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Brilliant_brown [7]
3 years ago
5

Provide the IUPAC names for

Chemistry
1 answer:
ollegr [7]3 years ago
6 0

<u>Answer:</u>

<u>For a:</u> The IUPAC name of the compound is N-ethylethaneamide.

<u>For b:</u> The IUPAC name of the compound is N,N-diethylmethaneamide.

<u>For c:</u> The IUPAC name of the compound is ethyl pentanoate

<u>Explanation:</u>

To name a compound, first look for the longest possible carbon chain.

  • <u>For a:</u>

Amide group is a type of functional group where an amine group is attached to a carbonyl group. The general formula of amide is R-CO-NH_2, where R is an alkyl or aryl group.

In part (a), the alkyl group has 2 carbon atoms and thus, the prefix used is 'eth-'

Also, an ethyl substituent is directly attached to N-atom. It is an alkane structured hydrocarbon thus, the suffix used will be '-ane'

Hence, the IUPAC name of the compound is N-ethylethaneamide.

  • <u>For b:</u>

Amide group is a type of functional group where an amine group is attached to a carbonyl group. The general formula of amide is R-CO-NH_2, where R is an alkyl or aryl group.

In part (b), the alkyl group has 1 carbon atoms and thus, the prefix used is 'meth-'

Also, two ethyl substituents are directly attached to N-atom. It is an alkane structured hydrocarbon thus, the suffix used will be '-ane'

Hence, the IUPAC name of the compound is N,N-diethylmethaneamide.

  • <u>For c:</u>

Esters are a kind of organic molecules having functional groups, R-COO-R' where R and R' are the alkyl or aryl groups. They are formed by the combination of alcohol and carboxylic acid.

These functional group compounds are named in two words which is alkyl alkanoates, where alkyl refers to the alcoholic part and alkanoate refers to the carboxylic acid part of the molecule. The numbering of the parent chain in esters is done from the carboxylic carbon. The alkyl part is not given any numbers.

In part (c), there are 5 carbon atoms present in a straight chain and thus, the prefix used is 'pent-'

Also, an ethyl group forms the alcoholic part.

Hence, the IUPAC name of the compound is ethyl pentanoate

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RideAnS [48]

The NaOH will be used What titrant  to titrate the 0. 02 m hcl phenol red solution.

Acid-base titrations may be the most typical titrations, although there are numerous more forms as well. Take a look at this illustration where sodium hydroxide is used to titrate a sample of hydrochloric acid (HCl) (NaOH). The titrant (NaOH), which is added gradually throughout the duration of the titration, has been added to the unknown solution.

Titrants are solutions with known concentrations that are added to solutions whose concentrations must be determined. The solution for whom the concentration needs to be determined is known as a titrant as well as analyte.

Therefore, the NaOH will be used as a titrant to titrate the 0. 02 m hcl phenol red solution.

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Balanced chemical equation: BaCl2(aq)+Na2SO4(aq)⟶BaSO4(s)+2NaCl(aq)
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Answer:

Part 1)  85.3 grams NaCl

Part 2)  8.79 x 10²³ formula units NaCl

Explanation:

<u>(Part 1)</u>

To find the mass of NaCl, you need to multiply the given value (1.46 moles) by the molar mass of NaCl. This measurement is the atomic masses of the elements times each of their quantities combined. In this case, there is only one mole of each element in the molecule. Moles should be located in the denominator of the conversion to allow for the cancellation of units. The final answer should have 3 sig figs to reflect the given value.

Molar Mass (NaCl): 22.99 g/mol + 35.45 g/mol
Molar Mass (NaCl): 58.44 g/mol

1.46 moles NaCl            58.44 g
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<u>(Part 2)</u>

I do not know which other question the second part is referring to, so I will just use the moles given in the first part. To find the formula units, you need to multiply the given value (1.46 moles NaCl) by Avogadro's Number. This conversion represents the number of formula units found in 1 mole of the sample. The moles should be in the denominator of the conversion to allow for the cancellation of units.

Avogadro's Number:

1 mole = 6.022 x 10²³ formula units

1.46 moles NaCl         6.022 x 10²³ units
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Vera_Pavlovna [14]

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The concentration after dilution is 1.4%.

We are aware that concentration and volume are related to each other by the formula -

C_{1} V_{1} = C_{2} V_{2}, where we have initial concentration and volume on Left Hand Side and final concentration and volume on Right Hand Side.

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Hence, the final concentration is 1.4%.

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The complete question is -

A 53.5 mL sample of an 5.4 % (m/v) KBr solution is diluted with water so that the final volume is 205.0 mL.

Calculate the final concentration and express your answer to two significant figures and include the appropriate units.

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