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pychu [463]
3 years ago
6

A nonaqueous solution has a solvent that is not water. Which is an example of a nonaqueous solution?

Chemistry
2 answers:
solmaris [256]3 years ago
5 0
Out of the given options, the best example of a solution containting a non-aqueous solvent is paint<span>. Steel is not a liquid solution, it is instead a solid-state solution so it is not considered. Orange juice and shaving cream both rely on water as a solvent. Paint is actually an emulsion, where polymer substances are spread throughout the oil-based solvent in which they are immiscible. The fact that paint thinners are also oil based and paint is unaffected by water also indicate that it uses a nonaqueous solvent.</span>
Kay [80]3 years ago
3 0
<span>The answer is paint. A dissoluble is a substance that breaks up a solute in the arrangement of an answer, and any dissoluble other than water is viewed as a non-fluid dissoluble. Some basic illustrations incorporate either, liquor, benzene, disulfide, carbon tetrachloride and CH3)2CO.</span><span />
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What is point group of allene?​
Fantom [35]

Allene (1,2-propadiene) has point group D2d, itself is achiral because it has two planes of symmetry. ... An allene with substituents on one terminal carbon atom are unlike and substituent on other terminal carbon atoms are same, allene will be achiral. It will have one symmetry plane.

Hope this helped :)

4 0
4 years ago
Read 2 more answers
2. How many moles of hydrogen atoms are there in 154 mL of 0.18 M H2SO4? Write your
katrin [286]

2. 0.05544 moles of hydrogen atom are present in 154 mL 0.18 M solution of H2SO4.

3. 15.2506 heat in Joules is absorbed by 150.0 mL of pure water that is heated from 21.2°C to  45.5°C.

4. 0.75 M is the concentration of Na+ ions in 25.0 mL of 1.50 M NaOH is reacted with 25.0 mL of  1.50 M HCI

Explanation:

Number of moles of H2SO4 can be calculated by the given volume and molarity from the formula:

molarity = \frac{number of moles}{volume of teh solution}

number of moles = molarity × volume of the solution of H2SO4

    number of moles = 0.18 × 0.154 litres

                                   = 0.02772 moles of H2SO4.

Since 1 mole of H2SO4 contains 2 moles of hydrogen

so, 0.02772 moles of H2SO4 will have x moles

\frac{1}{2} = \frac{0.0272}{x}

2 × 0.02772 = x

0.05544 moles of hydrogen atom are present in 154 mL 0.18 M solution of H2SO4.

3. The heat absorbed is calculated from the formula:

ΔH = cp × m × ΔT   ( ΔT = change in temperature in Kelvin, m in kg, cp= specific heat of water)

putting the values in formula:

ΔH = 4.184 × 0.15 × ( 318.65-294.35)

     =  4.184 × 0.15× 24.3

        = 15.2506 Joules of heat is absorbed.

4. The concentration of Na+ ions

the balanced equation is

NaOH + HCl⇒ NaCl + H20

from one mole of NaCl 1 mole of NaCl i.e one 1 mole of Na+ ions is formed.

number of moles of NaOH is calculated by the formula:

Molarity = \frac{number of moles}{volume of the solution}

number of moles = 0.025L × 1.50M

                             = 0.0375 moles of NaOH

so 1 mole of NaOH produces 1 mole of Na= ions

hence, 0.0375 moles produces x moles of Na+ ions

\frac{1}{1} = \frac{0.0375}{x}

moles of Na+ is produced.  

concentration of NaOH in 50 ml solution because NaOH and HCl of 25 ml reacted.

applying the molarity formula from above

Molarity =  \frac{0.0375}{0.05}

              =  0.75 M

6 0
3 years ago
How could you separate a mixture of sulfur, gold and ethanol? <br>​
stira [4]

Answer: Dissolve it  

Explanation:

3 0
3 years ago
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If a patient\'s blood pressure is 145 over 85 mmHg, what is it in atmospheres (atm)?
yan [13]
In 1(atm) the patients blood is 760 mmHg
6 0
3 years ago
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A piece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °C and dropped into 84 g of water (with an
laila [671]

The specific heat of metal is c = 3.433 J/g*⁰C.

<h3>Further explanation</h3>

Given

mass of metal = 68.6 g

t metal = 100 °C

mass water = 84 g

t water = 20 °C

final temperature = 52.1  °C

Required

The specific heat

Solution

Heat can be formulated :

Q = m.c.Δt

Q absorbed by water = Q released by metal

84 x 4.184 x (52.1-20)=68.6 x c x (100-52.1)

11281.738=3285.94 x c

c = 3.433 J/g*⁰C.

7 0
3 years ago
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