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3 years ago

When 13.0 g of a hydrocarbon are burned in excess O2, 9.0 g of H2O are formed. What is the formula of the hydrocarbon?

Chemistry

1 answer:

givi [52]3 years ago

6 0

Answer:

Explanation:

When a hydrocarbon burns, there are only two products. These are hydrogen and carbon iv oxide. The general equation for burning a hydrocarbon is shown below:

Fuel + O2 → CO2 + H2O

Since 9g Of water is formed, we can get the number of moles of hydrogen and the mass of hydrogen.

We now need to get the actual moles of carbon and hydrogen in the hydrocarbon to know what the chemical formula is.

The number of moles of hydrogen and its mass can be obtained from that of water. The number of moles of water is simply the mass of water divided by the molar mass. The molar mass of water is 18g/mol. The number of moles is thus 9/18 which is 0.5. Since there are two atoms of hydrogen in one molecule of water, the number of moles of hydrogen is thus 2 * 0.5 which equals one mole.

The mass of hydrogen is thus the number of moles of hydrogen * the molar mass of hydrogen which is

1 * 1 = 1g

Hence, the mass of the carbon in the hydrocarbon is 13 - 1 = 12g

The number of moles of carbon present is thus 12/12 which equals 1.

Since the mole ratio is 1 to 1 , the formula of the hydrocarbon is thus CH

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In the reaction, A → Products, the rate constant is 3.6 × 10−4 s−1. If the initial concentration of A is 0.548 M, what will be t

Arada [10]

Answer:

$\large\boxed{\large\boxed{0.529M}}$

Explanation:

Since the rate constant has units of s⁻¹, you can tell that the order of the reaction is 1.

Hence, the rate law is:

$r=d[A]/dt=-k[A]$

Solving that differential equation yields to the well known equation for the rates of a first order chemical reaction:

$[A]=[A]_0e^{-kt}$

You know [A]₀, k, and t, thus you can calculate [A].

$[A]=0.548M\times e^{-3.6\cdot 10^{-4}/s\times99.2s}$

$[A]=0.529M$

7 0

3 years ago

Which two subatomic particles have the greatest mass?

denpristay [2]

Answer:

PROTON AND NEUTRON

Explanation:

The mass of proton is : $1.67*10^{-27}kg$

A proton is one of the main particles that make up the atom . The other two particles are neutron and electron. Protons are found in the nucleus of the atom.This is a tiny , dense region at the centre of the atom. Protons have a positive charge of one (+1) and a mass of 1 atomic mass unit ( amu ) , which is about $1.67*10^{-27}kg$ . Together with neutrons , they make up virtually all of the mass of an atom.

The mass of neutron is also approximately: $1.67*10^{-27}kg$ but a little more than that .

Atoms of all elements - except Hydrogen , have neutrons in their nucleus . Unlike protons and electrons , these have no charge - they are electrically neutral . The mass of a neutron is slightly greater than the mass of a proton but not very significant

6 0

3 years ago

Since children of the first union/marriage have the trait the great- grandparents do not have the trait they must have been carr

stira [4]

Answer:

a.true

Explanation:

6 0

2 years ago

Read 2 more answers

A 1.450 g sample of an unknown organic compound , X, is dissolved in 15.0 g of toluene

muminat

Answer:

Molecular weight of the compound = 372.13 g/mol

Explanation:

Depression in freezing point is related with molality of the solution as:

$\Delta T_f = K_f \times m$

Where,

$\Delta T_f$ = Depression in freezing point

$K_f$ = Molal depression constant

m = Molality

$\Delta T_f = K_f \times m$

$1.33 = 5.12 \times m$

m = 0.26

Molality = $\frac{Moles\ of\ solute}{Mass\ of\ solvent\ in\ kg}$

Mass of solvent (toluene) = 15.0 g = 0.015 kg

$0.26 = \frac{Mole\ of\ compound}{0.015}$

Moles of compound = 0.015 × 0.26 = 0.00389 mol

$Mol = \frac{Mass\ in\ g}{Molecular\ weight}$

Mass of the compound = 1.450 g

$Molecular\ weight = \frac{Mass\ in\ g}{Moles}$

Molecular weight = $\frac{1.450}{0.00389} = 372.13\ g/mol$

4 0

3 years ago

Why is a cis-1,3-disubstituted cyclohexane more stable thanits trans isomer?

anzhelika [568]

Answer:

Explanation has been given below

Explanation:

In diaxial conformation of cis-1,3-disubstituted cyclohexane, 4 gauche-butane interactions along with syn-diaxial interaction are present. Hence it readily gets converted to diequitorial conformation where no such gauche-butane interaction is present
In two possible conformations of trans-1,3-disubstituted cyclohexane, 2 gauche-butane interactions are present in each of them.
Hence cis-1,3-disubstituted cyclohexane exists almost exclusively in diequitorial form. But trans-1,3-disubstituted cyclohexane has no such option.

Trans-1,3-disubstituted cyclohexane experiences gauche butane interaction in each of the two conformations.
Therefore cis-1,3-disubstituted cyclohexane is more stable than trans conformation

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3 years ago