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Write a balanced chemical equation for the complete combustion of C3H7BO3, a gasoline additive. The products of combustion are C

nexus9112 [7]

Answer:

2C₃H₇BO₃ + 8O₂ → 6CO₂ + 7H₂O + B₂O₃.

Explanation:

For balancing a chemical equation, we should apply the law of conversation of mass. It states that the no. of atoms in the reactants side is equal to that of the products side.

So, the balanced equation:

2C₃H₇BO₃ + 8O₂ → 6CO₂ + 7H₂O + B₂O₃.

It is clear that 2.0 moles of C₃H₇BO₃ is completely burned in 8 m oles of oxygen and produce 6 moles of CO₂, 7 moles of H₂O and 1 mole of B₂O₃.

7 0

3 years ago

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

Equal but opposite forces acting on an object results in what?

forsale [732]

Answer:

Action given and reaction taken

Also known as

Newton's third law of motion

Explanation:

An action will be done such as bouncing a ball on the wall

- You throw the ball (Action)

- The ball bounces back (Reaction)

Hope this Helps

3 0

2 years ago

How to balance the question?

umka21 [38]

Answer:

CH4 + 2O2 —> CO2 + 2H2O

CH4 + 2O2

C = 1

H = 4

O = 2 ×2 = 4

__o_o___

CO2 + 2H20

C = 1

H = 2 × 2 = 4

O = 2 + 2 = 4

So ; 1 C , 4 H , 4 O

I hope I helped you^_^

4 0

3 years ago

1.Why mixture of stone pebbles and water is considered a suspension? Write few properties of the suspension. How suspension diff

sergey [27]

Explanation:

1.A mixture of stone, pebbles, and water is an example of a suspension. Because the particle size is more than 1000nm.

Properties of suspensions:

Suspension is a heterogeneous mixture. The constituents of the mixture are clearly visible.

The particle size will be more than 1000nm. If the suspension is left undisturbed then the particles will settle down.

Properties of Colloids:

Whereas in colloidal solutions, the particle size is in between 1nm-1000nm.

The particles will never settle down in colloid.

They show the Tyndall effect.

2.The solubility of salt at 20^oC is 30. That means the maximum amount of salt that is present in 100 g of water is 30 g.

If the temperature of a saturated solution is increased then, its solubility increases or decreases based on the nature of the solute.

For example if we take sodium chloride salt ,then its solubility is not effected by the change in temperature.

For cerium sulphate salt if temperature increases, then its solubility in water decreases.

For sugar (sucrose) its solubility increases with increase in temperature.

8 0

3 years ago