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nasty-shy [4]
2 years ago
14

You have a 16.0-oz. (473-mL) glass of lemonade with a concentration of 2.66 M. The lemonade sits out on your counter for a coupl

e of days, and 150. mL of water evaporates from the glass. What is the new concentration of the lemonade?
Chemistry
1 answer:
salantis [7]2 years ago
7 0

<u>Answer:</u> The new concentration of lemonade is 3.90 M

<u>Explanation:</u>

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 mL)}}     .....(1)

Molarity of lemonade solution = 2.66 M

Volume of solution = 473 mL

Putting values in equation 1, we get:

2.66M=\frac{\text{Moles of lemonade}\times 1000}{473}\\\\\text{Moles of lemonade}=\frac{2.66\times 473}{1000}=1.26mol

Now, calculating the new concentration of lemonade by using equation 1:

Moles of lemonade = 1.26 moles

Volume of solution = (473 - 150) mL = 323 mL

Putting values in equation 1, we get:

\text{New concentration of lemonade}=\frac{1.26\times 1000}{323}\\\\\text{New concentration of lemonade}=3.90M

Hence, the new concentration of lemonade is 3.90 M

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Draw all of the constitutional isomers of the molecule with formula C3H5Br. Ignore geometric and stereoisomers.
Kay [80]

Answer:

-) 3-bromoprop-1-ene

-) 2-bromoprop-1-ene

-) 1-bromoprop-1-ene

-) bromocyclopropane

Explanation:

In this question, we can start with the <u>I.D.H</u> (<em>hydrogen deficiency index</em>):

I.D.H~=~\frac{(2C)+2+(N)-(H)-(X)}{2}

In the formula we have 3 carbons, 5 hydrogens, and 1 Br, so:

I.D.H~=~\frac{(2*3)+2+(0)-(5)-(1)}{2}~=~1

We have an I.D.H value of one. This indicates that we can have a cyclic structure or a double bond.

We can start with a linear structure with 3 carbon with a double bond in the first carbon and the Br atom also in the first carbon (<u>1-bromoprop-1-ene</u>). In the second structure, we can move the Br atom to the second carbon (<u>2-bromoprop-1-ene</u>), in the third structure we can move the Br to carbon 3 (<u>3-bromoprop-1-ene</u>). Finally, we can have a cyclic structure with a Br atom (<u>bromocyclopropane</u>).

See figure 1

I hope it helps!

7 0
1 year ago
For a pure substance, the liquid and gaseous phases can only coexist for a single value of the pressure at a given temperature.
anastassius [24]

Answer:

No, it is not.

Explanation:

Most solutions do not behave ideally. Designating two volatile  substances as A and B, we can consider the following two cases:

Case 1: If the intermolecular forces between A and B molecules are weaker than  those between A molecules and between B molecules, then there is a greater tendency  for these molecules to leave the solution than in the case of an ideal solution. Consequently,  the vapor pressure of the solution is greater than the sum of the vapor  pressures as predicted by Raoult’s law for the same concentration. This behavior gives  rise to the positive deviation.

Case 2: If A molecules attract B molecules more strongly than they do their own  kind, the vapor pressure of the solution is less than the sum of the vapor pressures as  predicted by Raoult’s law. Here we have a negative deviation.

The benzene/toluene system is an exception, since that solution behaves ideally.

8 0
2 years ago
How many grams are there in 7.5 X10^23 molecules of H2SO4?
mash [69]
<span>Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles.

 </span><span>7.5 X10^23 molecules of H2SO4 ( 1 mol / 6.022x10^23 molecules ) (98.08 g / 1 mol ) = 122.15 grams H2SO4</span>
6 0
2 years ago
Read 2 more answers
How much heat is required to heat 9.61g of ethanol (CH3CH2OH) from 24.10C to 67.30C?
Licemer1 [7]

Answer:

a.)  

To warm the liquid from 35°C to 78°C:

(2.3 J/g-K) x (42.0 g) x (78 - 35) = 4154 J

To vaporize the liquid at 78°C:

(38.56 kJ/mol) x (42.0 g C2H5OH / 46.06867 g C2H5OH/mol) = 35.154 kJ

Total:

4.154 kJ + 35.154 kJ = 39.3 kJ

b.)  

To warm the solid from -155°C to -114°C:

(0.97 J/g-K) x (42.0 g) x (-114°C - (-155°C)) = 1670 J

To melt the solid at -114°C:

(5.02 kJ/mol) x (42.0 g C2H5OH / 46.06867 g C2H5OH/mol) = 4.5766 kJ  

To warm the liquid from -114°C to 78°C:

(2.3 J/g-K) x (42.0 g) x (78 - (-114)) = 18547 J

To vaporize the liquid at 78°C:

35.154 kJ  (as in part a.)

Total:

1.670 kJ + 4.5766 kJ + 18.547 kJ + 35.154 kJ = 59.9 kJ

Explanation:

8 0
2 years ago
"A sample of silicon has an average atomic mass of 28.084amu. In the sample, there are three isotopic forms of silicon. About 92
adoni [48]

<u>Answer:</u> The percentage abundance for _{14}^{30}\textrm{Si} isotope is 3.09 %.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i    .....(1)

We are given:

Let the fractional abundance for _{14}^{28}\textrm{Si} isotope be 'x'

  • <u>For _{14}^{28}\textrm{Si} isotope:</u>

Mass of _{14}^{28}\textrm{Si} isotope = 27.9769 amu

Percentage abundance of _{14}^{28}\textrm{Si} isotope = 92.22 %

Fractional abundance of _{14}^{28}\textrm{Si} isotope = 0.9222

  • <u>For _{14}^{29}\textrm{Si} isotope:</u>

Mass of _{14}^{28}\textrm{Si} isotope = 28.9764 amu

Percentage abundance of _{14}^{28}\textrm{Si} isotope = 4.68%

Fractional abundance of _{14}^{28}\textrm{Si} isotope = 0.0468

  • <u>For _{14}^{30}\textrm{Si} isotope:</u>

Mass of _{14}^{30}\textrm{Si} isotope = 29.9737 amu

Fractional abundance of _{14}^{30}\textrm{Si} isotope = x

  • Average atomic mass of silicon = 28.084 amu

Putting values in equation 1, we get:

28.084=[(27.9769\times 0.9222)+(28.9764\times 0.0468)+(29.9737\times x)]\\\\x=0.0309

Converting this fractional abundance into percentage abundance by multiplying it by 100, we get:

\Rightarrow 0.0309\times 100=3.09\%

Hence, the percentage abundance for _{14}^{30}\textrm{Si} isotope is 3.09 %.

6 0
2 years ago
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