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3 years ago

When substances combine to form mixtures they keep their properties. O True O False

Chemistry

2 answers:

Gnesinka [82]3 years ago

8 0

Answer:

True

Explanation:

Rudik [331]3 years ago

7 0

Answer:

True

Explanation:

This is true because When two or more elements or compounds are blended without combining chemically, you've got a mixture. Each substance in a mixture keeps its own properties, and mixtures can be separated using physical or mechanical means. Solids, liquids and gases can all be blended into mixtures.

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the specific temperature at which a substance changes from a liquid to a gas is an example of a a. chemical bond. b. chemical ch

9966 [12]

Physical property because going from a liquid to a gas doesn't change it's chemical composition only it's physical state

4 0

4 years ago

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Assuming the materials don't mix, show how the materials would "stack up" in a graduated cylinder. (Based on their densities) (J

o-na [289]

Explanation:

air

wood

ice

corn oil

water

glycerin

rubber

corn syrup

steel

I believe?

6 0

3 years ago

What is the percent composition of N H S O in (NH4)2SO4

Lena [83]

Answer:

The percent composition is 21% N, 6% H, 24% S and 49% O.

Explanation:

1st) The molar mass of (NH4)2SO4 is 132g/mol, and it represents the 100% of the mass composition.

In 1 mole of (NH4)2SO4, there are:

- 2 moles of N.

- 8 moles of H.

- 1 mole of S.

- 4 moles of O.

2nd) It is necessary to calculate the mass of each element, multiplying its molar mass by the number of moles:

- 2 moles of N (14g/mol) = 28g

- 8 moles of H (1g/mol) = 8g

- 1 mole of S (32g/mol) = 32g

- 4 moles of O (16g/mol) = 64g

3rd) With a mathematical rule of three we can calculate the percent composition of each element in the molecule of (NH4)2SO4:

$\begin{gathered} \text{ Nitrogen:} \\ 132g-100\% \\ 28g-x=\frac{28g*100\%}{132g} \\ x=21\% \end{gathered}$ $\begin{gathered} \text{ Hydrogen:} \\ 132g-100\operatorname{\%} \\ 8g-x=\frac{8g*100\operatorname{\%}}{132g} \\ x=6\% \end{gathered}$ $\begin{gathered} \text{ Sulfur:} \\ 132g-100\operatorname{\%} \\ 32g-x=\frac{32g*100\operatorname{\%}}{132g} \\ x=24\% \end{gathered}$ $\begin{gathered} \text{ Oxygen:} \\ 100\%-21\%-6\%-24\%=49\% \\ \\ \end{gathered}$

In this case, we can calculate the percent composition of Oxygen by subtracting the other percentages, since the total must be 100%.

So, the percent composition is 21% N, 6% H, 24% S and 49% O.

8 0

1 year ago

A saline solution contains 1.1% NaCl by mass. How much NaCl is present in 96.3 g of this solution

timurjin [86]

Answer:

= 1.0593 g

Explanation:

1.1% NaCl by mass, means;

1.1 g per 100 g = 0.011

Therefore; the amount of NaCl present in 93.6 g of the solution is;

= 0.011 × 96.3 g

= 1.0593 g

8 0

4 years ago

A solution is prepared by dissolving 15.0 g of nh3 in 250.0 g of water. the density of the resulting solution is 0.974 g/ml. the

iris [78.8K]

To solve this problem, first we assume the volume is purely additive. The density of the mixture can then be calculated by the summation of mass fraction of each component divided by its individual density:

1 / ρ mixture = (x NH3 / ρ NH3) + (x H2O / ρ H2O) ---> 1

Calculating for mass fraction of NH3:

x NH3 = 15 g / (15 g + 250 g)

x NH3 = 0.0566

Therefore the mass fraction of water is:

x H2O = 1 – x NH3 = 1 – 0.0566

x H2O = 0.9434

Assuming that the density of water is 1 g / mL and substituting the known values back to equation 1:

1 / 0.974 g / mL = [0.0566 / (ρ NH3)] + [0.9434 / (1 g / mL)]

ρ NH3 = 0.680 g / mL

Given the density of NH3, now we can calculate for the volume of NH3:

V NH3 = 15 g / 0.680 g / mL

V NH3 = 22.07 mL

The number of moles NH3 is: (molar mass NH3 is 17.03 g / mol)

n NH3 = 15 g / 17.03 g / mol

n NH3 = 0.881 mol

Therefore the molarity of NH3 in the solution is:

Molarity = 0.881 mol / [(22.07 mL + 250 mL) * (1L / 1000 mL)

M = 3.238 mol/L = 3.24 M

8 0

3 years ago

Read 2 more answers