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lesantik [10]
1 year ago
5

ethanol and benzene dissolve in each other. when ml of ethanol is dissolved in l of benzene, what is the mass of the mixture?

Chemistry
1 answer:
dsp731 year ago
8 0

The total mass of the mixture will be "958.5 g".

Calculation ,

According to the question, the density of both the mixtures will be:

Ethanol = 0.785 g/mL ( given )

Benzene = 0.880 g/mL ( given )

The volume of both the mixtures,

Ethanol = 100 mL ( given )

Benzene = 1.00 L or 1000 mL  ( given )

As we know,

density = mass/volume

now,

The mass of Ethanol will be:

mass = density × volume =  0.785 g/mL × 100 mL

mass  = 78.5 g

The mass of Benzene will be:

mass = density × volume =   0.880 g/mL  × 1000 mL

mass  = 880 g

hence,

The total mass of the mixture = 78.5 g + 880 g = 958.5 g

Missing data ,

density of Ethanol = 0.785 g/mL

density of Benzene = 0.880 g/mL

The volume of both the mixtures,

Ethanol = 100 mL

Benzene = 1.00 L

Learn more about mass of the mixture here:

brainly.com/question/15189339

#SPJ4

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stellarik [79]
The correct answer is <span>Fusion Curve</span>
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3 years ago
The ionization constant (Ka) of HF is 6.7 X 10 ^-4. Which of the following is true in a 0.1 M solution of this acid?
spin [16.1K]
The ionization equation is:

HF ⇄ H(+) + F(-)

The ionization constant is Ka = [H(+)] * [H(-)] / [HF]

=> [H(+)] * [F(-)] = Ka * [HF]

Given that Ka < 1

[H(+)] * [F(-)] < [HF]

Which is [HF] >  [H(+)] * [F(-)] the option a. fo the list of choices.
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3 years ago
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Vinegar is a solution of acetic acid (the solute) in water (the solvent) with a solution density of 1010 g/L. If vinegar is 0.80
Damm [24]

Answer:

4.8 %

Explanation:

We are asked the concentration in % by mass, given the molarity of the solution and its density.

0.8 molar solution means that we have 0.80 moles of acetic acid in 1 liter of solution. If we convert the moles of acetic acid to grams, and the 1 liter solution to grams, since we are given the density of solution, we will have the values necessary to calculate the % by mass:

MW acetic acid = 60.0 g/mol

mass acetic acid (the solute) = 0.80 mol x 60 g / mol = 48.00 g

mass of solution = 1000 cm³ x 1.010 g/ cm³      (1l= 1000 cm³)

                            = 1010 g

% (by mass) = 48.00 g/ 1010 g  x 100 = 4.8 %

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ss7ja [257]
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3 years ago
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How many grams of a stock solution that is 92.5 percent H2SO4 by mass would be needed to make 250 grams of a 35.0 percent by mas
IrinaVladis [17]

94.6 g.  You must use 94.6 g of 92.5 % H_2SO_4 to make 250 g of 35.0 % H_2SO_4.

We can use a version of the <em>dilution formula</em>

<em>m</em>_1<em>C</em>_1 = <em>m</em>_2<em>C</em>_2

where

<em>m</em> represents the mass and

<em>C</em> represents the percent concentrations

We can rearrange the formula to get

<em>m</em>_2= <em>m</em>_1 × (<em>C</em>_1/<em>C</em>_2)

<em>m</em>_1 = 250 g; <em>C</em>_1 = 35.0 %

<em>m</em>_2 = ?; _____<em>C</em>_2 = 92.5 %

∴ <em>m</em>_2 = 250 g × (35.0 %/92.5 %) = 94.6 g

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