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wel
3 years ago
11

Assuming that the density of vinegar is 1.005 g/mL, calculate the molatrity of acetic acid in vinegar from an average of 4.6% fo

r the mass percentage of acetic acid in vinegar.
Chemistry
1 answer:
posledela3 years ago
5 0
Acetic acid is 60.05 grams/mole. In 1 liter of vinegar or 1000 ml  there would be 0.046% of acetic acid = 46 ml x 1.005g/ml = 46.23 grams/60.05 grams= 0.77 moles per litre of vinegar.This then would be the concentration of acetic acid in for example 1 liter of vinegar. 
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How many electrons does the oxygen atom need to become stable?
NemiM [27]

Answer:

2 electrons

Explanation:

Oxygen has 6 valence electrons and to be stable it needs 8. That means it needs 2 more electrons to have a full octet.

4 0
3 years ago
Read 2 more answers
This question has multiple parts. Work all the parts to get the most points. An aqueous antifreeze solution is 39.0% ethylene gl
nlexa [21]

Answer:

Molality = <u>10.300 m</u>

<u>Molarity = 6.5970 M</u>

<u>mole fraction = </u>0.156549

Explanation:

39.0 % = ethylene glycol

61.0 % = water

imagine the total mass = 100g

39.0% ethylene glycol = 39g

61.0 % water = 61g

1) Molality = number of moles / mass of solvent (kg)

Molar mass of ethylene glycol = 62.07g/mole

mole = 39g / 62.07g/mole = 0.6283 moles

Molality = moles / mass of solvent = 0.6283 moles / 0.061kg = <u>10.300 m</u>

<u />

<u>2) Molarity</u> = number of moles / volume of solution

Since we know the density  of the solution = 1.05g /ml

⇒ volume = 100g / 1.05g /mL   = 95.24 mL = 0.09524 L

Molarity = 0.6283 moles / 0.09524 L = <u>6.5970 M</u>

3) Mole fraction

moles water = 61g / 18.02g/mole  = 3.38513 moles

Total number of moles = moles of ethylene glycol + moles of water = 0.6283 + 3.38513 = 4.0134276 moles

Mole fraction = 0.6283/ 4.0134276 = 0.156549

3 0
3 years ago
For each row in the table below, decide whether the pair of elements will form a molecular or ionic compound. If they will, then
Olin [163]

Answer:

\begin{array}{cccll}\textbf{Element 1} & \textbf{ Element 2} &\textbf{Compound?} &\textbf{Formula} &\textbf{Type}\\\text{Ar}&\text{Xe} &\text{No} &\text{None}&\text{Neither}\\\text{F}& \text{Cs} &\text{Yes} &\text{CsF} &\text{Ionic}\\\text{N} &\text{Br} &\text{Yes} & \text{NBr}_{3}&\text{molecular} \\\end{array}

Explanation:

You look at the type of atom and their electronegativity difference.

If ΔEN <1.6, covalent; if ΔEN >1.6, ionic

Ar/Xe: Noble gases; no reaction

F/Cs: Non-metal + metal; ΔEN = |3.98 – 0.79| = 3.19; Ionic

N/Br: Two nonmetals; ΔEN = |3.04 - 2.98| = 0.

4 0
3 years ago
Solid NaBr is slowly added to a solution that is 0.073 M in Cu+ and 0.073 M in Ag+.Which compound will begin to precipitate firs
saul85 [17]

Answer :

AgBr should precipitate first.

The concentration of Ag^+ when CuBr just begins to precipitate is, 1.34\times 10^{-6}M

Percent of Ag^+ remains is, 0.0018 %

Explanation :

K_{sp} for CuBr is 4.2\times 10^{-8}

K_{sp} for AgBr is 7.7\times 10^{-13}

As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgBr has a smaller than CuBr then AgBr should precipitate first.

Now we have to calculate the concentration of bromide ion.

The solubility equilibrium reaction will be:

CuBr\rightleftharpoons Cu^++Br^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Cu^+][Br^-]

4.2\times 10^{-8}=0.073\times [Br^-]

[Br^-]=5.75\times 10^{-7}M

Now we have to calculate the concentration of silver ion.

The solubility equilibrium reaction will be:

AgBr\rightleftharpoons Ag^++Br^-

The expression for solubility constant for this reaction will be,

K_{sp}=[Ag^+][Br^-]

7.7\times 10^{-13}=[Ag^+]\times 5.75\times 10^{-7}M

[Ag^+]=1.34\times 10^{-6}M

Now we have to calculate the percent of Ag^+ remains in solution at this point.

Percent of Ag^+ remains = \frac{1.34\times 10^{-6}}{0.073}\times 100

Percent of Ag^+ remains = 0.0018 %

3 0
3 years ago
PLEASE HELP ITS DUE TODAY!
Yuri [45]

Answer:

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5 0
2 years ago
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