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ladessa [460]
3 years ago
12

Look at the picture of two different substances above. Which substance is a mixture? A. Y and Z

Chemistry
2 answers:
Murrr4er [49]3 years ago
7 0

Answer:

Option B

Explanation:

A mixture is two or more substances combined by physical means. Substance Y is composed of many different types of substances mixed together. Therefore, it is a mixture. Substance Z is composed of all the same types of particles. Therefore, it is not a mixture.

Feliz [49]3 years ago
5 0

Answer:

option B is correct.

Y is a mixture in the picture.

Explanation:

Different types of substances combine physically and form a mixture.  

Mixture formed by the combination of two or more substances that that combine in same physical state.

For example mixture of copper sulphate and solid chloride in dry form and that will look like white and blue crystal in a mixture.

The picture Z showing dots of same color and is not a mixture, while picture Y shows dots of different colors clearly indicating that it is picture for mixture.

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One of the commercial uses of sulfuric acid is the production of calcium sulfate and phosphoric acid. If 26.8 g of Ca₃(PO₄)₂ rea
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Answer:

The percent yield reaction is 64.3%

Explanation:

This is the ballanced reaction

Ca₃(PO₄)₂ (s) + 3H₂SO₄ (aq) → 2H₃PO₄ (aq) + 3CaSO₄ (aq)

Let's determine the moles of our reactants:

Mass / Molar mass = Mol

26.8 g / 310.18 g/m = 0.0864 moles of phosphate.

54.3 g / 98.06 g/m = 0.554 moles of sulfuric

1 mol of phosphate reacts with 3 mol of sulfuric so

0.0864 mol of PO₄⁻³ will react with (0.0864 .3)/1 = 0.259 moles

I have 0.554 of sulfuric, so this is the reactant in excess.

The limiting reagent is the Phosphate.

1 mol of PO₄⁻³ produces 2 mol of phosphoric

0.0864 of PO₄⁻³ will produce the double amount (0.0864 .2) = 0.173 moles

Mol . molar mass = Mass

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A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to . He'll do this by add
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Answer:

0.295 L

Explanation:

It seems your question lacks the final concentration value. But an internet search tells me this might be the complete question:

" A chemist must dilute 47.2 mL of 150. mM aqueous sodium nitrate solution until the concentration falls to 24.0 mM. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in liters. Be sure your answer has the correct number of significant digits. "

Keep in mind that if your value is different, the answer will be different as well. However the methodology will remain the same.

To solve this problem we can<u> use the formula</u> C₁V₁=C₂V₂

Where the subscript 1 refers to the concentrated solution and the subscript 2 to the diluted one.

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And <u>converting into L </u>becomes:

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