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3 years ago

13

An article about half-lives describes a parent isotope. What is a parent isotope?

1 answer:

Vikki [24]3 years ago

5 0

A parent isotope plays an important role in the decay chain process. The parent isotope is the one who undergoes the decay process in the nuclear reaction in order to form its daughter isotope. The daughter isotope may decay on its own and may have its own daughter isotope as well.

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What was the result of heating the mixture? All BUT ONE choice is correct.

snow_tiger [21]

Answer:

w gang alright

Explanation:

ay its b alright

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3 years ago

which two will help you measure how different colors absorb the energy in sunlight is it beaker or thermometer

KATRIN_1 [288]

Beaker does thermometer measures the thermal energy in the air

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3 years ago

In a solid to liquid mixture and increase in temperature means?

Gekata [30.6K]

Answer:

Increase in temperature will not affect a solid - liquid mixture. Example can be taken of salt and water

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3 years ago

What is the molality (m) of a solution that contains 76.5 g of KCl dissolved in 85.0 g of<br> water?

ladessa [460]

Answer:

76.5g KCl/74.55 grams per mole Kcl = x

molality= x/.085 kg H2O

Explanation:

well remember molality is moles of solute/kilograms of solvent. So it's the moles of KCl over 85 g of h20 converted into kg. if this makes sense.

5 0

3 years ago

When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr

Amanda [17]

<u>Answer:</u> The value of $K_c$ for the given reaction is 1.435

<u>Explanation:</u>

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

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of $N_2O_4$ = 9.2 g

Molar mass of $N_2O_4$ = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M$

For the given chemical equation:

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

<u>Initial:</u> 0.20

<u>At eqllm:</u> 0.20-x 2x

We are given:

Equilibrium concentration of $N_2O_4$ = 0.057

Evaluating the value of 'x'

$\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$[NO_2]_{eq}=2x=(2\times 0.143)=0.286M$

$[N_2O_4]_{eq}=0.057M$

Putting values in above expression, we get:

$K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435$

Hence, the value of $K_c$ for the given reaction is 1.435

6 0

3 years ago