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3 years ago

What is the IUPAC name of this compund?

Chemistry

2 answers:

enot [183]3 years ago

5 0

Answer:

The IUPAC name of the compound is 3-ethyl, 4-methyl, heptane.

Explanation:

IUPAC stands for International Union of Pure and Applied Chemistry.

An IUPAC name is a internationally used name of a compound with same name throughout the world. This compound is named according to the IUPAC rules with the following steps-

1. Identification of longest carbon chain.

2. Estimating the number of carbon in the longest carbon chain that is seven carbon means heptane.

3. The side chains are identified, that are branched off from the parent chain.

4. The parent chains is labelled from the side of nearest side chain.

5. The nearest side chain that is the third carbon contains ethyl as the side chain at 3 position of parent carbon. Thus, so named 3-ethyl.

6. The fourth carbon of parent chain contain a methyl group, thus named as 4-methyl.

The naming is based on the parent carbon chain and side chain naming are written first. Thus, the compound is 3-ethyl, 4-methyl, heptane.

mr_godi [17]3 years ago

4 0

Answer:

3-ethyl-4-methyl Heptane

Explanation:

longest chain is straight .

we start assigning number from where we are close to the branch to this mean its from left hand side. and we saw two chain 1 is ethyl and other is methyl . In alphabetical order e comes first as compare to m so we write ethyl first after that methyl and than in last we write longest chain name with ane (bcz we only have single vonds).

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1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.

Andrej [43]

Answer:

a. HCl.

b. 0.057 g.

c. 1.69 g.

d. 77 %.

Explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

$Mg+2HCl\rightarrow MgCl_2+H_2$

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

$n_{H_2}^{by\ HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl} =0.0284molH_2\\\\n_{H_2}^{by\ Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2$

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.

b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

$m_{H_2}=0.0284molH_2*\frac{2.02gH_2}{1molH_2}=0.057gH_2$

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

$m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg} =0.690gMg$

Thus, the mass of excess magnesium turns out:

$m_{Mg}^{excess}=2.38g-0.690g=1.69gMg$

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

$Y=\frac{0.044g}{0.057g} *100\%\\\\Y=77\%$

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2 years ago

What is the IUPAC name of this compund?​

What is the IUPAC name of this compund?