A cual grupo pertenecen las tierras raras y los elementos representativos??

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

How many grams of mercury can be produced if 18.0 g of mercury (11) oxide decomposes?

NARA [144]

Answer:

18.0 g of mercury (11) oxide decomposes to produce 9.0 grams of mercury

Explanation:

Mercury oxide has molar mass of 216.6 g/ mol. It gas a molecular formula of HgO.

The decomposition of mercury oxide is given by the chemical equation below:

2HgO ----> 2Hg + O₂

2 moles of HgO decomposes to produce 1 mole of Hg

2 moles of HgO has a mass of 433.2 g

433.2 g of HgO produces 216.6 g of Hg

18.0 of HgO will produce 18 × 216.6/433.2 g of Hg = 9.0 g of Hg

Therefore, 18.0 g of mercury (11) oxide decomposes to produce 9.0 grams of mercury

3 0

3 years ago

HELP ASAP! Chemical reactions involving liquids only will be influenced by ALL of the following EXCEPT: a) temperature of the sy

Rom4ik [11]

Chemical reactions involving only liquids will be influenced by ALL of the following EXCEPT: c) size of reactant particles d) pressure of the system.

Reason:
1) As per Arrhenius theory, rate of chemical reactions is influence by temperature. Mathematically it is expressed as: k = A exp(-Ea/RT)
where, k = rate constant, A = collision frequency, Ea = activation energy, T = temp. From above relation, it can be seen that, chemical reactions involving only liquids will be influenced temperature.

2) Also, if know that, rates of chemical reactions is mathematically expressed as: $Rate = [A]^{x} B^{y}....$ ,
where A, B, .. are the reactants.
From above relation, it can be seen that, chemical reactions involving only liquids will be influenced concentration i.e. number of particle.

3) However, since all the reactant and catalyst used (if any) is in liquid state, particle size of same will not influence the reaction.

4) Also, since there are no gas-phase reactant, pressure will not affect the reaction.

8 0

3 years ago

Watered down koolaid is an example of

N76 [4]

The answer is A . Saturated solution

5 0

3 years ago

If iodine-131 has a half-life of 8 days, how much of a 64.0 g sample of iodine-131 will remain after 32 day?

Mariulka [41]

Answer:

4 g OF IODINE-131 WILL REMAIN AFTER 32 DAYS.

Explanation:

Half life (t1/2) = 8 days

Original mass (No) = 64 g

Elapsed time (t) = 32 days

Mass remaining (Nt) = ?

Using the half life equation we can obtain the mass remaining (Nt)

Nt = No (1/2) ^t/t1/2

Substituting the values, we have;

Nt = 64 * ( 1/2 ) ^32/8

Nt = 64 * (1/2) ^4

Nt = 64 * 0.0625

Nt = 4 g

So therefore, 4 g of the iodine-131 sample will remain after 32 days with its half life of 8 days.

8 0

3 years ago

A cual grupo pertenecen las tierras raras y los elementos representativos??​

A cual grupo pertenecen las tierras raras y los elementos representativos??