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2 years ago

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To 225 mL of a 0.80M solution of KI, a student adds enough water to make 1.0L of a more dilute KI solution. What is the molarity

of the new solution? A.180M B. 2.8M C. 0.35M D. 0.18M

2 answers:

Makovka662 [10]2 years ago

8 0

Hope this helps you!

Lilit [14]2 years ago

8 0

<u>Answer:</u> The correct answer is Option D

<u>Explanation:</u>

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted solution

We are given:

Conversion factor: 1 L = 1000 mL

$M_1=0.80M\\V_1=225mL\\M_2=?M\\V_2=1L=1000mL$

Putting values in above equation, we get:

$0.80\times 225=M_2\times 1000\\\\M_2=0.18M$

Hence, the correct answer is Option D

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Answer:

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In this case, mercury (II) oxide (HgO) is obtained via the reaction:

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2 years ago

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Explanation:

It is known that 1 gram contains 1000 milligrams. And, mathematically we can represent it as follows.

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2 years ago

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Answer:

0.3023 M

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$H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

Initial: 0.52 0 0

Change: - x + x + x

Equilibrium: 0.52 - x + x + x

Given that;

acid dissociation constant ( $K_a$ ) = 0.42

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$0.42 = \frac{[x]^2}{0.52-x}}$

0.42(0.52-x) = x²

0.2184 - 0.42x = x²

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Using the quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$ ; ( where +/- represent ± )

= $\frac{-0.42+/-\sqrt{(0.42)^2-4(1)(-0.2184)} }{2*1}$

= $\frac{-0.42+/-\sqrt {0.1764+0.8736} }{2}$

= $\frac{-0.42+\sqrt {1.0496} }{2}$ OR $\frac{-0.42-\sqrt {1.0496} }{2}$

= $\frac{-0.42+1.0245}{2}$ OR $\frac{-0.42-1.0245}{2}$

= $\frac{0.6045}{2}$ OR $-\frac{1.4445}{2}$

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So, we go by the +ve integer that says:

x = 0.30225

x = [ $H_3}O^+$ ] = [ $Picric^-$ ] = 0.3023 M

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2 years ago

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