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3 years ago

An isopropyl solution contains 70 ml of isopropyl alcohol in 1.8 L of water. What is the percent by volume of the alcohol

Chemistry

1 answer:

xenn [34]3 years ago

5 0

Answer:

3.74%

Explanation:

We express a solution's volume by volume percent concentration, % v/v,

Take the ratio of the isopropyl alcohol (IPA) volume to the total volume of the solution, which is 1800 mL of water+ 70 mL of IPA,

and multiply by 100 to get the percentage: 70/(1800+70) *100 = 0.0374*100 = 3.74%

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What would be the formula of the precipitate that forms when pb(no3)2 (aq) and k2so4 (aq) are mixed?

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The formula of the ppt. formed is PbSo4 , which is inslouble.

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Iron has a density of 7.9 g/cm3. The volume of a regular cylinder is V =πr2h. An iron cylinder has a height of 3.00 m and a mass

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Pure hydrogen (H₂) is a hazardous substance. Thus, safer and more cost effective techniques have been developed to store it as a

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Answer: hello your question has some missing details attached below is the missing details

answer : 25.5% ( B )

Explanation:

<u>Determine percentage yield </u>

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4 0

3 years ago

in a simulation mercury removal from industrial wastewater, 0.020 L of 0.10 M sodium sulfide reacts with 0.050 L of 0.010 M merc

Margarita [4]

Answer: 0.1161 grams of mercury(II) sulfide) form.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$ .....(1)

a) Molarity of $Na_2S$ solution = 0.10 M

Volume of solution = 0.020 L

Putting values in equation 1, we get:

$0.10M=\frac{\text{Moles of }Na_2S}{0.020L}\\\\\text{Moles of Na_2S}={0.10mol/L\times 0.020}=0.002mol$

$\text {Moles of}Na_2S=0.10M\times 0.020L=0.002mol$

b) Molarity of $Hg(NO_3)_2$ solution = 0.010 M

Volume of solution = 0.050 L

Putting values in equation 1, we get:

$0.010M=\frac{\text{Moles of }Hg(NO_3)_2}{0.050L}\\\\\text{Moles of }Hg(NO_3)_2={0.010mol/L\times 0.050}=0.0005mol$

$Na_2S+Hg(NO_3)_2\rightarrow HgS+2NaNO_3$

According to stoichiometry :

1 mole of $Hg(NO_3)_2$ reacts with 1 mole of $Na_2S$

Thus 0.0005 moles of $HgNO_3$ reacts with= $\frac{1}{1}\times 0.0005=0.0005$ moles of $Hg(NO_3)_2$

Thus $Hg(NO_3)_2$ is the limiting reagent and $Na_2S$ is the excess reagent.

According to stoichiometry :

1 mole of $Hg(NO_3)_2$ forms= 1 mole of $Hg_2S$

Thus 0.0005 moles of $Hg(NO_3)_2$ forms= $\frac{1}{1}\times 0.0005=0.0005$ moles of $Hg_2S$

mass of $H_2S=moles\times {\text {Molar mass}}=0.0005mol\times 232.2g/mol=0.1161g$

Thus 0.1161 grams of mercury(II) sulfide) form.

5 0

3 years ago

Te second period at the bottom of the table is called

katovenus [111]

Answer:

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Explanation:

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3 years ago