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3 years ago

If you have 177 grams of C4H80, how many moles do you have?

Chemistry

1 answer:

quester [9]3 years ago

4 0

I think it’s 2.45, not 100% sure

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How many atoms of hydrogen are there in 3.0kg of ethane

kvasek [131]

Ethane has the formula C2H6.
From the periodic table:
molecular mass of carbon = 12 grams
molecular mass of hydrogen = 1 gram
Therefore:
molar mass of ethane = 2(12) + 6(1) = 30 grams

This means that each 30 grams of ethane contains 6 grams of hydrogen. To know how many grams of hydrogen are in 3000 grams of ethane (3 kg), we will simply use cross multiplication as follows:
mass of hydrogen = (3000 x 6) / 30 = 600 grams

7 0

3 years ago

When a catalyst is added to a reversible reaction in equilibrium state the value of equilibrium constant

Sergio [31]

Explanation:

The value of equilibrium constant doesn't change when a catalyst is added.

Equilibrium constant depends on Concentration of reactants , Pressure and Temperature.

6 0

2 years ago

Determine the molar mass of CuSO4 (the solute) in a 1.0M aqueous solution of CuSO4

inna [77]

Answer:

See explanation.

Explanation:

Hello,

In this case, we could have two possible solutions:

A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:

$M_{CuSO_4}=m_{Cu}+m_{S}+4*m_{O}=63.546 g/mol+32.00g/mol+4*16.00g/mol\\\\M_{CuSO_4}=159.546g/mol$

That is the mass of copper (II) sulfate contained in 1 mol of substance.

B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:

$M=\frac{n_{solute}}{V_{solution}}$

So you can solve for the moles of the solute:

$n_{solute}=M*V_{solution}$

Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:

$n_{solute}=1mol/L*1.5L=1.5mol$

But this is just a supposition.

Regards.

4 0

2 years ago

The bonds in the compound MgSO4 can be described as

Butoxors [25]

C. Sulfur and oxygen (non metals) forms a covalent bond while the magnesium (a metal) will react with both non metals to form an ionic bond

7 0

3 years ago

?Al + ?H2SO4 → ?Al2(SO4)3 +3H2

AfilCa [17]

The answer is: 27 grams of aluminium.

Balanced chemical reaction: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂.

n(H₂) = 1.5 mol; amount of hydrogen.

Form chemical reaction: n(Al) : n(H₂) = 2 : 3.

n(Al) = 2 · 1.5 mol ÷ 3.

n(Al) = 1.0 mol; amount of aluminium.

m(Al) = n(Al) · M(Al).

m(Al) = 1 mol · 27 g/mol.

m(Al) = 27 g; mass of aluminium.

5 0

3 years ago