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2 years ago

Please help, will give brainliest

Chemistry

1 answer:

sleet_krkn [62]2 years ago

5 0

Answer:

Mass = 61.6 g

Explanation:

Given data:

Mass of ammonia produced = 75.0 g

Mass of nitrogen needed = ?

Solution:

Chemical equation:

N₂ + 3H₂ → 2NH₃

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 75 g/ 17 g/mol

Number of moles = 4.4 mol

No we will compare the moles of ammonia with nitrogen.

NH₃ : N₂

2 : 1

4.4 : 1/2×4.4 =2.2 mol

Mass of N₂:

Mass = number of moles × molar mass

Mass = 2.2 mol × 28 g/mol

Mass = 61.6 g

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Why is it important to have the correct bond angles of the different atoms and the shape of the molecule? (20 points NEED ANSWER

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It is important to have the correct bond angles of the different atoms and the shape of the molecule due to following reasons;

Among other properties the polarity of compounds mainly depend upon the shape and bond angles of that particular compound. For example, considering the molecule of water, we already know that it is a polar molecule with partially positive hydrogen atoms and partially negative oxygen atoms and acts as universal solvent. The bond angle in water is about 104.5° with a Bent geometry. Unlike carbon dioxide (CO₂) which has Linear structure with bond angle 180° and is non-polar in nature therefore, the bent geometry in water is responsible for the polarity.

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3 years ago

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2 years ago

Which substance is not capable of acting as an arrhenius acid in aqueous solution

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2 years ago

What changes occur in O + O = O2

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They went from atoms to oxygen molecule.

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3 years ago

A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2

a_sh-v [17]

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

Hence, the empirical formula for the given compound is $CHO_2$

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:

Mass of molecular formula = 90.04 g/mol

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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3 years ago