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4 years ago

8

The side chain (r group) of the amino acid serine is ch2oh. The side chain of the amino acid leucine is ch2ch(ch3)2. Where would

you expect to find these amino acids in a globular protein in aqueous solution?

1 answer:

wel4 years ago

8 0

The answer is leucine would be in the interior, and serine would be on the exterior of the globular protein.

The side chain (R group) of the amino acid serine is CH₂OH. The side chain of the amino acid leucine is CH₂CH(CH₃)₂. In globular protein, leucine found in the interior, and serine found on the exterior. The nature of side chain decides the amino acid position in the globular protein , as CH₂CH(CH₃)₂ this is hydrophobic and CH₂OH is hydrophlic.

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How can you use the periodic table to find the number of protons in one atom of an element?

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A is your answer.
On the periodic table the atomic number is the number of protons inside the nucleus.

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Which structural formula represents an unsaturated hydrocarbon?

bija089 [108]

Answer:

4

Explanation:

your chosen answer is right as unsaturated mean contains double bond and 1st and 3rd is Wrong as Carbon only can have 4 bonds and 2nd is Wrong as it is saturated.

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3 years ago

A scientist measures the standard enthalpy change for the following reaction to be -17.2 kJ : Ca(OH)2(aq) 2 HCl(aq)CaCl2(s)

elixir [45]

Answer: $\Delta H^{0}=-173.72$ kJ/mol

Explanation: <u>Enthalpy</u> <u>Change</u> is the amount of energy in a reaction - absorption or release - at a constant pressure. So, <u>Standard</u> <u>Enthalpy</u> <u>of</u> <u>Formation</u> is how much energy is necessary to form a substance.

The standard enthalpy of formation of HCl is calculated as:

$\Delta ^{0}=\Sigma H_{products}-\Sigma H_{reactants}$

$Ca(OH)_{2}_{(aq)}+2HCl_{(aq)}$ → $CaCl_{2}_{(s)}+2H_{2}O_{(l)}$

Standard Enthalpy of formation for the other compounds are:

Calcium Hydroxide: $\Delta H^{0}=$ -1002.82 kJ/mol

Calcium chloride: $\Delta H^{0}=$ -795.8 kJ/mol

Water: $\Delta H^{0}=$ -285.83 kJ/mol

Enthalpy is given per mol, which means we have to multiply by the mols in the balanced equation.

Calculating:

$-17.2=[-795.8+2(285.85)]-[-1002.82+2\Delta H]$

$-17.2=-1367.46+1002.82-2\Delta H$

$2\Delta H=17.2-364.64$

$\Delta H=-173.72$

So, the standard enthalpy of formation of HCl is -173.72 kJ/mol

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3 years ago

A 3.50 g sample of an unknown compound containing only C , H , and O combusts in an oxygen‑rich environment. When the products h

statuscvo [17]

Explanation:

First, calculate the moles of $CO_{2}$ using ideal gas equation as follows.

PV = nRT

or, n = $\frac{PV}{RT}$

= $\frac{1 atm \times 4.41 ml}{0.0821 Latm/mol K \times 293 K}$ (as 1 bar = 1 atm (approx))

= 0.183 mol

As, Density = $\frac{mass}{volume}$

Hence, mass of water will be as follows.

Density = $\frac{mass}{volume}$

0.998 g/ml = $\frac{mass}{3.26 ml}$

mass = 3.25 g

Similarly, calculate the moles of water as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{3.25 g}{18.02 g/mol}$

= 0.180 mol

Moles of hydrogen = $0.180 \times 2$ = 0.36 mol

Now, mass of carbon will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

0.183 mol = $\frac{mass}{12 g/mol}$

= 2.19 g

Therefore, mass of oxygen will be as follows.

Mass of O = mass of sample - (mass of C + mass of H)

= 3.50 g - (2.19 g + 0.36 g)

= 0.95 g

Therefore, moles of oxygen will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{0.95 g}{16 g/mol}$

= 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

C H O

No. of moles: 0.183 0.36 0.059

On dividing: 3.1 6.1 1

Therefore, empirical formula of the given compound is $C_{3}H_{6}O$ .

Thus, we can conclude that empirical formula of the given compound is $C_{3}H_{6}O$ .

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3 years ago

kykrilka [37]

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