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horrorfan [7]
3 years ago
14

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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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>

molar mass = 35 g/mol

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product  LiH(s) = 8.0 g ( actual yield )

theoretical yield ( LiH ) =  4 * 7.95  = 31.8 g

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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
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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

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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:

The second row elements are called Actinidies . abbreviated as Ac

Explanation:

5 0
3 years ago
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