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2 years ago

All solutionsare mixtures; butall mixtures arenot solutions.why?

1 answer:

5 0

All solutions are mixtures of two or more substances, but unless the mixture has a homogeneous distribution of solutes in the solvent, then the mixture is not a solution. Therefore, all mixtures are not solutions.

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During an experiment, 95 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of

almond37 [142]

Answer:

Actual yield: 86.5 grams.

Explanation:

How many moles of formula units in 95 grams of calcium carbonate $\rm CaCO_3$ ?

Refer to a modern periodic table for relative atomic mass data:

Ca: 40.078;
C: 12.011;
O: 15.999.

Formula mass of $\rm CaCO_3$ :

$M(\mathrm{CaCO_3}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{1\times 12.011}_{\rm C} + \underbrace{3\times 15.999}_{\rm O} = \rm 100.086\;g\cdot mol^{-1}$ .

$\displaystyle n(\mathrm{CaCO_3}) = \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} = \rm \frac{95\;g}{100.086\;g\cdot mol^{-1}} = 0.949184\;mol$ .

How many moles of $\rm CaCl_2$ will be produced?

The coefficient in front of $\rm CaCO_3$ in the chemical equation is the same as that in front of $\rm CaCl_2$ . That is:

$\displaystyle \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = 1$ .

$\displaystyle n(\mathrm{CaCl_2}) = n(\mathrm{CaCO_3})\cdot \frac{n(\rm CaCl_2)}{n(\rm CaCO_3)} = n(\mathrm{CaCO_3}) = \rm 0.949184\;mol$ .

What's the theoretical yield of calcium chloride? In other words, what's the mass of $\rm 0.949184\;mol$ of $\rm CaCl_2$ ?

Again, refer to a periodic table for relative atomic data:

Ca: 40.078;
Cl: 35.45.

$M(\mathrm{CaCl_2}) = \underbrace{1\times 40.078}_{\rm Ca} + \underbrace{2\times 35.45}_{\rm Cl} = \rm 110.978\;g\cdot mol^{-1}$ .

$\begin{aligned}m(\mathrm{CaCl_2}) &= n(\mathrm{CaCl_2})\cdot M(\mathrm{CaCl_2})\\ &= \rm 0.949184\;mol\times 110.978\;g\cdot mol^{-1}\\ &= \rm 105.339\; g\end{aligned}$ .

What's the actual yield of calcium chloride?

$\displaystyle \text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%$ .

$\displaystyle \begin{aligned}\text{Actual Yield} &= \text{Theoretical Yield}\cdot \frac{\text{Percentage Yield}}{100\%}\\ &=\rm 105.339\; g \times \frac{82.15\%}{100\%}\\&= \rm 86.5\;g \end{aligned}$ .

8 0

3 years ago

When sodium and oxygen react to form an ionic compound, the mass of the compound ____ the sum of the masses of the individual el

Anna007 [38]

Hi Lara

The answer is : C

Is equal to

I hope that's help!

5 0

3 years ago

Read 2 more answers

9. A student wished to prepare ethylene gas by dehydration of ethanol at 140oC using sulfuric acid as the dehydrating agent. A l

Arte-miy333 [17]

We have that the the liquid is

C_2H_5OH (ethanol
And at a condition of H_2SO4 as catalyst and temp 170

From the question we are told

A student wished to prepare <em>ethylene </em>gas by <em>dehydration </em>of ethanol at 140oC using sulfuric acid as the <em>dehydrating </em>agent.
A low-boiling liquid was obtained instead of ethylene.
What was the liquid, and how might the reaction conditions be changed to give ethylene

<h3>Ethylene formation</h3>

Generally the equation is

2C_2H_5OH------CH3CH_2O-CH_2CH_3+H_20

Therefore

with ethanol at 140oC

The product is diethyl ethen

The reaction at 170 ethylene will give

C_2H_5OH-------CH_2=CH_2+H_2O( at a condition of H_2SO4 as catalyst and temp 170)

Therefore

The the liquid is

C_2H_5OH (ethanol

For more information on Ethylene visit

brainly.com/question/20117360

8 0

1 year ago

Which of the following statements is true about one formula unit of RuF2?<br>

valina [46]

For me the answer is E. It is composed of one Ru2+ ion and two F- ions.

6 0

2 years ago

BRAINLIEST

mars1129 [50]

I believe that it most likely would be C,

6 0

2 years ago

All solutionsare mixtures; butall mixtures arenot solutions.why?​

All solutionsare mixtures; butall mixtures arenot solutions.why?