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4 years ago

Element X had a mass of 300g and a 1/2 life of 10 years. how many grams will remain after 40 years?

Chemistry

1 answer:

kiruha [24]4 years ago

3 0

<h3>Answer:</h3>

18.75 grams

<h3>Explanation:</h3>

Half-life refers to the time taken by a radioactive material to decay by half of the original mass.
In this case, the half-life of element X is 10 years, which means it takes 10 years for a given mass of the element to decay by half of its original mass.
To calculate the amount that remained after decay we use;

Remaining mass = Original mass × (1/2)^n, where n is the number of half-lives

Number of half-lives = Time for the decay ÷ Half-life

= 40 years ÷ 10 years

= 4

Therefore;

Remaining mass = 300 g × (1/2)⁴

= 300 g × 1/16

= 18.75 g

Hence, a mass of 300 g of an element X decays to 18.75 g after 40 years.

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<u>Answer:</u> The correct option is C) 1.68 mol/L

<u>Explanation:</u>

Molarity is defined as the amount of solute expressed in the number of moles present per liter of solution. The units of molarity are mol/L. The formula used to calculate molarity:

$\text{Molarity of solution}=\frac{\text{Given mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (L)}}$ .....(1)

Given values:

Given mass of $C_6H_{12}O_6$ = 150 g

Molar mass of $C_6H_{12}O_6$ = 180 g/mol

Volume of the solution = 0.50 L

Putting values in equation 1, we get:

$\text{Molarity of solution}=\frac{150}{180\times 0.50}\\\\\text{Molarity of solution}=1.68M$

Hence, the correct option is C) 1.68 mol/L

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3 years ago

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3 years ago

Consider the chemical equation for the ionization of CH3NH2 in water. Estimate the percent ionization of CH3NH2 in a 0.050 M CH3

anyanavicka [17]

Answer: The percent ionization of $CH_3NH_2$ in a 0.050 M $CH_3NH_2(aq)$ solution is 8.9 %

Explanation:

$CH_3NH_2+H_2O\rightarrow OH^-+CH_3NH_3^+$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_b=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= concentration = 0.050 M and $\alpha$ = degree of ionisation = ?

$K_b=4.4\times 10^{-4}$

Putting in the values we get:

$4.4\times 10^{-4}=\frac{(0.050\times \alpha)^2}{(0.050-0.050\times \alpha)}$

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3 years ago

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3 years ago

Read 2 more answers

determine the empirical and molecular formula of a compound composed of 18.24 g carbon, 0.51 g hydrogen, and 16.91 g fluorine an

Vika [28.1K]

Answer: The empirical formula for the given compound is $C_3HF_2$ and molecular formula for the given compound is $C_{24}H_8F_{16}$

Explanation : Given,

Mass of C = 18.24 g

Mass of H = 0.51 g

Mass of F = 16.91 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{18.24g}{12g/mole}=1.52moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.51g}{1g/mole}=0.51moles$

Moles of Fluorine = $\frac{\text{Given mass of Fluorine}}{\text{Molar mass of Fluorine}}=\frac{16.91g}{19g/mole}=0.89moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.51 moles.

For Carbon = $\frac{1.52}{0.51}=2.98\approx 3$

For Hydrogen = $\frac{0.51}{0.51}=1$

For Fluorine = $\frac{0.89}{0.51}=1.74\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : F = 3 : 1 : 2

The empirical formula for the given compound is $C_3H_1F_2=C_3HF_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Molar mass = 562.0 g/mol

Mass of empirical formula = 3(12) + 1(1) + 2(19) = 75 g/eq

Putting values in above equation, we get:

$n=\frac{562.0}{75}=7.49\approx 8$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_3HF_2=(C_3HF_2)_n=(C_3HF_2)_8=C_{24}H_8F_{16}$

Thus, the molecular formula for the given compound is $C_{24}H_8F_{16}$

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3 years ago