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2 years ago

Is hydrogen gas a mixture or pure substance?

Chemistry

1 answer:

Akimi4 [234]2 years ago

8 0

Answer:

pure hydrogen is a pure substance even though it consists of many different types of molecules. what makes it pure substance is that it is free from contamination.

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Which event is an example of melting?

avanturin [10]

Answer:

A. Wax drips down the side of a lot candle.

Explanation:

The chemical change from solid to liquid. This is a combustion reaction, so carbon dioxide gas and water vapour is also produced but you can't see them

8 0

3 years ago

stomach acid contains dilute hydrochloric acid write a chemical equation to show the neutralization of stomach acid by milk of m

Eddi Din [679]

Answer:

Mg(OH)2(s) + 2HCl(aq) → 2H2O(l) + MgCl2(aq)

Explanation:

4 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

A gas mixture with 4 mol of Ar, x moles of Ne, and y moles

maks197457 [2]

Answer:

a) $\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

b) $x=4$

c) $\Delta G_{max}=-2721.9 J/mol$

Explanation:

Gas mixture:

$n_{Ar}= 4 mol$

$n_{Ne}= x mol$

$n_{Xe}= y mol$

$n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}$

$n_{Ne} + n_{Xe}=2*n_{Ar}$

$x + y=8 mol$

$y=8 mol- x$

Mol fractions:

$x_{Ar}=\frac{4 mol}{12 mol}=1/3$

$x_{Ne}=\frac{x mol}{12 mol}=x/12$

$x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12$

Expression of $\Delta G_{mixing}$

$\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)$

$\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]$

$\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

Expression of $\Delta G_{max}$

$\frac{d \Delta G_{mixing}}{dx}=0$

$\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]$

$0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)$

$0=[ln (x/12)-ln ((8-x)/12)$

$ln (x/12)=ln ((8-x)/12)$

$x=(8-x)$

$x=4$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]$

$\Delta G_{max}=-2721.9 J/mol$

4 0

3 years ago

Is sugar ( in water) an example of an element or compound

tresset_1 [31]

Answer:

A compound

Example:

Sugar contains carbon, hydrogen and oxygen. Water (H2O) has two hydrogen atoms (H2) and one oxygen atom (O). So both sugar and water are chemical compounds. Together they are a mix of chemical compounds.

8 0

3 years ago