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3 years ago

Please help Show your workings

Chemistry

1 answer:

Trava [24]3 years ago

3 0

Answer:

jbgnbvjbvn v ncbncjbgnbhjfjhfjhfjbfihfjh

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What are two examples of a chemical change

lyudmila [28]

Answer:

1.Cooking an egg

2. Burning wood

Hope it helps :)

6 0

3 years ago

Read 2 more answers

A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of

vlada-n [284]

Answer: The molecular of the compound is, $C_2H_3O$

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 $CO_2=3.095g$

Mass of $H_2O=1.902g$

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 $3.095g$ of carbon dioxide, $\frac{12}{44}\times 3.095=0.844g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in $1.902g$ of water, $\frac{2}{18}\times 1.092=0.121g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.621)-[(0.844)+(0.121)]=0.656g$

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles$

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 $0.041$ moles.

For Carbon = $\frac{0.0703}{0.041}=1.71\approx 2$

For Hydrogen = $\frac{0.121}{0.041}=2.95\approx 3$

For Oxygen = $\frac{0.041}{0.041}=1$

Step 3: Taking the mole ratio as their subscripts.

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

Hence, the empirical formula for the given compound is $C_2H_3O_1=C_2H_3O$

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{46.06}{43}=1$

Molecular formula = $(C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O$

Therefore, the molecular of the compound is, $C_2H_3O$

6 0

3 years ago

What is the volume of 14.0g of nitrogen gas at STP?

lozanna [386]

Answer:

The volume of 14.0 g of nitrogen gas at STP is 11.2 liter.

Explanation:

STP stands for standard pressure and temperature.

The International Institute of of Pure and Applied Chemistry, IUPAC changed the definition of standard temperature and pressure (STP) in 1982:

Before the change, STP was defined as a temperature of 273.15 K and an absolute pressure of exactly 1 atm (101.325 kPa).

After the change, STP is defined as a temperature of 273.15 K and an absolute pressure of exactly 105 Pa (100 kPa, 1 bar).

Using the ideal gas equation of state, PV = nRT you can calculate the volume of one mole (n = 1) of gas. With the former definition, the volume of a mol of gas at STP, rounded to 3 significant figures, was 22.4 liter. This is classical well known result.

With the later definition, the volume of a mol of gas at STP is 22.7 liter.

I will use the traditional measure of 22.4 liter per mole of gas.

1) Convert 14.0 g of nitrogen gas to number of moles:

n = mass in grams / molar mass

Atomic mass of nitrogen: 14.0 g/mol

Nitrogen gas is a diatomic molecule, so the molar mass of nitrogen gas = molar mass of N₂ = 14.0 × 2 g/mol = 28.0 g/mol

n = 14.0 g / 28.0 g/mol = 0.500 mol

2) Set a proportion to calculate the volume of nitrogen gas:

22.4 liter / mol = x / 0.500 mol

Solve for x: x = 0.500 mol × 22.4 liter / mol = 11.2 liter.

Conclusion: the volume of 14.0 g of nitrogen gas at STP is 11.2 liter.

6 0

3 years ago

Consider the balanced equation for the following reaction:

Bad White [126]

Answer: The theoretical yield of the lithium chlorate is 1054.67 grams

Explanation:

To calculate the mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Actual moles of lithium chlorate = 9.45 moles

Molar mass of lithium chlorate = 90.4 g/mol

Putting values in above equation, we get:

$9.45mol=\frac{\text{Actual yield of lithium chlorate}}{90.4g/mol}\\\\\text{Actual yield of lithium chlorate}=(9.45mol\times 90.4g/mol)=854.28g$

To calculate the theoretical yield of lithium chlorate, we use the equation:

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Actual yield of lithium chlorate = 854.28 g

Percentage yield of lithium chlorate = 81.0 %

Putting values in above equation, we get:

$81=\frac{854.28g}{\text{Theoretical yield of lithium chlorate}}\times 100\\\\\text{Theoretical yield of lithium chlorate}=\frac{854.28\times 100}{81}=1054.67g$

Hence, the theoretical yield of the lithium chlorate is 1054.67 grams

7 0

3 years ago

How many multiple bonds in ch4

chubhunter [2.5K]

It has 8 bonding electrons

7 0

3 years ago