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kiruha [24]
3 years ago
11

What is C2H8 ? please hurry (CH4=methane) (C2H2=acetylene)

Chemistry
2 answers:
swat323 years ago
8 0

Answer:

Carbon carbon | C2H8

Explanation:

happy to help

JulijaS [17]3 years ago
6 0

Answer:

Carbon carbon | C2H8 - PubChem.

Explanation:

Brainliest plz :)

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4.50 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 1
Anna11 [10]

Answer:

\mathbf{C_{10}H_8}   ( Naphthalene )

Explanation:

Given that:

4.50 g of a Compound X is made up of Carbon , Hydron and Oxygen

It's molecular molar mass = 128 g/mol

Compound X undergoes combustion reaction and the product yield :

CO_2 with mass 15.47g and :

H_2O with mass 2.53 g

The objective is to use this information to determine the molecular formula of X.

We all know that ; number of moles = mass/molar mass

where the molar mass of H_2O is 18 g/mol

number of moles of H_2O product = 2.53 g/18 g/mol

number of moles of H_2O product = 0.1406 moles

Also; the molar mass of CO_2 product = 44 g/mol

number of moles of CO_2 product = 15.47g/ 44 g/mol

number of moles of CO_2 product =  0.3516 moles

number of moles of Compound X in the reactant side= 4.50 g /128 g/mol

number of moles of Compound X n the reactant side= 0.03516 moles

Now; number number of moles of CO_2 in reactant = 0.3516 moles/0.03516 moles

Now; number number of moles of CO_2 in reactant = 10

number of moles of H_2O reactant = 0.1406 moles × 2/0.03516

number of moles of H_2O reactant = 7.997 ≅ 8

Since we said the Compound X is known to be made of Carbon C , Hydrogen H and Oxygen O

Then the molecular formula can be written as :

\mathbf{C_{10}H_8O_{x}}

In order to find the x; we have

128  = (12 × 10 + 1 × 8 + 16 × x)

128 = 120 + 8 + 16x)

128  =  128 + 16 x

128 - 128 = 16 x

0 = 16 x

x = 0/16

x = 0

As x = 0 ; hence there are no oxygen present in the reaction

Thus; the molecular formula of Compound X = \mathbf{C_{10}H_8} which is also known as Naphthalene

5 0
3 years ago
Easy Help Please!!!!!!
IRISSAK [1]
False it can only be one to take place
5 0
3 years ago
Please help!
Len [333]
It will slowly rotten and turn brown
7 0
4 years ago
In a titration experiment a student uses 1.4 m hbr solution and the indicator phenolphthalein to determine the concentration of
Likurg_2 [28]
The question is incomplete. Complete question is attached below:
...........................................................................................................................

Answer: 
Given: conc. of HBr = 1.4 M
Volume of HBr = 15.4 mL
Volume of KOH = 22.10 mL

We know that, M1V1 = M2V2
                        (HBr)      (KOH)

Therefore, M2 = M1V1/V2
                        = 1.4 X 15.4/22.10
                        = 0.9756 M

Concentration of KOH is 0.9756 M.

3 0
3 years ago
Read 2 more answers
HELPP
slamgirl [31]

Answer:

In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.

Explanation:

tally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:

\displaystyle \%\text{H}=\frac{\text{mass H}}{\text{mass compound}}\times 100\%%H=

mass compound

mass H

×100%

\displaystyle \%\text{C}=\frac{\text{mass C}}{\text{mass compound}}\times 100\%%C=

mass compound

mass C

×100%

If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:

\displaystyle \%\text{H}=\frac{2.5\text{g H}}{10.0\text{g compound}}\times 100\%=25\%%H=

10.0g compound

2.5g H

×100%=25%

\displaystyle \%\text{C}=\frac{7.5\text{g C}}{10.0\text{g compound}}\times 100\%=75\%%C=

10.0g compound

7.5g C

×100%=75%

7 0
3 years ago
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