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3 years ago

7

PLEASE HELP ME NOW I NEED HELP What will happen if a single crystal is introduced into a super-saturated solution? a. The crysta

l will dissolve c. The solute will begin to form crystals b. Nothing d. The solution will vaporize Please select the best answer from the choices provided A B C D

1 answer:

Klio2033 [76]3 years ago

4 0

Answer:

c

It excess out the crystal

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What is molarity of 47.0 g KCl dissolved in enough water to give 375 mL of solution?

natita [175]

This question provides us –

Weight of $\bf KCl$ is = 47 g
Volume, V = 375 mL

__________________________________________

Molar Mass of $\bf KCl$ –

$\qquad$ $\twoheadrightarrow\bf 39.0983 \times 35.453$

$\qquad$ $\twoheadrightarrow\bf 74.5513$

<u>Using formula</u> –

$\qquad$ $\purple{\twoheadrightarrow\bf Molarity _{(Solution)} = \dfrac{ W\times 1000}{MV}}$

$\qquad$ $\twoheadrightarrow\bf Molarity _{(Solution)} = \dfrac{ 47 \times 1000}{74.5513\times 375}$

$\qquad$ $\twoheadrightarrow\bf Molarity _{(Solution)} = \dfrac{47000}{27956.7375}$

$\qquad$ $\twoheadrightarrow\bf Molarity _{(Solution)} = \cancel{\dfrac{47000}{27956.7375}}$

$\qquad$ $\twoheadrightarrow\bf Molarity _{(Solution)} = 1.68117M$

$\qquad$ $\pink{\twoheadrightarrow\bf Molarity _{(Solution)} = 1.7M}$

Henceforth, Molarity of the solution is = 1.7M

___________________________________________

6 0

2 years ago

Match the following aqueous solutions with the appropriate letter from the column on the right.1. 0.19 m AgNO3 2. 0.17 m CrSO4 3

vichka [17]

Answer:

0.13 m of $Mn(NO_3)_2$ → Highest boiling point

0.19 m of $AgNO_3$ → Second Highest boiling point

0.17 m of $CrSO_4$ → Third highest boiling point

0.31 m Sucrose (nonelectrolyte) → Lowest boiling point

Explanation:

Elevation in boiling is given by :

$\Delta T_b=i\times k_b\times m$

Where :

i = van't Hoff factor

$k_b$ = Molal Elevation constant of solvent

m = molaity of the solution

1) 0.19 m of $AgNO_3$

$AgNO_3\rightarrow Ag^++NO_3^{-}$

i = 2 (electrolyte)

Molality of the solution = 0.19

Elevation is boiling point of solution:

$\Delta T_b=2\times k_b\times 0.19 m$

$\Delta T_b=0.38 m\times k_b$

2) 0.17 m of $CrSO_4$

$CrSO_4\rightarrow Cr^{2+}+SO_4^{2-}$

i = 2 (electrolyte)

Molality of the solution = 0.17

Elevation is boiling point solution :

$\Delta T_b=2\times k_b\times 0.17 m$

$\Delta T_b=0.34 m\times k_b$

3) 0.13 m of $Mn(NO_3)_2$

$Mn(NO_3)_2\rightarrow Mn^{2+}+2NO_3^{-}$

i = 3 (electrolyte)

Molality of the solution = 0.13

Elevation is boiling point solution :

$\Delta T_b=3\times k_b\times 0.13 m$

$\Delta T_b=0.39 m\times k_b$

4) 0.31 m Sucrose (nonelectrolyte)

i = 1 ( non electrolyte)

Molality of the solution = 0.31 m

Elevation is boiling point solution :

$\Delta T_b=1\times k_b\times 0.31 m$

$\Delta T_b=0.31 m\times k_b$

Higher the value of elevation in temperature higher will be the boiling point of the solution .

The decreasing order of solution from highest boiling point to lowest boiling point is :

$0.39 m\times k_b>0.38 m\times k_b>0.34 m\times k_b>0.31 m\times k_b$

0.13 m of $Mn(NO_3)_2$ → Highest boiling point

0.19 m of $AgNO_3$ → Second Highest boiling point

0.17 m of $CrSO_4$ → Third highest boiling point

0.31 m Sucrose (nonelectrolyte) → Lowest boiling point

6 0

3 years ago

I need help with number 8 please help ASAP.

Rzqust [24]

Answer:

33.33% = 33%

Explanation:

MgCO3(s) + 2HCl (aq) --> MgCl2(aq) + H20(l) + CO2(g)

1 mole of MCO3 will produce → 1 mole of CO2

We need to get the number of mole of CO2:

and when we have 0.22 g of CO2, so number of mole = mass / molar mass

Moles = 0.22 g / 44 g/mol = 0.005 mole

Moles of Mg = moles of CO2 = 0.005 mole

Mass of Mg = moles * molar mass

= 0.005 * 84 /mol = 0.42 g

Percent of MgCO3 by mass of Mg = 0.42 g / 1.26 * 100

=33.33 %

8 0

3 years ago

Giving brainiest to whoever answers correctly.

Digiron [165]

Answer:250

Explanation:

5 0

3 years ago

A cleaning bottle contains 83.1 g of ammonia. How many molecules of ammonia are in the bottle?

Tasya [4]

Answer:

2.94 x $10^2^4$

Explanation:

First we need to find out how many moles of ammonia there are, using the formula: Mass = mr x moles.

We know the mass is 83.1g, now we need to find the mR of ammonia - NH3.

N = 14, H = 1, so 14 + (3x1) = an mr of 17.

Moles = mass/ mr = 83.1/17 = 4.8882

Now we can multiply the moles by avogadro's constant to find the number of molecules:

4.8882 x (6.02 x $10^2^3$ ) = 2.94 x $10^2^4$ molecules of ammonia

4 0

3 years ago