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3 years ago

Write a balanced chemical equation for the reaction that occurs when heptane, C7H16(l) , burns in air.

1 answer:

8 0

Start with Unbalanced Equation and balance it, so...
C7H16+O2--->CO2+H2O
There are 7 C atoms on the left-hand side, so we need 7 C atoms on the right-hand side. Add a 7 in front of the CO2...7CO2+H2O on right side now.
We have fixed 16 H atoms on the left-hand side, so we need 16 H atoms on the right-hand side. Add an 8 in front of H2O to make 16 (8x2)...7CO2+8H2O on right side now.
There are 22 O atoms on the right-hand side: 14 from the CO2 and 8 from the H2O. Add an 11 in front of the O2 on the left side to make 22 (11x2).
Every formula now has a fixed coefficient. You should have a balanced equation of...
C7H16+11O2--->7CO2+8H2O

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This is my almost last question plsss help

Morgarella [4.7K]

The third one is correct, not sure abt another one

7 0

2 years ago

Read 2 more answers

Calculate the molarity of a solution made by adding 120 g of naoh (40.00 g/mol) to enough water to make 500.0 ml of solution. g

sertanlavr [38]

An: Calculate the molarity of a solution made by adding 120 g of NaOH (40.00 g/mol) to enough water to make 500.0 mL of solution. a) 4.0 M b) 6.0 M c) 1.0 ...

Explanation:

7 0

1 year ago

what volume of co2 is produced at stp when 270g of glucose are consumed in the following reaction? c6h12o6 + 6o2(g) -> 6co2 (

lana [24]

Answer:

202 L

Explanation:

Step 1: Write the balanced equation

C₆H₁₂O₆ + 6 O₂(g) ⇒ 6 CO₂(g) + 6 H₂O(l)

Step 2: Calculate the moles corresponding to 270 g of C₆H₁₂O₆

The molar mass of C₆H₁₂O₆ is 180.16 g/mol.

270 g × 1 mol/180.16 g = 1.50 mol

Step 3: Calculate the moles of CO₂ generated from 1.50 moles of glucose

The molar ratio of C₆H₁₂O₆ to CO₂ is 1:6. The moles of CO₂ formed are 6/1 × 1.50 mol = 9.00 mol

Step 4: Calculate the volume of 9.00 moles of CO₂ at STP

The volume of 1 mole of an ideal gas at STP is 22.4 L.

9.00 mol × 22.4 L/mol = 202 L

4 0

3 years ago

For each solute, identify the better solvent: water or carbon tetrachloride

Olin [163]

Do you mean which one out of water and carbon tetrachloride
if so it will be water

3 0

3 years ago

Three 1.0-l flasks, maintained at 308 k, are connected to each other with stopcocks. initially the stopcocks are closed. one of

Licemer1 [7]

Answer:

0.6103 atm.

Explanation:

We need to calculate the vapor pressure of each component after the stopcocks are opened.

Volume after the stopcocks are opened = 3.0 L.

1) For N₂:

P₁V₁ = P₂V₂

P₁ = 1.5 atm & V₁ = 1.0 L & V₂ = 3.0 L.

P₂ of N₂ = P₁V₁ / V₂ = (1.5 atm) (1.0 L) / (3.0 L) = 0.5 atm.

2) For H₂O:

Pressure of water at 308 K is 42.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of H₂O = (1.0 atm x 42.0 mmHg) / (760.0 mmHg) = 0.0553 atm.

We must check if more 2.2 g of water is evaporated,

n = PV/RT = (0.0553 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.00656 mole.

m = n x cmolar mass = (0.00656 mole) (18.0 g/mole) = 0.118 g.

It is lower than the mass of water in the flask (2.2 g).

3) For C₂H₅OH:

Pressure of C₂H₅OH at 308 K is 102.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of C₂H₅OH = (1.0 atm x 102.0 mmHg) / (760.0 mmHg) = 0.13421 atm.

We must check if more 0.3 g of C₂H₅OH is evaporated,

n = PV/RT = (0.13421 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.01594 mole.

m = n x molar mass = (0.01594 mole) (46.07 g/mole) = 0.7344 g.

It is more than the amount in the flask (0.3 g), so the pressure should be less than 0.13421 atm.

We have n = mass / molar mass = (0.30 g) / (46.07 g/mole) = 0.00651 mole.

So, P of C₂H₅OH = nRT / V = (0.00651 mole) (0.082 L.atm/mole.K) (308.0 K) / (3.0 L) = 0.055 atm.

So, total pressure = P of N₂ + P of H₂O + P of C₂H₅OH = 0.5 atm + 0.0553 atm + 0.055 atm = 0.6103 atm.

3 0

2 years ago