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3 years ago

What happens to a glass of sugar solution when sugar is added to it?

Chemistry

2 answers:

ipn [44]3 years ago

8 0

Answer:

c: the molarity of the solution increases

galina1969 [7]3 years ago

7 0

Answer:

The molarity of the solution increases.

Explanation:

Molarity is the measure of the concentration of the solute in the solution. In this case, the solvent is the sugar solution and the solute is the sugar.

If sugar is ADDED to the already sugary solution, then there would be more sugar. Therefore, the sugar (solute) would increase in number.

This means that the answer is the third choice: The molarity of the solution increases.

The answer would not be the first or second choice because there isn't anything in the question that implies water. It just says sugar solution.

The answer is not the last choice because the sugar concentration does not decrease after you have added more sugar to it. It increases.

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You were instructed to add 1.0 mL out of 4.0 mL of an undiluted sample to 99 mL of sterile diluent. Instead, you add the entire

finlep [7]

Answer:

A.) Intended dilution factor : 0.01

B.) Actual dilution factor: 0.039

Explanation:

Dilution factor(DF) is ratio of the volume of the undiluted sample to total volume of diluted solution

A.) What was your intended dilution factor

This would be the the addition 1.0ml from 4.0ml to 99ml of sterile

Volume of sterile diluent = 99ml

Volume of Undiluted sample = 1.0ml

Total volume of diluted solution = 99ml + 1m = 100ml

Dilution factor = Volume of undiluted sample ÷ Total volume of diluted solution

Dilution factor = 1:100

= 1ml ÷ 100ml

= 0.01

B.) What was your actual dilution factor?

This would be the entire addition of 4.0ml to 99ml sterile diluent

Volume of sterile diluent = 99ml

Volume of Undiluted sample = 4.0ml

Total volume of diluted solution = 99ml + 4.0ml = 103ml

Dilution factor = Volume of undiluted sample ÷ Total volume of diluted solution

Dilution factor

= 4:103

= 4ml ÷ 103ml

= 0.039

8 0

3 years ago

What group will group 16 most likely bond with?

ikadub [295]

They will most likey group with the second group. Since they have 6 electrons and want to have a full outer shell so therefore would group with the second group.

7 0

3 years ago

A 1.2 kg sample of Th-228 has a half-life of 1.9 years. How many grams of Th is left after 13.3 years?

Bumek [7]

.009375 kg or 9.375 grams will remain

7 0

3 years ago

In part A of the experiment you will need to make a 4.80 x 10-4 M KSCN solution starting with a 2.00 x 10-3M KSCN solution. You

ryzh [129]

Answer:

Answer to you need to make a 6.00 x 10-4 M KSCN solution starting with a 2.00 x 10-3 M KSCN solution. You will be making up the so. ... X 10-3 M KSCN Solution. You Will Be Making Up The Solution In A 25 ML Volumetric Flask And Using 0.5 M HNO3 As The Diluent. What Volume Of 2.00 X 10-3 M KSCN Will You Need?

Explanation:

7 0

3 years ago

1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.

Andrej [43]

Answer:

a. HCl.

b. 0.057 g.

c. 1.69 g.

d. 77 %.

Explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

$Mg+2HCl\rightarrow MgCl_2+H_2$

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

$n_{H_2}^{by\ HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl} =0.0284molH_2\\\\n_{H_2}^{by\ Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2$

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.

b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

$m_{H_2}=0.0284molH_2*\frac{2.02gH_2}{1molH_2}=0.057gH_2$

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

$m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg} =0.690gMg$

Thus, the mass of excess magnesium turns out:

$m_{Mg}^{excess}=2.38g-0.690g=1.69gMg$

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

$Y=\frac{0.044g}{0.057g} *100\%\\\\Y=77\%$

Best regards!

8 0

3 years ago