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ivanzaharov [21]
4 years ago
9

Brass is an alloy of copper and zinc. The two metals are melted, mixed, and then solidified. How would you BEST classify brass?

A) homogeneous mixture B) heterogeneous mixture C) pure substance: mixture D) pure substance: compound
Chemistry
2 answers:
leonid [27]4 years ago
7 0

Answer:

homogeneous

Explanation:

Although made of two solids, brass is a homogeneous mixture or a solution. When melted and mixed, the two metals are evenly distributed throughout the entire mixture. The amounts of copper and zinc may vary, depending on the type of brass desired.

Lady bird [3.3K]4 years ago
3 0

Answer:

The Answer is A) homogeneous mixture...............

Explanation:

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If pure water boils at 99.8 degrees celcius, what is the expected elevated boiling point of a solution of 2.50g of CaCl2, in 50.
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INFORMATION:

We know that:

- pure water boils at 99.8 degrees celcius

And we must calculate the expected elevated boiling point of a solution of 2.50g of CaCl2, in 50.0mL (i.e., 50.0g) of H2O

STEP BY STEP EXPLANATION:

To calculate it, we need to use that:

Boiling point of solution = boiling point of pure solvent + boiling point elevation (ΔTb)

The elevation in boiling point (ΔTb) is proportional to the concentration of the solute in the solution. It can be calculated via the following equation.

ΔTb=i\times k_b\times m

Where,

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From given information, we know that:

- i = 3

Now, the ebullioscopic constant (Kb) is often expressed in terms of °C * kg * mol^-1. The value of Kb for water is 0.512.

So, kb = 0.512 °C * kg * mol^-1

Then, we must calculate the molality

\begin{gathered} Molality=\frac{\text{ moles of solute}}{\text{ kg of solvent}} \\ Molality=\frac{\frac{2.5g}{110.98\frac{g}{mol}}}{0.05kg}=0.45\frac{mol}{kg} \end{gathered}

So, m = 0.45 mol/kg

Replacing the values in the formula for ΔTb

\begin{gathered} ΔT_b=3\times0.512\frac{\degree C\cdot kg}{mol}\times0.45\frac{mol}{kg} \\ ΔT_b=0.69\degree C \end{gathered}

Finally, the expected elevated boiling point of the solution would be

\text{ Boling point of solution}=99.8\degree C+0.69\degree C=100.49\degree C

ANSWER:

The expected elevated boiling point of the solution is 100.49 °C

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