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Luda [366]
3 years ago
15

Ethanol is used widely as a solvent in laboratories for various chemical reactions. A laboratory technician takes 50.0 mL of eth

anol and dilutes it to 100.0 mL with distilled water. What is the percentage concentration by volume of the resulting solution?
Chemistry
2 answers:
puteri [66]3 years ago
8 0
This question is quite vague, as the initial concentration of ethanol is not provided. However, from experience I can tell you that most laboratory work is done with 98% ethanol, and not absolute ethanol (100%). So in order to calculate the final concentration, we need to take the given values, which includes the initial concentration (98%), the initial volume (50.0mL) and the final volume (100.0mL). We apply the following equation to calculate the final concentration:

C1V1 = C2V2
C1 = Initial concentration
C2 = Final concentration
V1 = Initial volume
V2 = Final volume

(98%)(50.0mL) = (C2)(100.0mL)
Therefore, the final concentration (C2) = 49%
borishaifa [10]3 years ago
8 0
The answer is B 50% just finished the test
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Kylee has been learning that engineering and technology are beneficial to people. She now sees how the things people synthesize
marysya [2.9K]

Answer:

They are synthesized from raw materials

Explanation:

All synthesized products goes through lots of processes. This processes involves the conversion of substances known as raw materials into a more useful and modified form.

These synthesized products are usually more useful and made in order to solve a given problem or offer a particular service. Raw materials are cheaper than the synthesized products and are common in all industries and fields.

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3 years ago
A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are
vaieri [72.5K]

Answer: Resulting solution will not be neutral because the moles of OH^-ions is greater. The remaining concentration of [OH^-]ions =0.0058 M.

Explanation:

Given,

[HCl]=0.100 M

[HNO_3] = 0.200 M

[Ca(OH)_2] =0.0100 M

[RbOH] =0.100 M

Few steps are involved:

Step 1: Calculating the total moles of H^+ ion from both the acids

moles of H^+ in HCl

HCl\rightarrow {H^+}+Cl^-

if 1 L of HClsolution =0.100 moles of HCl

then 0.05L of HCl solution= 0.05 \times0.1 moles= 0.005 moles    (1L=1000mL)

moles of H^+ in HCl = 0.005 moles

Similarliy

moles of H^+ in HNO_3

HNO_3\rightarrow H^++NO_3^-}

If 1L of HNO_3 solution= 0.200 moles

Then 0.1L of HNO_3 solution= 0.1 \times 0.200 moles= 0.02 moles

moles of H^+ in HNO_3 =0.02 moles

so, Total moles of H^+ ions  = 0.005+0.02= 0.025 moles     .....(1)

Step 2: Calculating the total moles of [OH^-] ion from both the bases

Moles of OH^-\text{ in }Ca(OH)_2

Ca(OH)_2\rightarrow Ca^2{+}+2OH^-

1 L of Ca(OH)_2= 0.0100 moles

Then in 0.5 L Ca(OH)_2 solution = 0.5 \times0.0100 moles = 0.005 moles

Ca(OH)_2 produces two moles of OH^- ions

moles of OH^- = 0.005 \times 2= 0.01 moles

Moles of OH^- in RbOH

RbOH\rightarrow Rb^++OH^-

1 L of RbOH= 0.100 moles

then 0.2 [RbOH] solution= 0.2 \times 0.100 moles = 0.02 moles

Moles of OH^- = 0.02 moles

so,Total moles of OH^- ions = 0.01 + 0.02=0.030 moles      ....(2)

Step 3: Comparing the moles of both H^+\text{ and }OH^- ions

One mole of H^+ ions will combine with one mole of OH^- ions, so

Total moles of H^+ ions  = 0.005+0.02= 0.025 moles....(1)

Total moles of OH^- ions = 0.01 + 0.02=0.030 moles.....(2)

For a solution to be neutral, we have

Total moles of H^+ ions = total moles of OH^- ions

0.025 moles H^+ will neutralize the 0.025 moles of OH^-

Moles of OH^- ions is in excess        (from 1 and 2)

The remaining moles of OH^- will be = 0.030 - 0.025 = 0.005 moles

So,The resulting solution will not be neutral.

Remaining Concentration of OH^- ions = \frac{\text{Moles remaining}}{\text{Total volume}}

[OH^-]=\frac{0.005}{0.85}=0.0058M

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tiny-mole [99]
True, but where is the question?
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mixer [17]

$8.09

Because you multiply and subtract to get the answer

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True or False
Softa [21]

Answer:

True

Explanation:

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