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4 years ago

Name the following compound:

Chemistry

2 answers:

devlian [24]4 years ago

7 0

Answer : The correct option is, 2-ethylpentane

Explanation :

The IUPAC nomenclature of alkanes are :

First select the longest possible carbon chain.
For the number of carbon atom, we add prefix as 'meth' for 1, 'eth' for 2, 'prop' for 3, 'but' for 4, 'pent' for 5, 'hex' for 6, 'sept' for 7, 'oct' for 8, 'nona' for 9 and 'deca' for 10.
A suffix '-ane' is added at the end of the name.
The numbering is done in such a way that first carbon of double bond gets the lowest number.
If two of more similar alkyl groups are present, then the words 'di', 'tri' 'tetra' and so on are used to specify the number of times these alkyl groups appear in the chain.

Hence, the name for given compound will be, 2-ethylpentane and the structure are shown below.

Marrrta [24]4 years ago

5 0

Answer:

3-methylhexane

Explanation:

The compound is 6 carbon atoms long, a hexane. It branches with a methyl group on the third carbon group.

This makes it a 3-methylhexane

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Ionic bonds form between which two types of elements?

Alla [95]

Answer:

Ionic bonds usually occur between metal and nonmetal ions. For example, sodium (Na), a metal, and chloride (Cl), a nonmetal, form an ionic bond to make NaCl.

Explanation:

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3 years ago

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How many grams of chromium are in 2.0 x 1024 atoms

Vaselesa [24]

Answer: 173 g ( 0.17 kg in right accuracy)

Explanation: Amount in moles is n = N/Na = 2.0·10^24 / 6.022·10^23 (1/mol).

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3 years ago

Why does the temperature of the water stay the same when it melts and boils?

krok68 [10]

The temperature stays the same when a solid changes to a liquid because energy is required to break the forces between particles of water therefore changing the state of matter and separating the particles away from each other.

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5 0

3 years ago

Find the volume of 56.0 grams of O2

N76 [4]

Answer:

39.2 L at STP

Explanation:

Convert the grams to moles first by dividing 56.0 by the molar mass of O2 (32.0) then convert to volume by multiplying by 22.4.

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5 0

3 years ago

A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.

irga5000 [103]

Answer:

1. pH = 1.23.

2. $H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

$H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+$

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

$pH=pKa+log(\frac{[base]}{[acid]} )$

Whereas the pKa is:

$pKa=-log(Ka)=-log(5.90x10^{-2})=1.23$

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

$pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23$

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

$n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol$

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

$H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)$

Which is also shown in net ionic notation.

Best regards!

4 0

3 years ago