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3 years ago

G what is the difference between the sidechains of leucine and isoleucine? select one:

Chemistry

1 answer:

statuscvo [17]3 years ago

4 0

c. Isoleucine has a carbon “branched” closer to the alpha carbon than does leucine.

The structure of leucine is CH3CH(CH3)CH2CH(NH2)COOH.

The structure of isoleucine is CH3CH2CH(CH3)CH(NH2)COOH.

In leucine, the CH3 group is two carbons away from the α carbon; in isoleucine, the CH3 group is on the carbon next to the α carbon.

Thus, isoleucine has the closer branched carbon.

“One is charged, the other is not” is incorrect. Both compounds are uncharged.

“One has more H-bond acceptors than the other” is incorrect. Each acid has two H-bond acceptors — the N in the amino and the O in the carbonyl group.

“They have different numbers of carbon atoms” is incorrect. They each contain six carbon atoms.

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The variable n represents

Vika [28.1K]

C is the correct answer

8 0

2 years ago

A beaker with 1.60×102 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and c

stich3 [128]

Answer:

The pH will change 0.16 ( from 5.00 to 4.84)

Explanation:

Step 1: Data given

volume of acetic acid buffer = 160 mL

The total molarity of acid and conjugate base in this buffer is 0.100 M

A student adds 7.10 mL of a 0.460 M HCl solution to the beaker.

The pKa of acetic acid is 4.740

pH = 5.00

Step 2: Calculate concentration of acid

Consider x = concentration acid

Consider y = concentration conjugate base

x + y = 0.100

5.00 = 4.740 + log y/x

5.00 - 4.740 = log y/x

0.26 = log y/x

10^0.26 =1.82 = y/x

1.82 x = y

Since x+y = 0.100

x + 1.82 x = 0.100

2.82 x = 0.100

x =0.0355 M = concentration acid

Step 3: Calculate concentration of conjugate base

y = 0.100 - x

0.100 - 0.0355 =0.0645 M= concentration conjugate base

Step 4: Calculate moles of acid

Moles = volume * molarity

moles acid = 0.160 L * 0.0355 M= 0.00568 moles

Step 5: Calculate moles of conjugate base

moles conjugate base = 0.0645 M * 0.160 L=0.01032 moles

Step 6: Calculate moles HCl

moles HCl = 7.10 * 10^-3 L * 0.460 M=0.003266 moles

Step 7: Calculate new moles

A- + H+ = HA

moles conjugate base = 0.01032 - 0.003266 =0.007054 moles

moles acid = 0.00568 + 0.003266=0.008946 moles

Step 8: Calculate the total volume

total volume = 160 + 7.10 = 167.1 mL = 0.1671 L

Step 9: Calculate the concentration of the acid

concentration acid = 0.008946/ 0.1671 =0.0535 M

Step 10: Calculate the concentration of conjugate base

concentration conjugate base = 0.007054/ 0.1671 =0.0422 M

Step 11: Calculate the pH

pH = 4.740 + log 0.0535/ 0.0422=4.84

change pH = 5.00 - 4.84=0.16

The pH will change 0.16

5 0

2 years ago

A 500.0-mL buffer solution is 0.100 M in HNO2 and 0.150 M in KNO2. Determine whether each addition would exceed the capacity of

Leviafan [203]

Answer:

None of the additions will exceed the capacity of the buffer.

Explanation:

As we know a buffer has the ability to resist pH changes when small amounts of strong acid or base are added.

The pH of the buffer is given by the Henderson-Hasselbach equation:

pH = pKa + log [A⁻] / [HA]

where A⁻ is the conjugate base of the weak acid HA.

Now we can see that what is important is the ratio [A⁻] / [HA] to resist a pH change brought about by the addition of acid or base.

It follows then that once we have consumed by neutralization reaction either the acid or conjugate base in the buffer, this will lose its ability to act as such and the pH will increase or decrease dramatically by any added acid or base.

Therefore to solve this question we must determine the number of moles of acid HNO₂ and NO₂⁻ we have in the buffer and compare it with the added acid or base to see if it will deplete one of these species.

Volume buffer = 500.0 mL = 0.5 L

# mol HNO₂ = 0.5 L x 0.100 mol/L = 0.05 mol HNO₂

# mol NO₂⁻ = 0.5 L x 0.150 mol/L = 0.075 mol NO₂⁻

a. If we add 250 mg NaOH (0.250 g)

molar mass NaOH =40 g/mol

# mol NaOH =0.250 g/ 40g/mol = 0.0063 mol

0.0063 mol NaOH will be neutralized by 0.0063 mol HNO₂ and we have plenty of it, so it would not exceed the capacity of the buffer.

b. If we add 350 mg KOH (0.350 g)

molar mass KOH =56.10 g

# mol KOH = 0.350 g/56.10 g/mol = 0.0062 mol

Again the capacity of the buffer will not be exceeded since we have 0.05 mol HNO₂ in the buffer.

c. If we add 1.25 g HBr

molar mass HBr = 80.91 g/mol

# mol HBr = 1.25 g / 80.91 g/mol = 0.015 mol

0.015 mol Hbr will neutralize 0.015 mol NO₂⁻ and we have to start with 0.075 mol in the buffer, therefore the capacity will not be exceeded.

d. If we add 1.35 g HI

molar mass HI = 127.91 g/mol

# mol HI = 1.35 g / 127.91 g/mol = 0.011 mol

Again the capacity of the buffer will not be exceed since we have plenty of it in the buffer after the neutralization reaction.

7 0

2 years ago

How does a sample of hydrogen at 10 °C compare to a sample of hydrogen at 350 K?

Aleks04 [339]

Answer: -

The hydrogen at 10 °C has slower-moving molecules than the sample at 350 K.

Explanation: -

Temperature of the hydrogen gas first sample = 10 °C.

Temperature in kelvin scale of the first sample = 10 + 273 = 283 K

For the second sample, the temperature is 350 K.

Thus we see the second sample of the hydrogen gas more temperature than the first sample.

We know from the kinetic theory of gases that

The kinetic energy of gas molecules increases with the increase in temperature of the gas. The speed of the movement of gas molecules also increase with the increase in kinetic energy.

So higher the temperature of a gas, more is the kinetic energy and more is the movement speed of the gas molecules.

Thus the hydrogen at 10 °C has slower-moving molecules than the sample at 350 K.

8 0

2 years ago

Read 2 more answers

Consider the molecules H2O, H2S, H2Se, and H2Te. Which do you expect to have the highest boiling point, and why?

STatiana [176]

Answer:

a. H20,because it experiences hydrogen bonding.

5 0

2 years ago