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Dovator [93]
1 year ago
9

stearic acid and linoleic acid stearic acid and linoleic acid stearic acid has the higher melting point, because it has two more

methylene groups (giving it a greater surface area and therefore more intermolecular van der waals interactions) than linoleic acid.
Chemistry
1 answer:
elena55 [62]1 year ago
4 0

Acid palmitic acid has higher melting point, because it has two more methylene groups.

\text { Myristic acid } \rightarrow \mathrm{CH}_3\left(\mathrm{CH}_2\right)_{12} \mathrm{COOH} \\& \text { Palmitic acid } \rightarrow \mathrm{CH}_3\left(\mathrm{CH}_2\right)_{14} \mathrm{COOH} \\

Acid palmitic acid has higher melting point, because it has two more methylene groups.

Giving it a greater surface area and therefore more intermolecular  van der waals interact than the myristic acid.

stearic arid \mathrm{CH}_3\left(\mathrm{CH}_2\right)_{16} \mathrm{COOH}

linoleic acid \mathrm{C}_{18} \mathrm{H}_{32} \mathrm{O}_2 (two double bond)

Stearic acid has higher Melting point, because it does not have any Carbon-Carbon double bonds, whereas linoleic acid has two cis double bonds which prevent the molecules from packing closely together.

Oleic Acid and Linoleic acid.

\mathrm{C}_{18} \mathrm{H}_{34} \mathrm{O}_2-one double bond (cis)

Acid palmitic acid has higher melting point, because it has two more methylene groups.

For more such question on methylene group.

brainly.com/question/4279223

#SPJ4

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1. A container holds four times as many moles of CO2 as O2. If the total pressure in the container is 20. atm,
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Answer:

Partial pressure of CO2 = 16 atm

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Partial pressure of CO2 = (4/5) × 20

Partial pressure of CO2 = 16 atm

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Convert each of the measurements given to the new unit stated using the factor label method (dimensional analysis). Show all wor
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Be sure to answer all parts. Consider the formation of ammonia in two experiments. (a) To a 1.00−L container at 727°C, 1.30 mol
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<u>Answer:</u> The value of K_c for 2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g) reaction is 5.13\times 10^2

<u>Explanation:</u>

We are given:

Initial moles of nitrogen gas = 1.30 moles

Initial moles of hydrogen gas = 1.65 moles

Equilibrium moles of ammonia = 0.100 moles

Volume of the container = 1.00 L

For the given chemical equation:

                N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)

<u>Initial:</u>            1.30       1.65

<u>At eqllm:</u>       1.30-x    1.65-3x             2x

Evaluating the value of 'x'

\Rightarrow 2x=0.100\\\\\Rightarrow x=0.050mol

The expression of K_c for above equation follows:

K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}

Equilibrium moles of nitrogen gas = (1.30-x)=(1.30-0.05)=1.25mol

Equilibrium moles of hydrogen gas = (1.65-x)=(1.65-0.05)=1.60mol

Putting values in above expression, we get:

K_c=\frac{(0.100)^2}{1.25\times (1.60)^3}\\\\K_c=1.95\times 10^{-3}

Calculating the K_c' for the given chemical equation:

2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)

K_c'=\frac{1}{K_c}\\\\K_c'=\frac{1}{1.95\times 10^{-3}}=5.13\times 10^2

Hence, the value of K_c for 2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g) reaction is 5.13\times 10^2

8 0
3 years ago
Read 2 more answers
Five students performed a Kjeldahl nitrogen analysis of a protein sample. The following weight % nitrogen values were determined
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Answer:

G_calculated = 1.756

The outlier should be rejected, as G_cal > G_tab (= 1.463) at 95 % confidence.

Explanation:

The Grubb's test is used for identifying an outlier in data, which is from the same population. For this, a statistical term, G, is calculated for the suspected outlier. If the calculated value is greater than the tabulated G value then the suspected value is rejected. This term is given as,

G_calculated = | suspect value - mean| / s

Here,  suspect value is 13.8, mean is to be taken of all the data (including suspected value). s is the standard deviation of the sample data.

s is calculated from the following formula:

s = (Σ(xi - x)²/(N-1))^1/2

Here, x is the mean, which is 15.24, xi is individual value and N is the total number of data (5).

From the above formula, s is found to be

Standard Deviation, s = 0.820

Now for G value,

G_calculated = | 13.8 - 15.24| / (0.820)

G_ calculated = 1.756

The tabulated G value at 95 % confidence and N -1 (5 - 1 = 4) degree of freedom is, 1.463.

As calculated G (1.756) is greater than the tabulated G (1.463), the value 13.8 is considered an outlier at 95 % confidence.  

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