What element has 42 neutrons

We can change a gas to liquid by the temperature and the pressure. NextReset

USPshnik [31]

We can change a gas to a liquid by INCREASING the temperature and DECREASING the pressure

7 0

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 $C_2H_2O_4$

3 0

4 years ago

In order to change a solid to a liquid or a liquid to a gas, what is needed?

Neporo4naja [7]

Heat; rather, or change of the molecules to make them move faster

6 0

3 years ago

Plants undergo photosynthesis to produce glucose according to the reaction below. What mass of water is required to produce 5.0g

solniwko [45]

Answer:

option a) 3 g

Explanation:

mass of Glucose = 5 g

Mass of H₂O = ?

Reaction Given:

6CO₂ + 6H₂O ----> C₆H₁₂O₆ + 6O₂

Solution:

First we have to find mass of glucose from balanced reaction.

So,

Look at the reaction

6CO₂ + 6H₂O -------> C₆H₁₂O₆ + 6O₂

6 mol 1 mol

As 6 mole of water (H₂O) give 1 mole of Glucose (C₆H₁₂O₆ )

Convert moles to mass

molar mass of C₆H₁₂O₆ = 6(12) + 12(1) + 6(16)

molar mass of C₆H₁₂O₆ = 72 + 12 + 96

molar mass of C₆H₁₂O₆= 180 g/mol

molar mass of H₂O = 2(1) + 16 = 18 g/mol

Now

6CO₂ + 6H₂O ---------> C₆H₁₂O₆ + 6O₂

6 mol (18 g/mol) 1 mol (180 g/mol)

108 g 180 g

108 g of water (H₂O) produce 180 g of glucose (C₆H₁₂O₆)

So

if 108 g of water (H₂O) produce 180 g of glucose (C₆H₁₂O₆) so how many grams of water (H₂O) will be required to produce 5 g of glucose (C₆H₁₂O₆).

Apply Unity Formula

108 g of water (H₂O) ≅ 180 g of glucose (C₆H₁₂O₆)

X g of water (H₂O) ≅ 5 g of glucose (C₆H₁₂O₆)

Do cross multiply

mass of water (H₂O) = 108 g x 5 g / 180 g

mass of water (H₂O) = 3 g

So 3 g of water is required to produce 5 g of glucose.

7 0

3 years ago

Who pass the grade quarter?.

Fed [463]

Answer:

me

Explanation:

8 0

3 years ago

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