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3 years ago

What is the molarity of a solution prepared by dissolving 10.0g of kno3 in 250 ml of solution

Chemistry

1 answer:

maw [93]3 years ago

8 0

Answer:

Mass of KNO3= 10g

Molar mass of KNO3 = 101.1032g/mol

Volume = 250ml = 0.25L

No of mole on of KNO3 = mass of KNO3/Molar mass of KNO3

no of mole of KNO3 = 10/101.1032

No of mole of KNO3 = 0.09891

molarity of KNO3 = no of mole of KNO3/Vol (L)

Molarity = 0.09891/0.25 = 0.3956M

Molarity of KNO3 = 0.3956M

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You start out with 800 moles of a radioactive substance. After 24 hours, 100 moles remain. What is the half-life of the substanc

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The half life of the substance is 8 hours , Option D is the correct answer.

What is Half Life ?

Half-life is the time required by an unstable element to reduce to half of its initial value.

The term is commonly used in nuclear physics.

The formula is

$\rm N(t) = \; N_{0} \;(\dfrac{1}{2} )^{\frac{t}{t_{1/2} } }$

N(t) = quantity of the substance remaining

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The data given in the question is N ₀ = 800 moles , N(t)= 100 , t= 24 hours

$\rm t_{1/2} =\; ?$

Substituting values in the equation

$\rm 100 = \; 800 \;(\dfrac{1}{2} )^{\frac{24}{t_{1/2} } }$

$\rm \dfrac{100}{800} = \;(\dfrac{1}{2} )^{\frac{24}{t_{1/2} } }\\\\\rm \dfrac{1}{8} = \;(\dfrac{1}{2} )^{\frac{24}{t_{1/2} } }\\\\\rm (\dfrac{1}{2})^{3} = \;(\dfrac{1}{2} )^{\frac{24}{t_{1/2} } }\\\\\rm 3=\dfrac{24}{t_{1/2} }\\\\t_{1/2}= \dfrac{24}{3}\\\\t_{1/2} = 8 hours$

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2 years ago

In degrees celsius, the boiling point of water is ___ degrees higher than the freezing point. (Apex)

mixer [17]

The freezing point of water in Celsius is 0 degrees and the boiling point is 100 degrees so the answer would be 100 degrees

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2 years ago

Read 2 more answers