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KengaRu [80]
3 years ago
9

BRAINLIESTTT ASAP! PLEASE HELP ME :)

Chemistry
1 answer:
Iteru [2.4K]3 years ago
7 0

Answer:

The mass of the products left in the test tube will be less than that of the original reactants.

Explanation

The equation for the reaction is

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

 1.0          3.0              3.9             0.1

Assume you started with 1.0 g of Mg.

It will react with 3.0 g of HCl to form 3.9 g of MgCl2 and 0.1 g of H2

.

Mass of reactants = mass of products

        1.0 g + 3.0 g = 3.9 g + 0.1 g

                    4.0 g = 4.0 g

The Law of Conservation of Mass is obeyed.

However, your test tube and its contents will weigh 0.1 g less than it did before the reaction.

Does that contradict the Law of Conservation of Mass? It does not.

One of the products was the gas, hydrogen, and it escaped from the test tube. You weren't measuring all the products, so test tube and its contents weighed less than before.

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3 years ago
Which formulas could represent the empirical formula and the molecular formula of a given compound?
Alina [70]

Answer:

d

Explanation:

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A chemist dissolves of pure hydroiodic acid in enough water to make up of solution. Calculate the pH of the solution. Be sure yo
Gre4nikov [31]

Answer:

1.76

Explanation:

There is some info missing. I think this is the original question.

<em>A chemist dissolves 660.mg of pure hydroiodic acid in enough water to make up 300.mL of solution. Calculate the pH of the solution. Be sure your answer has the correct number of significant digits.</em>

<em />

Step 1: Calculate the molarity of HI(aq)

M = mass of solute / molar mass of solute × liters of solution

M = 0.660 g / 127.91 g/mol × 0.300 L

M = 0.0172 M

Step 2: Write the acid dissociation reaction

HI(aq) ⇄ H⁺(aq) + I⁻(aq)

HI is a strong acid, so [H⁺] = 0.0172 M

Step 3: Calculate the pH

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5 0
4 years ago
A chemist must dilute of aqueous aluminum chloride solution until the concentration falls to . He'll do this by adding distilled
marshall27 [118]

Answer:

0.257 L

Explanation:

The values missing in the question has been assumed with common sense so  that the concept could be applied

Initial volume of the AICI3 solution =23.1 \mathrm{mL}

Initial Molarity of the solution =833 \mathrm{mM}

Final molarity of the solution =75.0 \mathrm{mM}

Final volume of the solution =?

From Law of Dilution, M_{f} V_{f}=M_{i} V_{i}

\Rightarrow V_{f}=\frac{M_{i} V_{i}}{M_{f}}=\frac{833 \mathrm{mM} \times 23.1 \mathrm{mL}}{75.0 \mathrm{mM}}=256.564 \mathrm{mL}=0.256564 \mathrm{L}=0.257 L

Final Volume of the solution =0.257

7 0
3 years ago
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