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hram777 [196]
2 years ago
15

Recall that your hypothesis is that these values are the fraction of atoms that are still radioactive after n half-life cycles.

Record in the appropriate blanks.

Chemistry
2 answers:
jolli1 [7]2 years ago
8 0

Answer : A= 0.5, B = 0.25 , C = 0.125, D = 0.015625 and E = 0.00390625

Explanation :

Half life of a substance is defined as the amount of time taken by the substance to reduce to half of its original amount.

Here n represents the number of half lives.

The amount of substance that remains after n half lives can be calculated using the given formula, 0.5^{n}

So when we have n =1,

Fraction of substance that remains = 0.5¹ = 0.5.

That means after first half life over, the amount of substance that remains is 0.5 times that of original.

Therefore we have A = 0.5

When n = 2, we have 0.5² = 0.25

So when 2 half lives are over, the amount of substance that remains is 0.25 times that of original

Therefore B = 0.25

When n = 3, we have 0.5³ = 0.125

So when 3 half lives are over, the amount of substance that remains is 0.125 times that of original.

Therefore we have C = 0.125

When n = 6 , we have 0.5⁶ = 0.015625

So D = 0.015625

When n = 8, we have 0.5⁸ = 0.00390625

Therefore E = 0.00390625

The values for A, B, C, D and E are 0.5, 0.25, 0.125, 0.015625 and 0.00390625 respectively.

9966 [12]2 years ago
7 0
A=0.5
B=0.25
C=0.125
D=0.015625
E=0.00390625
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<u>Answer:</u> The mass of 97 % of NaOH solution required is 114.33 g

<u>Explanation:</u>

To calculate mass of a substance, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

We are given:

Density of 10 % solution = 1.109 g/mL

Volume of 10% solution = 1 L = 1000 mL     (Conversion factor:  1 L = 1000 mL)

Putting values in above equation, we get:

1.109g/mL=\frac{\text{Mass of }10\%\text{ solution}}{1000mL}\\\\\text{Mass of }10\%\text{ solution}=1109g

The mass of 10 % solution is 1109 g.

To calculate the mass of concentrated solution, we use the equation:

c_1m_1=c_2m_2

where,

c_1\text{ and }m_1 are the concentration and mass of concentrated solution.

c_2\text{ and }m_2 are the concentration and mass of diluted solution.

We are given:

c_1=97\%\\m_1=?g\\c_2=10\%\\m_2=1109g

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2 years ago
ethylene glycol used in automobile antifreeze and in the production of polyester. The name glycol stems from the sweet taste of
Luden [163]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is CH_3O and C_4H_{12}O_4

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

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Mass of H_2O=5.58g

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Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

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In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 9.06 g of carbon dioxide, \frac{12}{44}\times 9.06=2.47g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 5.58 g of water, \frac{2}{18}\times 5.58=0.62g of hydrogen will be contained.

  • Mass of oxygen in the compound = (6.38) - (2.47 + 0.62) = 3.29 g

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

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.47g}{12g/mole}=0.206moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.62g}{1g/mole}=0.62moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{3.29g}{16g/mole}=0.206moles

  • <u>Step 2:</u> 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 0.206 moles.

For Carbon = \frac{0.206}{0.206}=1

For Hydrogen  = \frac{0.62}{0.206}=3

For Oxygen  = \frac{0.206}{0.206}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 3 : 1

Hence, the empirical formula for the given compound is C_1H_{3}O_1=CH_3O

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}}

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n=\frac{124g/mol}{31g/mol}=4

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Hello,

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