One hundred students were surveyed about who they would vote for. What is

Chemistry

1 answer:

GuDViN [60]3 years ago

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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

When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to

Reika [66]

Answer:

$0.4167\ \text{M}^{-1}\text{min}^{-1}$

Explanation:

Half-life

${t_{1/2}}A=2\ \text{min}$

${t_{1/2}}B=4\ \text{min}$

Concentration

${[A]_0}_A=1.2\ \text{M}$

${[A]_0}_B=0.6\ \text{M}$

We have the relation

$t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}$

$\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}$

Comparing the exponents we get

$1=n-1\\\Rightarrow n=2$

The order of the reaction is 2.

$t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}$