation:

Calculate the standard entropy of reaction at 298 K for the reaction Hg(liq) + Cl2(g) → HgCl2(s) The standard molar entropies of the species at that temperature are: Sºm (Hg,liq) = 76.02 J / (K mol) ; Sºm (Cl2,g) = 223.07 J / (K mol) ; Sºm (HgCl2,s) = 146.0 J / (K mol)

Hg(liq) + Cl2(g) → HgCl2(s)

Given that;

The standard molar entropies of the species at that temperature are:

Sºm (Hg,liq) = 76.02 J / (K mol) ;

Sºm (Cl2,g) = 223.07 J / (K mol) ;

Sºm (HgCl2,s) = 146.0 J / (K mol)

The standard molar entropies of reaction = Sºm[products] - Sºm [ reactants]

= 146.0 J / (K mol) – [76.02 J / (K mol) +223.07 J / (K mol) ]

= -153.09 J / (K mol)

= or -153.1 J / (K mol)

Hence the answer is −153.1 J / (K mol)

5 0

4 years ago

At which point does a planet move most slowly in its orbit , at aphelion or perihelion

SVEN [57.7K]

Answer:

aphelion

Explanation:

A planet moves with constantly changing speed as it moves about its orbit. The fastest a planet moves is at perihelion (closest) and the slowest is at aphelion (farthest).

7 0

2 years ago