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BaLLatris [955]
3 years ago
10

What is the IUPAC name of the following compound?

Chemistry
1 answer:
vaieri [72.5K]3 years ago
4 0

Answer:

2,2,3-trimethylbutane

Step-by-step explanation:

Here are the rules for naming alkanes.

1. Find the longest continuous chain of carbon atoms.

There is a <em>four-carbon chain</em> (blue).

2. Name the main chain.

The chain has four C atoms, so its base name is <em>butane</em>.

3. Identify and name all the substituents.

There are three CH₃ groups (<em>methyl</em><em>,</em> blue bonds).

4. Number the main chain from the end that gives the lowest possible locants to the substituents.

From the left-hand end, the locants are 2,2,3. From the right-hand end, they are 2,3,3. The first point of difference is the second locant. 2 < 3, so we <em>number from the left</em>.

5. Put the names of the substituents with their locants in alphabetical order in front of the base name with no spaces.

Use numbers to connect numbers and hyphens to connect numbers and letters.

The IUPAC name is 2,2,3-trimethylbutane.

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The answer is (3) metallic. Cobalt is a transition metal, so it can't be covalent bonds, which bond non-metals, therefore eliminating choice 1 and 2. Ionic bonds are between metals and non metals, but solid cobalt does not have a non metal, eliminating choice 4 as well. Metallic bonds are bonds between metals, therefore the answer is (3) metallic.
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3 years ago
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Calculate the percentage by mass of oxygen in trioxonitrate (v) acid, (HNO3). [ H=1, N=14, O=16]​
Vadim26 [7]

Answer:

Percentage by mass of oxygen = 76.20% (Approx)

Explanation:

Given:

HNO3

H=1, N=14, O=16]

Find:

Percentage by mass of oxygen

Computation:

HNO3

Total mass = 1 + 14 + 3(16)

Total mass = 63

Mass of oxygen = (3)(16) = 48

Percentage by mass of oxygen = [48/63]100

Percentage by mass of oxygen = 76.20% (Approx)

4 0
3 years ago
65. Which of the following heavy metal is not
creativ13 [48]
I believe it is ni . Hope this helps
3 0
2 years ago
Combustion analysis of a 13.42-g sample of estriol (which contains only carbon, hydrogen, and oxygen) produced 36.86 g CO2 and 1
meriva

Answer:

C18H24O3

Explanation:

Step 1:

Data obtained from the question. This include the following:

Mass of estriol = 13.42g

Mass of CO2 = 36.86g

Mass of H2O = 10.06g

Molar mass of estriol = 288.38g/mol

Step 2:

Determination of the mass of Carbon (C), Hydrogen (H) and Oxygen (O) present in the compound. This is illustrated below:

For Carbon, C:

Molar mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 36.86 = 10.05g

For Hydrogen, H:

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 10.06 = 1.12g

For Oxygen, O:

Mass of O = 13.42 – (10.05 + 1.12) = 2.25g

Step 3:

Determination of the empirical formula for estriol. This is illustrated below:

C = 10.05g

H = 1.12g

O = 2.25g

Divide by their molar mass

C = 10.05/12 = 0.8375

H = 1.12/1 = 1.12

O = 2.25/16 = 0.1406

Divide by the smallest i.e 0.1406

C = 0.8375/0.1406 = 6

H = 1.12/0.1406 = 8

O = 0.1406/0.1406 = 1

Therefore, the empirical formula for estriol is C6H8O

Step 4:

Determination of the molecular formula for estriol. This is illustrated below:

Molecular formula is simply a multiple of the empirical formula i.e

Molecular formula => [C6H8O]n

[C6H8O]n = 288.38g/mol

[(12x6) + (8x1) + 16]n = 288.38

[72 + 8 + 16]n = 288.38

96n = 288.38

Divide both side by 96

n = 288.38/96 = 3

Molecular formula => [C6H8O]n

=> [C6H8O]n

=> [C6H8O]3

=> C18H24O3

Therefore, the molecular formula for estriol is C18H24O3

4 0
3 years ago
Complete combustion of 8.60 g of a hydrocarbon produced 26.5 g of CO2 and 12.2 g of H2O. What is the empirical formula for the h
amid [387]

The empirical formula of the hydrocarbon is C_2H_3 if combustion of 8.60 g of a hydrocarbon produced 26.5 g of CO_2 and 12.2 g of H_2O.

<h3>What is an empirical formula?</h3>

A chemical formula showing the simplest ratio of elements in a compound rather than the total number of atoms in the molecule CH_2O is the empirical formula for glucose.

1 mole of carbon dioxide contains a mass of 44 g, out of which 12 g are carbon.

Hence, in this case the mass of carbon in 8.46 g of CO_2:

(\frac{12}{44}) × 8.46 = 2.3073 g

1 mole of water contains 18 g, out of which 2 g is hydrogen;

Therefore, 2.6 g of water contains;

(\frac{2}{18} × 2.6 = 0.2889 g of hydrogen.

Therefore, with the amount of carbon and hydrogen from the hydrocarbon, we can calculate the empirical formula.

We first calculate the number of moles of each,

Carbon = \frac{2.3073}{12}  = 0.1923 moles

Hydrogen = \frac{0.2889}{1}= 0.2889 moles

Then, we calculate the ratio of Carbon to hydrogen by dividing by the smallest number value;

            Carbon : Hydrogen

               \frac{0.1923}{0.1923} : \frac{0.2889}{0.1923}

                      1 :  1.5

                     (1 : 1.5) 2

                    = 2 : 3

Hence, the empirical formula of the hydrocarbon is C_2H_3.

Learn  more about the empirical formula here:

brainly.com/question/14044066

#SPJ1

8 0
2 years ago
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