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RUDIKE [14]
4 years ago
8

A beaker contains 175.32g of salt, NaCl(58.44g/mol). Water is added until the final volume is 2 liters. The solution should be d

escribed as
A. 87.66M

B. 29.22M

C. 3M

D. 1M

E. 1.5M
Chemistry
1 answer:
cricket20 [7]4 years ago
8 0

Answer:

E. 1.5M

Explanation:

because there is total 175.32 g of NaCl dissolved in 2 liter of water, and 58.44 g = 1 mol

so total miles = 175.32/58.32 = 3 moles

now 3 moles are in 2 liter of water, hence one liter contains 3/2= 1.5

Therefore answer is E. 1.5M

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Could the mechanism for the reaction of crystal violet with sodium hydroxide consist of one elementary step? Explain.
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Answer:

Yes

Explanation:

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CV⁺+ OH⁻ ⟶ CVOH

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7 0
3 years ago
The osmotic pressure of an aqueoussolution of urea at 300 K is
den301095 [7]

<u>Answer:</u> The freezing point of solution is -0.09°C

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=imRT

where,

\pi = osmotic pressure of the solution = 120 kPa

i = Van't hoff factor = 1 (for non-electrolytes)

m = concentration of solute in terms of molality = ?

R = Gas constant = 8.31\text{L kPa }mol^{-1}K^{-1}

T = temperature of the solution = 300 K

Putting values in above equation, we get:

120kPa=1\times m\times 8.31\text{ L kPa }mol^{-1}K^{-1}\times 300K\\\\m=0.05m

  • To calculate the depression in freezing point, we use the equation:

\Delta T=i\times K_f\times m

where,

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

K_f = molal freezing point depression constant = 1.86°C/m

m = molality of solution = 0.05 m

Putting values in above equation, we get:

\Delta T=1\times 1.86^oC/m\times 0.05m\\\\\Delta T=0.09^oC

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.  

  • The equation used to calculate freezing point of solution is:

\Delta T=\text{freezing point of water}-\text{freezing point of solution}

where,

\Delta T = Depression in freezing point = 0.09 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

0.09^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-0.09^oC

Hence, the freezing point of solution is -0.09°C

4 0
3 years ago
What is the number of valence electrons in an atom of Al?<br> a. 13<br> b. 3<br> c. 10<br> d. 8
Svet_ta [14]
There are three valence electrons in an atom of Aluminum (Al).
                     ·
                   ·  ·
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Al ·  or    :    Al    :  ·
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The answer is B.
7 0
3 years ago
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