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kati45 [8]
3 years ago
10

How is a mixture different from a compound? 1. Particles of a mixture are combined chem2. Components of a mixture can be sepa- r

ated by physical means. 3. Composition of a mixture may be con- stant. 4. Components of a mixture can only be separated chemically.
Chemistry
1 answer:
wariber [46]3 years ago
3 0

Answer:

2. Components of a mixture can be separated by physical means

Explanation:

The Compound is defined as the substances that are chemically combined while mixtures are the substances in which two or more compounds are physically combined together.

The difference between compound and mixture from the given options is that mixture can be separated physically unlike compounds.

In a mixture, two or more compounds can be mixed in any proportion that can be separated by physical means. For example: salt from seawater (mixture) can be separated evaporation and nails from the mixture of sand and iron can be separated through magnetic separation.

Hence, the correct answer is "Components of a mixture can be separated by physical means".

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P4 +6Br2 --> 4PBr3
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Answer:

1 mole of P₄ consumed:4 moles of PBr₃ produced

Explanation:

Look at the coefficients in the balanced chemical equation to determine the ratios. For every 1 mole of P₄ consumed (P₄ has a coefficient of 1), 4 moles of PBr₃ would be produced (PBr₃ has a coefficient of 4).

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2.00 L of 0.800 M NaNO3 must be prepared from a solution known to be 2.50 M in concentration.How many mL are required? Plus don'
Morgarella [4.7K]

Answer:

1.36 × 10³ mL of water.

Explanation:

We can utilize the dilution equation. Recall that:

\displaystyle M_1V_1= M_2V_2

Where <em>M</em> represents molarity and <em>V</em> represents volume.

Let the initial concentration and unknown volume be <em>M</em>₁ and <em>V</em>₁, respectively. Let the final concentration and required volume be <em>M</em>₂ and <em>V</em>₂, respectively. Solve for <em>V</em>₁:

\displaystyle \begin{aligned} (2.50\text{ M})V_1 &= (0.800\text{ M})(2.00\text{ L}) \\ \\ V_1 & = 0.640\text{ L} \end{aligned}

Therefore, we can begin with 0.640 L of the 2.50 M solution and add enough distilled water to dilute the solution to 2.00 L. The required amount of water is thus:
\displaystyle 2.00\text{ L} - 0.640\text{ L} = 1.36\text{ L}

Convert this value to mL:
\displaystyle 1.36\text{ L} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 1.36\times 10^3\text{ mL}

Therefore, about 1.36 × 10³ mL of water need to be added to the 2.50 M solution.

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