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2 years ago

Which statement describes heterogeneous mixtures?

2 answers:

5 0

The right answer is the option A. They are made up of two or more pure substances that are not chemically bonded. A classicall expample ot heterogeneous mixtures are sugar and salt. Sugar is a pure substance, salt is also a pure substance, when you mixe them you form mixture where salt is still salt and sugar is still sugar, that is what "they are not chemically bonded" means. So you can separate them by physical media.

Artemon [7]2 years ago

4 0

The righ answer is the option A. They are made up of two or more pure substances that are not chemically bonded. A classicall expample ot heterogeneous mixtures are sugar and salt. Sugar is a pure substance, salt is also a pure substance, when you mixe them you form mixture where salt is still salt and sugar is still sugar, that is what "they are not chemically bonded" means. So you can separate them by physical media.

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Calculate the mass, in grams, of Ag2CrO4 that will precipitate when 50.0mL of 0.20M AgNO3 solution is mixed with 40.0mL of 0.10M

Darina [25.2K]

Answer:

1.327 g Ag₂CrO₄

Explanation:

The reaction that takes place is:

2AgNO₃(aq) + K₂CrO₄(aq) → Ag₂CrO₄(s) + 2KNO₃(aq)

First we need to identify the limiting reactant:

We have:

0.20 M * 50.0 mL = 10 mmol of AgNO₃

0.10 M * 40.0 mL = 4 mmol of K₂CrO₄

If 4 mmol of K₂CrO₄ were to react completely, it would require (4*2) 8 mmol of AgNO₃. There's more than 8 mmol of AgNO₃ so AgNO₃ is the excess reactant. That makes K₂CrO₄ the limiting reactant.

Now we calculate the mass of Ag₂CrO₄ formed, using the limiting reactant:

4 mmol K₂CrO₄ * $\frac{1mmolAg_2CrO_4}{1mmolK_2CrO_4} *\frac{331.73mg}{1mmolAg_2CrO_4}$ = 1326.92 mg Ag₂CrO₄

1326.92 mg / 1000 = 1.327 g Ag₂CrO₄

7 0

2 years ago

Someone help me please, I will give you brainliest

Nikitich [7]

Answer:

conserve mass.

Explanation:

3 0

3 years ago

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vlabodo [156]

Left side metals right side non metals. arrganged in groups (columns) and periods (rows), each group tells you amount of valence electrons

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2 years ago

An object was measured by a worker as

Gre4nikov [31]

Answer:

$\boxed{2.8 \, \%}$

Explanation:

$\text{Percent error} = \dfrac{\lvert \text{Measured - Actual}\lvert}{ \text{Actual}} \times100 \, \%$

Data:

Predicted = 17.4 cm

Actual = 17.9 cm

Calculation:

$\text{Percent error} = \dfrac{\lvert 17.4 - 17.9\lvert}{17.9} \times 100 \, \% \\\\= \dfrac{\lvert-0.5\lvert}{17.9} \times 100 \, \% = \dfrac{0.5}{17.9} \times 100 \, \% \\\\= 0.028 \times 100 \, \% = \textbf{2.8 \%}\\\\\text{The percent error in the measurement is } \boxed{\textbf{2.8 \%}}$

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2 years ago

Give the electronegativity for the following elements of LI

amid [387]

Answer:

0.98

Explanation:

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2 years ago