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2 years ago

In the chemical equation 3c2h4 how many atoms of carbon are represented

1 answer:

5 0

An example.
water is H2O

2 hydrogen, 1 oxygen

so the number to the right means how much of what is on the left.

so it looks like 2, because C2, but look at the 3 at the beginning. that means
3 (c2h4)

so 6 carbons, 12 hydrogen

the ratio of c2 to h4 doesn't change it's always 1:2.

but the 3 at the front is a different number relating to how much you have

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2 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:

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

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6 0

3 years ago

Does anyone know the answer to these?

Rashid [163]

Answer:

The answer to your question is below

Explanation:

Data

mass of CaCO₃ = 155 g

mass of HCl = 250 g

mass of CaCl₂ = 142 g

reactants = CaCO₃ + HCl

products = CaCl₂ + CO₂ + H₂O

1.- Balanced chemical reaction

CaCO₃ + 2HCl ⇒ CaCl₂ + CO₂ + H₂O

2.- Limiting reactant

molar mass of CaCO₃ = 40 + 12 + 48 = 100 g

molar mass of HCl = 2[1 + 35.5 ] = 73 g

theoretical proportion CaCO₃ /HCl = 100 / 73 = 1.37

experimental proportion CaCO₃ /HCl = 155 / 250 = 0.62

As the experimental proportion was lower than the theoretical proportion the limiting reactant is CaCO₃

3.-

Calculate the molar mass of CaCl₂

CaCl₂ = 40 + 71 = 111 g

100 g of CaCO₃ ------------------ 111 g of CaCl₂

155 g of CaCO₃ ----------------- x

x = (155 x 111) / 100

x = 17205 / 100

x = 172.05 g of CaCl₂

4.- percent yield

Percent yield = 142 / 172.05 x 100 = 82.5 %

5.- Excess reactant

100 g of CaCO₃ -------------------- 73 g of HCl

155 g of caCO₃ ------------------- x

x = (155 x 73)/100

x = 133.15 g

Mass of HCl = 250 - 133.15

= 136.9 g

4 0

2 years ago