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3 years ago

Can someone pls help me with this question

Chemistry

1 answer:

artcher [175]3 years ago

5 0

Answer:

The answer to your question is 24.325

Explanation:

Data

Magnesium-24 Abundance = 78.70%

Magnesium-25 Abundance = 10.13%

Magnesium-26 Abundance = 11.17%

Process

1.- Convert the abundance to decimals

Magnesium-24 Abundance = 78.70/100 = 0.787

Magnesium-25 Abundance = 10.13/100 = 0.1013

Magnesium-26 Abundance = 11.17/100 = 0.1117

2.- Write an equation

Average atomic mass = (Atomic mass-1 x Abundance 1) + (Atomic mass 2 x

Abundance-2) + (Atomic mass 3 x Abundance 3)

3.- Substitution

Average atomic mass = (24 x 0.787) + (25 x 0.1013) + (26 x 0.1117)

4.- Simplification

Average atomic mass = 18.888 + 2.533 + 2.904

5.- Result

Average atomic mass = 24.325

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Answer: The amount of heat released is 84 calories.

Explanation:

The equation used to calculate the amount of heat released or absorbed, we use the equation:

$Q= m\times c\times \Delta T$

where,

Q = heat gained or released = ? Cal

m = mass of the substance = 10g

c = specific heat of aluminium = 0.21 Cal/g ° C

Putting values in above equation, we get:

$\Delta T={\text{Change in temperature}}=(10-50)^oC=-40^oC$

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3 years ago

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Answer: The empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

Explanation:

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.

We are given:

Mass of $CO_2=13.00g$

Mass of $H_2O=2.662g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in 13.00 g of carbon dioxide, $\frac{12}{44}\times 13.00=3.54g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 2.662 g of water, $\frac{2}{18}\times 2.662=0.296g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

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 0.295 moles.

For Carbon = $\frac{0.295}{0.295}=1$

For Hydrogen = $\frac{0.296}{0.295}=1$

For Oxygen = $\frac{0.603}{0.295}=2.044\approx 2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

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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 :

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Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

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

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

Hence, the empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

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