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2 years ago

7

What is the normality of a solution prepared by dissolving 75.0 g citric acid, a triprotic acid with molar mass of 192.14 g, in

water to make 250. mL solution?
a. 0.0576 N
b. 0.293 N
c. 4.68 N
d. 0.173 N
e. 1.56 N

1 answer:

Mademuasel [1]2 years ago

5 0

It’s C. 4.68 N

I hope this helped out! Have a nice day :)

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Chemical digestion would not be possible without _____, which speed up the rate of chemical reactions in your body?

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Enzymes in our body increases the speed of chemical digestion.

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3 years ago

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How many moles of PC15 can be produced from 58.0 g of Cl₂ (and excess<br> P4)?

ludmilkaskok [199]

0.3268 moles of PC15 can be produced from 58.0 g of Cl₂ (and excess

P4)

<h3>How to calculate moles?</h3>

The balanced chemical equation is

$P_{4} + 10Cl_{2} = 4PCl_{5}$

The mass of clorine is m( $Cl_{2}$ ) = 58.0 g

The amount of clorine is n( $Cl_{2}$ ) = m( $Cl_{2}$ )/M( $Cl_{2}$ ) = 58/70.906 = 0.817 mol

The stoichiometric reaction,shows that

10 moles of $Cl_{2}$ yield 4 moles of $PCl_{5}$ ;

0.817 of $Cl_{2}$ yield x moles of $PCl_{5}$

n( $PCl_{5}$ ) = 4*0.817/10 = 0.3268 mol

To know more about stoichiometric reaction, refer:

brainly.com/question/14935523

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1 year ago

A red blood cell is placed into each of the following solutions. Indicate whether crenation, hemolysis, or neither will occur. S

Burka [1]

Answer:

Solution A: crenation

Solution B: hemolysis

Solution C: hemolysis

Solution D: crenation

Solution E: crenation

Explanation:

Hemolysis is the rupturing of red blood cells, which results in the release of hemoglobin (from within the red blood cells) into the plasma. If a red blood cell is placed in a hypotonic solution, water will flow into the cell, the cell will swell and hemolysis will.

Crenation: when a red blood cell is placed in a <em>hypertonic solution (</em>such as highly saline solution), the red blood cell will lose water(osmosis) and will shrink in size. The red blood cell has undergone crenation.

In order for a red blood cell to prevent from undergoing hemolysis or crenation, the cell must be placed in an<em> isotonic solution, </em>i.e either in <u>0.9% (m/v) NaCl solution</u> or <u>5% glucose solution</u>

Solution B and Solution C are hypotonic solution, thus red blood cell has undergone hemolysis.
Solution A, D and E are hypertonic solution. thus red blood cell has undergone crenation

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3 years ago

When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing poin

LenaWriter [7]

The question is incomplete, the complete question is:

When 177. g of alanine $(C_3H_7NO_2)$ are dissolved in 800.0 g of a certain mystery liquid X, the freezing point of the solution is $5.9^oC$ lower than the freezing point of pure X. On the other hand, when 177.0 g of potassium bromide are dissolved in the same mass of X, the freezing point of the solution is $7.2^oC$ lower than the freezing point of pure X. Calculate the van't Hoff factor for potassium bromide in X.

<u>Answer:</u> The van't Hoff factor for potassium bromide in X is 1.63

<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

$\Delta T_f=i\times K_f\times m$

OR

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

<u>When alanine is dissolved in mystery liquid X:</u>

$\Delta T_f=5.9^oC$

i = Vant Hoff factor = 1 (for non-electrolytes)

$K_f$ = freezing point depression constant

$m_{solute}$ = Given mass of solute (alanine) = 177. g

$M_{solute}$ = Molar mass of solute (alanine) = 89 g/mol

$w_{solvent}$ = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

$5.9=1\times K_f\times \frac{177\times 1000}{89\times 800}\\\\K_f=\frac{5.9\times 89\times 800}{1\times 177\times 1000}\\\\K_f=2.37^oC/m$

<u>When KBr is dissolved in mystery liquid X:</u>

$\Delta T_f=7.2^oC$

i = Vant Hoff factor = ?

$K_f$ = freezing point depression constant = $2.37^oC/m$

$m_{solute}$ = Given mass of solute (KBr) = 177. g

$M_{solute}$ = Molar mass of solute (KBr) = 119 g/mol

$w_{solvent}$ = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

$7.2=i\times 2.37\times \frac{177\times 1000}{119\times 800}\\\\i=\frac{7.2\times 119\times 800}{2.37\times 177\times 1000}\\\\i=1.63$

Hence, the van't Hoff factor for potassium bromide in X is 1.63

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3 years ago

Explain why the Chemicals properties of Mg 24 12 and Mg 26 12 are the same

Volgvan

Answer:

both Mg isotopes have same atomic number which is equal to number of protons , and the chemical properties of an element depends upon the the number of protons in its atom . and both Mg 24 12 and Mg 26 12 have same number of protons hence they exhibits similar chemical

okay

Explanation:

i hope it helped you .

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3 years ago