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3 years ago

9

3CuCl2 + 2Al = 2AlCl3 + 3Cu using the molar ratio between the limiting reactant (aluminum) and the product copper, determine the

theoretical yield of copper.

1 answer:

inn [45]3 years ago

4 0

Answer:

0.014

Explanation:

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Which statement about Niels Bohr’s atomic model is true?

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3 years ago

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If 5.58 g of iron reacts with sulfur to produce 8.79 g of iron sulfide, what is the mass of reacting sulfur? A) 3215 B) 14:37 C)

gogolik [260]

Answer: A) 3.21 g

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side.

$Fe+S\rightarrow FeS$

We are given:

Mass of iron = 5.58 g

Mass of iron sulphide = 8.79 g

Mass of sulphur = x g

Total mass on reactant side = 5.58 + x

Total mass on product side = 8.79 g

Applying law of conservation of mass, we get:

$5.58+x=8.79\\\\x=3.21g$

Hence, the mass of reacting sulfur is 3.21 g.

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3 years ago

My professor gave me two questions to solve using the Van Der Waals Equation. She told us to solve for P and the second one we h

Fed [463]

Answer:

P=atm

$b=\frac{L}{mol}$

Explanation:

The problem give you the Van Der Waals equation:

$(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT$

First we are going to solve for P:

$(P+\frac{n^{2}a}{V^{2}})=\frac{nRT}{(V-nb)}$

$P=\frac{nRT}{(V-nb)}-\frac{n^{2}a}{v^{2}}$

Then you should know all the units of each term of the equation, that is:

$P=atm$

$n=mol$

$R=\frac{L.atm}{mol.K}$

$a=atm\frac{L^{2}}{mol^{2}}$

$b=\frac{L}{mol}$

$T=K$

$V=L$

where atm=atmosphere, L=litters, K=kelvin

Now, you should replace the units in the equation for each value:

$P=\frac{(mol)(\frac{L.atm}{mol.K})(K)}{L-(mol)(\frac{L}{mol})}-\frac{(mol^{2})(\frac{atm.L^{2}}{mol^{2}})}{L^{2}}$

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

$P=\frac{L.atm}{L-L}-atm$

Then operate the fraction subtraction:

P= $P=\frac{L.atm-L.atm}{L}$

$P=\frac{L.atm}{L}$

And finally you can find the answer:

P=atm

Now solving for b:

$(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT$

$(V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}$

$nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}$

$b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}$

Replacing units:

$b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}$

Multiplying and dividing units,(please see the second photo below), we have:

$b=\frac{L-\frac{L.atm}{atm}}{mol}$

$b=\frac{L-L}{mol}$

$b=\frac{L}{mol}$

7 0

3 years ago

An oxide of aluminum contains 0.545 g of Al and 0.485 g of O. Find the empirical formula

uysha [10]

Answer:

The answer is: <u>Al2O3</u>

Explanation:

The data they give us is:

0.545 gr Al

0.485 gr O.

To find the empirical formula without knowing the grams of the compound, we find it per mole:

0.545 g Al * 1 mol Al / 27 g Al = 0.02 mol Al

0.485 g O * 1 mol O / 16 g O = 0.03 mol O

Then we must divide the results obtained by the lowest result, which in this case is 0.02:

0.02 mol Al / 0.02 = 1 Al

0.03 mol O / 0.02 = 1.5 O

Since both numbers have to give an integer, multiply by 2 until both remain integers:

1Al * 2 = 2Al

1.5O * 2 = 3O

Now the answer is given correctly:

Al2O3

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3 years ago

Explain what causes air pressure:

Sidana [21]

Gravity pulling the mass of the atmosphere toward the center of the earth. Atmosphere includes air, water vapor, etc.

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