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Julie takes a cold glass of clear, colorless liquid out of the refrigerator and puts it on the table. As the liquid warms up to

marysya [2.9K]

The bubbles indicate that there is a gas dissolved in the liquid, which may remain dissolved when the liquid is very cold, inside the refrigerator.

When the temperature of the liquid increases, it reduces its ability to dissolve the gas and the gas escapes (bubbles) from the liquid.

At the end, the behavior of the liquid permits you to conclude that the liquid is a mixture because it has a gas dissolved in it.

8 0

3 years ago

Read 2 more answers

Of the isomeric C5H11+ carbocations which one is the most stable?

Studentka2010 [4]

Methyl groups are electron-donating, so the more CH3 groups attached to the +ve carbon, the more stable the carbocation becomes.

6 0

3 years ago

Why is u-235 a good isotope of uranium for creating chain reactions?

likoan [24]

under specific conditions it can promptly be taken apart easily, yielding a great deal of energy . It is subsequently said to be 'fissile' and we utilize the articulation 'atomic splitting'. In the interim, similar to every single radioactive isotope, they rot.

5 0

2 years ago

Calculate the empirical formula of a compound that is 42.9% Carbon and 57.1% Oxygen.

Marrrta [24]

Answer:

CO.

Explanation:

Assuming the given percentages are by mass, we can solve this problem via imagining we have 100 g of the compound, if that were the case we would have:

42.9 g of C

57.1 g of O

Now we convert those masses into moles, using the elements' respective molar masses:

42.9 g of C ÷ 12 g/mol = 3.57 mol C

57.1 g of O ÷ 16 g/mol = 3.58 mol O

As the number of C moles and O moles is roughly the same, the empirical formula for the compound is CO.

4 0

2 years ago

Shtirlitz [24]

Answer: $6.4\times 10^{-7}M$ .

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n= moles of solute

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{0.6597g}{158g/mol}=0.004mole$

$V_s$ = volume of solution in ml = 500 ml

$Molarity=\frac{0.004\times 1000}{500}=0.008M$

According to the dilution law:

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = 0.008 M

$V_1$ = volume of stock solution = 2 ml

$M_2$ = molarity of resulting solution = ?

$V_2$ = volume of resulting solution = 1000 ml

$0.008\times 2 ml=M_2\times 1000ml$

$M_2=0.000016M$

According to the dilution law:

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = 0.000016 M

$V_1$ = volume of stock solution = 10 ml

$M_2$ = molarity of resulting solution = ?

$V_2$ = volume of resulting solution = 250 ml

$0.000016\times 10 ml=M_2\times 250ml$

$M_2=0.00000064M=6.4\times 10^{-7}M$

Scientific notation is defined as the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.

Therefore, the concentration of of the final solution is $6.4\times 10^{-7}M$ .

7 0

2 years ago