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pychu [463]
3 years ago
6

When aluminum is mixed with iron ii oxide iron metal and aluminum oxide?

Chemistry
1 answer:
Alexandra [31]3 years ago
6 0
Are produced along with a large quantitu of heat
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A student ran the following reaction in the laboratory at 311 K:CH4(g) + CCl4(g) 2CH2Cl2(g)When she introduced 4.10×10-2 moles o
statuscvo [17]

<u>Answer:</u> The value of K_{eq} is 0.044

<u>Explanation:</u>

We are given:

Initial moles of methane = 4.10\times 10^{-2}mol=0.0410moles

Initial moles of carbon tetrachloride = 6.51\times 10^{-2}mol=0.0651moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}

So, concentration of methane = \frac{0.0410}{1.00}=0.0410M

Concentration of carbon tetrachloride = \frac{0.0651}{1.00}=0.0651M

The given chemical equation follows:

                    CH_4(g)+CCl_4(g)\rightleftharpoons 2CH_2Cl_2(g)

<u>Initial:</u>          0.0410    0.0651

<u>At eqllm:</u>     0.0410-x   0.0651-x       2x

We are given:

Equilibrium concentration of carbon tetrachloride = 6.02\times 10^{-2}M=0.0602M

Evaluating the value of 'x', we get:

\Rightarrow (0.0651-x)=0.0602\\\\\Rightarrow x=0.0049M

Now, equilibrium concentration of methane = 0.0410-x=[0.0410-0.0049]=0.0361M

Equilibrium concentration of CH_2Cl_2=2x=[2\times 0.0049]=0.0098M

The expression of K_{eq} for the above reaction follows:

K_{eq}=\frac{[CH_2Cl_2]^2}{[CH_4]\times [CCl_4]}

Putting values in above expression, we get:

K_{eq}=\frac{(0.0098)^2}{0.0361\times 0.0603}\\\\K_{eq}=0.044

Hence, the value of K_{eq} is 0.044

8 0
3 years ago
2. (Exercise 4.1.6) A liquid adhesive consists of a polymer dissolved in a solvent. The amount of polymer in the solution is imp
Lilit [14]

Answer:

800 lb of pure solvent , 1700 lb of 20% solution and 500 lb of 10% solution will be mixed to form 3000 lb of 13 % solution .

Explanation:

3000 lb of 13% solution is required .

Total adhesive in weight = 3000 x .13 = 390 lb of adhesive

Available = 500 lb of 10% solution = 50 lb of adhesive

Rest = 390 - 50 = 340 lb required .

rest mass of solution = 3000 - 500 = 2500 lb

mass of adhesive required = 340 lb

Let the mass  of 20% required be V

mass of adhesive = .20 V

.20 V = 340

V = 1700

rest of the volume = 2500 - 1700 = 800 lb which will be of pure solvent

So 800 lb of pure solvent , 1700 lb of 20% solution and 500 lb of 10% solution will be mixed to form 3000 lb of 13 % solution .

6 0
3 years ago
State whether the following statements are true or false (with justification). (a) 1 mol of N2 has more molecules than 1 mol of
MAVERICK [17]

Answer:

A. False.

Every substance contains the same number of molecules i.e 6.02x10^23 molecules

B. False.

Mass conc. = number mole x molar Mass

Mass conc. of 1mole of N2 = 1 x 28 = 28g

Mass conc. of 1mol of Ar = 1 x 40 = 40g

The mass of 1mole of Ar is greater than the mass of 1mole of N2

C. False.

Molar Mass of N2 = 2x14 = 28g/mol

Molar Mass of Ar = 40g/mol

The molar mass of Ar is greater than that of N2.

Explanation:

7 0
2 years ago
In some areas of the Earth, the crust is squeezed and pushed upward. This is a _______ process in that it directly forms _______
NemiM [27]
So..... I believe this is a Convergent boundary and mountains..
7 0
2 years ago
Read 2 more answers
How many PO43? ions are in a mole of K3PO4?
adoni [48]

K₃PO₄ → 3K⁺ (aq) + PO₄³⁻(aq)

One mole of PO₄³⁻ ion gets dissociated from one mole of K₃PO₄

As per the definition of Avogadro's number, 1 mole = 6.022 x 10²³ ions

One mole of PO₄³⁻ ions x  (6.022 x 10²³ ions/ 1 mole of PO₄³⁻ ions )

= 6.022 x 10²³ ions

Therefore , there are 6.022 x 10²³ PO₄³⁻ ions in a mole of K₃PO₄.

7 0
2 years ago
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