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3 years ago

The three factors that determine the characteristics of a solution are

Chemistry

1 answer:

Ahat [919]3 years ago

7 0

Answer:

Instrinsic Solubility, Temperature, and Sample Size.

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a helium balloon has a volume of 2.95 liters at 25 c. the volume of the ballon decreased to 2.25 l after it is placed outside on

OleMash [197]

Answer:

The outside temperature is -45.8°C

Explanation:

When a gas keeps on constant its moles and its pressure, we can assume that volume will be increased or decreased as the T° (absolute T° in K).

V1 / T1 = V2 / T2

2.95L/298K = 2.25L / T2

(2.95L/298K ) . T2 = 2.25L

T2 = 2.25L . 298K / 2.95L

T2 = 227.2K

T°K - 273 = T°C

227.2K - 273 = -45.8°C

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3 years ago

To separate a mixture of iron filings and salt, the<br> most efficient method would be

CaHeK987 [17]

Iron filings can be attracted by the magnet whereas salt can not. So the mixture can be separated by a magnet

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4 years ago

Read 2 more answers

A mbxture of iron and sulfur can be separated by attracting the Iron particles with a magnet. If the iron and sulfur mixture is

Alex777 [14]

Answer:

C. homogeneous mixture

4 0

4 years ago

1 point

inysia [295]

Answer:

The fungus has grown larger

Explanation:

Because where the orange is in the fridge and even normally you out oranges on the counter or in a bowl, where it's in the fridge it got old faster.

8 0

3 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0

3 years ago

The three factors that determine the characteristics of a solution are​

The three factors that determine the characteristics of a solution are