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Octane has a density of 0.703 g/ml. Calculate the mass of CO2(g) produced by burning one

STatiana [176]

The mass of CO₂ gas produced during the combustion of one gallon of octane is 8.21 kg.

The given parameters:

Density of the octane, ρ = 0.703 g/ml
Volume of octane, v = 3.79 liters

The mass of the octane burnt is calculated as follows;

$m = \rho V\\\\m = 0.703 \ \frac{g}{ml} \times 3.79 \ L \ \frac{1000 \ ml}{L} \\\\m = 2,664.37 \ g$

The combustion reaction of octane is given as;

$2C_8H_{18} + \ 25O_2 \ --> \ 16CO_2 \ + \ 18H_2O$

From the reaction above:

228.46 g of octane -------------------> 704 g of CO₂ gas

2,664.37 of octane --------------------> ? of CO₂ gas

$= \frac{2,664.37 \times 704}{228.46} \\\\= 8,210.3 \ g\\\\= 8.21 \ kg$

Thus, the mass of CO₂ gas produced during the combustion of one gallon of octane is 8.21 kg.

Learn more about combustion of organic compounds here: brainly.com/question/13272422

8 0

2 years ago

In a synthesis reaction you start with 1.7L of Hydrogen how many liters of water will be produced?

jolli1 [7]

15.3 litres of water will be produced if we take 1.7 litres of Hydrogen

Explanation:

Let's take a look over synthesis reaction;

$H_{2}+ O_{2}$ $H_{2}O$

Balancing the chemical reaction;

$2H_{2} +O_{2}$ $2H_{2}O$

Thus, 2 moles of hydrogen molecules are required to form 2 moles of water molecules.

Equating the molarity;

$\frac{1.7*1}{2*2}$ = $\frac{x*1}{2*18}$

(Since, the molecular mass of hyd and water is 2 and 18 respectively)

x= $\frac{1.7*2*18}{2*2}$

x= 15.3 litres.

Thus,15.3 L of water will be produced if we take 1.7 litres of Hydrogen in a synthesis reaction.

6 0

2 years ago

I need the answer for this question

ch4aika [34]

I want to say A only... Hope I helped!!

5 0

2 years ago

A 20.0 g piece of a metal is heated and place into a calorimeter containing 250.0 g of water initially at 25.0 oC. The final tem

BartSMP [9]

Answer:

$Q_{metal} = -6799\,J$

Explanation:

By the First Law of Thermodynamics, the piece of metal and water reaches thermal equilibrium when water receives heat from the piece of metal. Then:

$Q_{metal} = - Q_{w}$

$Q_{metal} = m_{w} \cdot c_{p,w}\cdot (T_{1}-T_{2})$

$Q_{metal} = (250\,g)\cdot \left(4.184\,\frac{J}{g\cdot ^{\textdegree}C} \right)\cdot (25\,^{\textdegree}C - 31.5\,^{\textdegree}C)$

$Q_{metal} = -6799\,J$

6 0

3 years ago

Read 2 more answers

A solution contains 0.115 mol h2o and an unknown number of moles of nacl. the vapor pressure of the solution at 30°c is 25.7 tor

Valentin [98]

According to Raoult's low:
We will use this formula: Vp(Solution) = mole fraction of solvent * Vp(solvent)
∴ mole fraction of solvent = Vp(Solu) / Vp (Solv)
when we have Vp(solu) = 25.7 torr & Vp(solv) = 31.8 torr
So by substitution:
∴ mole fraction of solvent = 25.7 / 31.8 =0.808
when we assume the moles of solute NaCl = X
and according to the mole fraction of solvent formula:
mole fraction of solvent = moles of solvent / (moles of solvent + moles of solute)
by substitute:
∴ 0.808 = 0.115 / (0.115 + X)
So X (the no.of moles of NaCl) = 0.027 m

8 0

2 years ago