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bagirrra123 [75]
3 years ago
7

How many secondary (2°) carbons are found in 5-ethyl-3,3,4-trimethylheptane??

Chemistry
2 answers:
Gelneren [198K]3 years ago
6 0
5-Ethyl-3,3,4-trimethylheptane has 3 secondary carbon. A primary carbon written as 1° is a carbon that has one carbon atom attached to it. A secondary carbon written as 2° is a carbon attached to two other carbons, while a tertiary carbon written as 3° is a carbon attached to three other carbons. 
Kipish [7]3 years ago
3 0

Answer:

3 secondary carbons are found in 5-ethyl-3,3,4-trimethylheptane.

Explanation:

5-ethyl-3,3,4-trimethylheptane's structure can be seen in the attached picture, in this structure, each vertex represents one carbon atom.

<em>A secondary carbon</em> is a carbon that is bonded to only 2 other carbon atoms, each secondary carbon is highlighted in blue in the attached picture.

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The molar mass of barium nitrate (ba(no3)2) is 261.35 g/mol. what is the mass of 5.30 × 1022 formula units of ba(no3)2? 0.0900 g
liberstina [14]

Molar mass is the ratio of the mass to that amount of the substance. The mass of the barium nitrate in the formula unit is 23.0 grams.

<h3>What is mass?</h3>

The mass of a substance is the product of the molar mass of the compound and the number of moles of the compound.

Given,

Molar mass of barium nitrate = 261.35 g/mol

If, 6.022 \times 10^{23} have a mass of 261.35 g/mol then, 5.30 \times 10^{22} formula units will have a mass of,

\begin{aligned}& = \dfrac{261.35 \times 5.30 \times 10^{22}}{6.022\times 10^{23}}\\\\&= 23.0\;\rm gm\end{aligned}

Therefore, option C. 23.0 gm is the mass of barium nitrate.

Learn more about mass here:

brainly.com/question/24958554

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A 25.00-mL aliquot of a nitric acid solution of unknown concentration is pipetted into a 125-mL Erlenmeyer flask and 2 drops of
Semenov [28]

Answer:

0.0611M of HNO3

Explanation:

<em>The concentration of the NaOH solution must be 0.1198M</em>

<em />

The reaction of NaOH with HNO3 is:

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<em>1 mole of NaOH reacts per mole of HNO3.</em>

That means the moles of NaOH used in the titration are equal to moles of HNO3.

<em>Moles HNO3:</em>

12.75mL = 0.01275L * (0.1198mol / L) = 0.0015274 moles NaOH = Moles HNO3.

In 25.00mL = 0.025L -The volume of the aliquot-:

0.00153 moles HNO3 / 0.025L =

<h3> 0.0611M of HNO3</h3>
7 0
3 years ago
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