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marissa [1.9K]
1 year ago
11

Combustion analysis of a hydrocarbon produced 33.01 g of CO2 and 6.75 g of H2O.Calculate the empirical formula of the hydrocarbo

n
Chemistry
1 answer:
Basile [38]1 year ago
7 0

Given, 33.019 mol of co₂ of and 6.75g of  h₂o

mol of co₂ = mass/molar mass = 33.01g/44gmol = 0.75 mol

Since 1mol have co₂ 1 mol c

Hence 0.75 mol have = 0.75 mol

Now,

mol of H₂O = 6.75g/18gmol = 0.375 mol

Since 1mol H20 have = 2 mol H

Hence 0.375 mol H20 have = 2x0.375 mol H = 0.75 mol H.

Now

mol of H = [0.75 mol x4] = 3mol

mol of c = [0.75 molx4] = 3 mol

Hence

Empirical Formula = CH.

This is done by dividing the atomic mass by the molecular mass and then multiplying it by the mass of the compound is made. Therefore, the molecular formula for this hydrocarbon is CH3 because it has a ratio of 1 carbon atom to 3 hydrogen atoms. The general formula for saturated hydrocarbons with one ring is CnH2n. Most compounds require information other than the chemical formula to elucidate the structure. However, the CH ratio in the chemical formula can provide information about the chemical structure.

Learn more about Hydrocarbon here:-brainly.com/question/19453390

#SPJ1

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a mineral a.has a chemical formula b. occurs naturally c. has a characteristic internal structure d.all of the above
Anuta_ua [19.1K]
Answer: option d. all of the above.

Explanation:

A mineral is an element or a inorganic compound that existes in nature as solid cristals; usually combined with other minerals in ores.

Some examples of minerals, among many, are titania, wich is TiO2, zirconia, which is ZrO2, silica, which is SiO2, gold, Au, silver, Ag.

As you see the definition and examples given meet the whole features included in the stament: a. the have a chemical formula, b they occur naturally, and c.have a characteristic internal structure (that is the way how the atoms are arranged in the specifi cristal).
4 0
3 years ago
Read 2 more answers
4.81*10^24 atoms of lithium
enyata [817]

Answer:

Mass = 55.52 g

Explanation:

Given data:

Number of atoms of Li = 4.81×10²⁴ atom

Number of grams = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

For Li:

4.81×10²⁴ atom × 1 mol / 6.022 × 10²³ atom

8 moles

Mass in gram:

Mass = number of moles × molar mass

Mass = 8 mol × 6.94 g/mol

Mass = 55.52 g

4 0
3 years ago
What is the mass of 975 mL of mercury? Its density is 13.5 g/mL
miss Akunina [59]

Answer:

72.22 g

Explanation:

975 mL Mercury× 13.5 g/mL = 72.22 g

7 0
3 years ago
If the volume of a gas container at 32 degrees Celsius changes from 1.55 L to 755 mL, what will the final temperature be?
QveST [7]
So to solve this you need to know Charles’s law which is: V1/T1=V2/T2. Where T1 and V1 is the initial volume and Temperature and V2 and T2 is the temperature and volume afterwards. So first plug in the numbers you are given. V1= 1.55L T1= 32C° V2= 755mL T2=?. Since your volumes are two different units you change 755mL to be in L so that would be 0.755 L. And since your temp isn’t in Kelvin you do 273+32= 305K°. You then would rearrange your equation to solve for T2 which is V2T1/V1. Then you plug in your numbers (0.755L)(305K)/1.55L. Then you solve and would be 148.5645161 —> 1.49 x 10^2 K
4 0
3 years ago
What is the wavelength of radiation emitted when an electron goes from the n = 7 to the n = 4 level of the Bohr hydrogen atom? G
Phantasy [73]

Answer:

the wavelength of radiation emitted  is \mathbf{\lambda= 2169.62 \ nm}

Explanation:

The energy of the Bohr's hydrogen atom can be expressed with the formula:

\mathtt{E_n =- \dfrac{13.6\ ev}{n^2}}

For n = 7:

\mathtt{E_7 =- \dfrac{13.6\ ev}{7^2}}

\mathtt{E_7 =-0.27755 \ eV}

For n = 4

\mathtt{E_4=- \dfrac{13.6\ ev}{4^2}}

\mathtt{E_4 =- 0.85\ eV}

The  electron goes from the n = 7 to the n = 4, then :

\mathtt{E_7-E_4 = (-0.27755 - (-0.85) ) \ eV}

\mathtt{= 0.57245\ eV}

Wavelength of the radiation emitted:

\mathtt{\lambda= \dfrac{hc}{0.57245 \ eV}}

where;

hc  = 1242 eV.nm

\mathtt{\lambda= \dfrac{1242 \ eV.nm }{0.57245 \ eV}}

\mathbf{\lambda= 2169.62 \ nm}

4 0
3 years ago
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