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DerKrebs [107]
3 years ago
13

Ideally, the van 't Hoff factor should be equal to the number of ions that make up a compound. In reality, van 't Hoff factors t

end to be lower due to ion-pairing. Select all conditions that would increase the effect of ion- pairing and decrease the observed van 't Hoff factor.
A. Lower charge on the ions
B. High concentration
C. Low concentration
D. Higher charge on the ions
Chemistry
1 answer:
Musya8 [376]3 years ago
3 0

Answer:

High concentration

Higher charge on the ions

Explanation:

Ideally, the Van't Hoff factor is defined as the ratio between the actual concentration of particles produced when the substance is dissolved in a solution and the concentration of a substance as calculated from its mass(Wikipedia).

The Van't Hoff factor is influenced by the concentration of ions in solution as well as the magnitude of charge on the ions in solution.

Highly charged ions tend to remain paired in solution thereby leading to a lower value of Van't Hoff factor. Also, in highly concentrated solutions, the Van't Hoff factor is lower than in dilute solutions due to the pairing of ions

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To make a 450g solution with a mass by mass concentration of 7%, how much salt and water do you need to mix?​
Yuri [45]

Explanation:

The given data is:

The mass% of a solution is 7.

Mass of solution =450g

The mass of salt required can be calculated as shown below:

mass percent =(mass of solute/mass of solution )x100\\

Substitute the given values in this formula to get the mass of solute that is salt:

mass percent =(mass of solute/mass of solution )x100\\\\7=\frac{mass of salt}{450g} x100\\=>mass of salt=7x450g /100\\mass of salt=31.5g

Mass of salt =31.5g

Mass of solvent that is water = 450g-31.5g=418.5g

4 0
3 years ago
Find the empirical and molecular formula for a molar mass of 60.10g/mol; 39.97% carbon 13.41% hydrogen: 46.62% nitrogen
PSYCHO15rus [73]
Assuming we have 100g, this means that

39.97g Carbon * 1 mol / 12 g = 3.33 mol Carbon
13.41g Hydrogen * 1 mol/1 g = 13.41 mol Hydrogen
46.62g Nitrogen * 1 mol / 14 g = 3.33 mol Nitrogen
Dividing everything by 3.33, we get

1 mol Carbon, 4.03 mol Hydrogen, 1 mol Nitrogen.

Empirical formula is CH4N

<span>The mass of the empirical formula is
12 + 4 + 14 = 30

Since the molar mass is double, we multiply all our subscripts

The molecular formula is C2H8N2

The answers to this question are </span><span>an empirical formula of CH4N</span> and a molecular formula of C2H8N2 .
5 0
3 years ago
A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of
Kobotan [32]

Answer:

pH =1 2.84

Explanation:

First we have to start with the <u>reaction</u> between HCl and KOH:

HCl~+~KOH->~H_2O~+~KCl

Now <u>for example, we can use a volume of 10 mL of HCl</u>. So, we can calculate the moles using the <u>molarity equation</u>:

M=\frac{mol}{L}

We know that 10mL=0.01L and we have the concentration of the HCl 0.282M, when we plug the values into the equation we got:

0.282M=\frac{mol}{0.01L}

mol=0.282*0.01

mol=0.00282

We can do the same for the KOH values (50mL=0.05L and 0.141M).

0.141M=\frac{mol}{0.05L}

mol=0.141*0.05

mol=0.00705

So, we have so far <u>0.00282 mol of HCl</u> and <u>0.00705 mol of KOH</u>. If we check the reaction we have a <u>molar ratio 1:1</u>, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an <u>excess of KOH</u>. This excess can be calculated if we <u>substract</u> the amount of moles:

0.00705-0.00282=0.00423mol~of~KOH

Now, if we want to calculate the pH value we will need a <u>concentration</u>, in this case KOH is in excess, so we have to calculate the <u>concentration of KOH</u>. For this, we already have the moles of KOH that remains left, now we need the <u>total volume</u>:

Total~volume=50mL+10mL=60mL

60mL=0.06L

Now we can calculate the concentration:

M=\frac{0.00423mol}{0.06L}

M=0.0705

Now, we can <u>calculate the pOH</u> (to calculate the pH), so:

pOH=-Log(0.0705)

pOH=1.15

Now we can <u>calculate the pH value</u>:

14=~pH~+~pOH

pH=14-1.15=12.84

8 0
3 years ago
The flavor of anise is due to anethole, a compound with the molecular forumal C10H12O. Combustion of one mole of anethole produc
AfilCa [17]

Answer:

Change in temperature of calorimeter is 4.52^{0}\textrm{C}

Explanation:

Molar mass of anethole = 148.2 g/mol

So, 0.950 g of anethole = \frac{0.950}{148.2}moles of anethole = 0.00641 moles of anethole

1 mol of anethole releases 5541 kJ of heat upon combustion

So, 0.00641 moles of anethole release (5541\times 0.00641)kJ of heat or 35.52 kJ of heat

7.854 kJ of heat increases 1^{0}\textrm{C} temperature of calorimeter.

So, 35.52 kJ of heat increases (\frac{1}{7.854}\times 35.52)^{0}\textrm{C} or 4.52^{0}\textrm{C} temperature of calorimeter

So, change in temperature of calorimeter is 4.52^{0}\textrm{C}

7 0
3 years ago
You spill a little bit of olive oil on your hands. You wash off the oil with soap and water. How does the soap get the oil off y
Serhud [2]
It doesnt, soap is comprised of water, and water and oil do not have the same molarity, therefore do not mix and will just continue to rub oil on your hand but not come off
8 0
3 years ago
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