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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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3 years ago
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katrin2010 [14]

Answer:

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Explanation:

4 0
3 years ago
Read 2 more answers
What is the name of this hydrocarbon? 2-dimethylyne 2-methylbutane 4-methylbutene 4,3-methylbutyne 4,4-methylbutane
Alisiya [41]
You have shown no structure, still we can work out for the answer :).

First Option: <span>2-dimethylyne: This name is incorrect, because it does not contain any parent chain name. 

Second Option: </span><span>2-methylbutane: I have drawn its structure below, compare it with your structure.

Third Option: </span>4-methylbutene:<span> I have drawn the structure for this name. But this name is also against IUPAC rules, and its correct name is 1-Pentene.

Fourth Option: </span><span>4,3-methylbutyne: Again incorrect name, 4,3 are two positions, but here only one substituent is given (methyl). I have drawn structure of 3,4-Dimethylbutene, this name is incorrect, and the correct name for this compound is 3-Methyl-1-pentene.

Fifth Option: </span><span>4,4-methylbutane: Again incorrect name, 4,4 means at 4 position two substituents, but in name only methyl is given. Anyhow lets make it two and draw a structure, for 4,4-Dimethylbutane. Ooops!! This name is also incorrect, and the correct name for this compound is 2-Methylpentane.

Result:
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8 0
3 years ago
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A 25.0 mL aliquot of 0.0680 M EDTA was added to a 59.0 mL solution containing an unknown concentration of V3 . All of the V3 pre
givi [52]

Answer:

\mathbf{0.02 M}

Explanation:

\text{So, from the given question:}

\text{EDTA will make complex with} V^{+3} \text{and the remaining EDTA will react with }Ga^{+3}

\text{Hence, the total concentration of} V^{+3} & Ga^{+3} \text{will be equivalent to EDTA concentration.}

V_{EDTA} = 25 \ mL

V_{V^{+3}} = 59.0 \ mL

V_{Ga^{+3}} = 13.0 \ mL

M_{EDTA} = 0.0680 \ M

M_{V^{+3}} = ???(unknown)

M_{Ga^{+3}} = 0.0400 \ M

V^{+3} + EDTA \to V[EDTA] + EDTA(Excess)  \to^{CoA} \ Ga[EDTA] _{complex}

M_{EDTA} \times V_{EDTA} = ( V_{V^+3}\times M_{V^{+3}}+ V_{Ga^{+3} }\times M_{Ga^{+3}}})

0.0680 \times 25 = (59\times x + 13 \times 0.040) \\ \\ 1.7 = 59x + 0.52\\ \\ 1.7 - 0.52 = 59x \\ \\ 59x = 1.18

x = \dfrac{1.18}{59}

\mathbf{x =0.02 \ M }

5 0
3 years ago
Two substances A and B, initially at different temperatures, are thermally isolated from their surroundings and allowed to come
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Answer:

B

Explanation:

For solving this we need a heat balance

Q_{a} = Q_{b}\\m_{a}*C_{a}*\Delta T_{a} = m_{b}*C_{b}*\Delta T_{b}

By changing the corresponding relations, we have

m_{a}*C_{a}*\Delta T_{a} = \frac{1}{2}m_{a}*4C_{a}*\Delta T_{b} \\\\\\

By cancelling similar factor, we obtain

\Delta T_{a} = 2 \Delta T_{b}\\\frac{\Delta T_{a}}{\Delta T_{b}} = 2\\

Which means that the change of temperature in A is twice the change of B

3 0
3 years ago
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