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in an experiment 3.5g of element A reacted with 4.0g of element G to form a compound Calculate the empirical formula for this co

kolezko [41]

Additional information

Relative atomic mass(Ar) : A=7, G=16

The empirical formula : A₂G

<h3>Further explanation</h3>

Given

3.5g of element A

4.0g of element G

Required

the empirical formula for this compound

Solution

The empirical formula is the smallest comparison of atoms of compound forming elements.

The empirical formula also shows the simplest mole ratio of the constituent elements of the compound

mol of element A :

$\tt mol=\dfrac{mass}{Ar}\\\\mol=\dfrac{3.5}{7}=0.5$

mol of element G :

$\tt mol=\dfrac{4}{16}=0.25$

mol ratio A : G = 0.5 : 0.25 = 2 : 1

4 0

2 years ago

Which flower parts make up a pistil?

liberstina [14]

Answer:

B

Explanation

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4 0

2 years ago

Electronegativity is :_______________

bixtya [17]

Answer:

c) the attraction of an atom for the electrons in a covalent bond.

Explanation:

Electronegativity, symbol χ, is a chemical property describing an atom's ability of to attract a shared pair of electrons to itself. It is influenced by the atomic number of the atom and the distance between the valence electrons and the charged nucleus. As the electronegativity number of atoms increases, the more the atom attracts electrons towards itself.

7 0

3 years ago

Read 2 more answers

A gaseous compound is 30.4 % N and 69.6% OF. A 5.25 g sample of the gas occupies a volume of 1.00 L and exerts a pressure of 958

Harman [31]

Answer:

The molecular formula = N2O4

Explanation:

Step 1: Data given

A gaseous compound is 30.4 % N and 69.6%

Mass of the compound = 5.25 grams

Volume of the gas = 1.00 L

Pressure of the gas = 958 mmHg = 1.26 atm

Temperature of the gas = -4 °C = 273 -4°C = 269 Kelvin

Molar mass of N = 14 g/mol

Molar mass of O = 16 g/mol

Step 2: Calculate mass of N

Mass of Nitrogen = 5.25 grams * 0.304 = 1.596 grams

Step 3: Calculate mass of O

Mass of Oxygen = 5.25 grams * 0.696 = 3.654 grams

Step 4: Calculate number of moles N

Number of moles N = Mass of N/ Molar mass of N

Moles of N = 1.596 grams / 14g/mol

Moles of N = 0.114 moles N

Step 5: Calculate moles of O

Moles O = 3.654 grams / 16 g/mol

Moles 0 = 0.2884 moles

Step 6: Calculate empirical formule

We calculate the empirical formule by dividing number of moles by the smallest number of mol

N : 0.114 / 0.114 = 1

O: 0.2284 / 0.114 = 2

Empirical formule = NO2

Step 7: Calculate number of moles of 5.25 g sample via gas law:

p*V = nRT

⇒ with p = the pressure = 1.26 atm

⇒ with v = 1.00 L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/ K*mol

⇒ with T = the temperature = 269 K

number of moles n = (p*V)/(R*T)

n = (1.26*1L)/(0.08206*269)

n = 0.057 mol

Step 8: Calculate molar mass of the compound

This means 5.25 grams of the gas = 0.057 moles

So 1 mol of the compound has a molar mass of: 5.25 / 0.057 = 92.11 g/mol

Step 9: Calculate molar mass of the empirical formula NO2

N = 14 g/mol

O = 16 g/mol

NO2 = 14 + 16 + 16 = 46 g/mol

The empirical formule NO2 has a molar mass of 46 g/mol

Step 10: Calculate molecular formula

92.11 / 46 = 2

This means the empirical formula should be multiplied by 2

2*(NO2) = N2O4

The molecular formula = N2O4

8 0

3 years ago

Pls help me i dont know this

r-ruslan [8.4K]

Answer:

1. 12.6 moles

2. 8.95 moles

3. 2A + 5B → 3C

4. 48 moles

Explanation:

1. 2Fe + 3Cl₂ → 2FeCl₃

We assume the chlorine in excess. Ratio is 2:2

2 moles of Fe, can produce 2 moles of chloride

12.6 moles of Fe will produce 12.6 moles of chloride.

2. 2Fe + 3Cl₂ → 2FeCl₃

For the same reaction, first of all we need to convert the mass to moles:

500 g . 1mol / 55.85 g = 8.95 mol

As ratio is 2:2, the moles we have are the same, that the produced

4. The reaction for the combustion is:

2C₂H₆ (g) + 7O₂ (g) → 4CO₂ (g) + 6H₂O (l)

We assume the oxygen in excess.

Ratio is 2:6, so 2 mol of ethane produce 6 moles of water

Therefore 16 moles of ethane may produce (16 .6) / 2 = 48 moles

3 0

3 years ago