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3 years ago

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Alpha helices are a type of secondary structure in proteins. What is the length of a 27.0 kDa single‑stranded α‑helical protein

segment? Assume a mean residue mass of 110 Da .

1 answer:

iogann1982 [59]3 years ago

8 0

Answer : The length of protein will be, 36.8 A⁰

Explanation :

First we have to calculate the amino acid residue.

$\text{Amino acid residue}=\frac{\text{Length of single strand of protein}}{\text{Mean residue mass}}$

Given:

Length of single strand of protein = 27.0 kDa

Mean residue mass = 110 Da

$\text{Amino acid residue}=\frac{27.0kDa}{110Da}=\frac{27000Da}{110Da}=245.4$

Now we have to calculate the number of turns in protein.

$\text{Number of turns in protein}=\frac{\text{Amino acid residue}}{3.6\text{amino acid per turn of the alpha-helix}}$

As there are 3.6 amino acid per turn of the alpha-helix.

$\text{Number of turns in protein}=\frac{245.4}{3.6}=68.17$

Now we have to calculate the length of the protein.

$\text{Length of protein}=\text{Number of turns in protein}\times \text{Length of each turn}$

As, we know that the length of each turn = 0.54 A⁰

$\text{Length of protein}=68.17turns\times 0.54\AA=36.8\AA$

Thus, the length of protein will be, 36.8 A⁰

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Calculate the solubility at 25°C of CuBr in pure water and in a 0.0120M CoBr2 solution. You'll find Ksp data in the ALEKS Data t

iragen [17]

Answer:

S = 7.9 × 10⁻⁵ M

S' = 2.6 × 10⁻⁷ M

Explanation:

To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.

CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)

I 0 0

C +S +S

E S S

The solubility product (Ksp) is:

Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²

S = 7.9 × 10⁻⁵ M

<u>Solubility in 0.0120 M CoBr₂ (S')</u>

First, we will consider the ionization of CoBr₂, a strong electrolyte.

CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)

1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.

Then,

CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)

I 0 0.0240

C +S' +S'

E S' 0.0240 + S'

Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')

In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.

S' = 2.6 × 10⁻⁷ M

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2 years ago

In each row check off the boxes that apply to the highlighted reactant. reaction The highlighted reactant acts as a... (check al

tekilochka [14]

The given question is incomplete. The complete question is :

In each row check off the boxes that apply to the underlined reactant. The underlined reactant acts as a... (check all that apply)

1. $HCH_3CO_2(aq)+NH_3(aq)\rightarrow CH_3COO^-(aq)+NH_4^+(aq)$

here underlined is $HCH_3CO_2$

A. Brønsted-Lowry acid

B. Brønsted-Lowry base

C. Lewis acid

D. Lewis base

2. $BH_3(aq)+NH_3(aq)\rightarrow BH_3NH_3(aq)$

Here underlined is $NH_3$

A. Brønsted-Lowry acid

B. Brønsted-Lowry base

C. Lewis acid

D. Lewis base

3. $HNO_2(aq)+C_2H_5NH_2(aq)\rightarrow NO_2^-(aq) + C_2H_5NH_3^+(aq)$

Here underlined is $C_2H_5NH_2$

A. Brønsted-Lowry acid

B. Brønsted-Lowry base

C. Lewis acid

D. Lewis base

Answer: 1. Brønsted-Lowry acid

2. Lewis base

3. Brønsted-Lowry base

Explanation:

According to the Bronsted Lowry conjugate acid-base theory, an acid is defined as a substance which donates protons and a base is defined as a substance which accepts protons.

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

1. $HCH_3CO_2(aq)+NH_3(aq)\rightarrow CH_3CO^{2-}(aq)+NH_4^+aq)$

As $HCH_3CO_2(aq)$ is donating a proton , it acts as a bronsted acid.

2. $BH_3(aq)+NH_3(aq)\rightarrow BH_3NH_3(aq)$

As $NH_3$ contains a lone pair of electron on nitrogen , it can easily donate electrons to $BH_3$ and act as lewi base.

3. $HNO_2(aq)+C_2H_5NH_2(aq)\rightarrow NO_2^-(aq) + C_2H_5NH_3^+(aq)$

As $C_2H_5NH_2(aq)$ is accepting a proton , it acts as a bronsted base.

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3 years ago

What is the melting point of water?

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11111nata11111 [884]

Answer: True

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2 years ago

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"Compounds A and B react to give a single product, C. Write the rate law for each of the following cases and determine the units

olasank [31]

Answer:

Part a: <em>Units of k is </em> $M^{-2}s^{-1}$ <em> where reaction is first order in A and second order in B</em>

Part b: <em>Units of k is </em> $M^{-1}s^{-1}$ <em> where reaction is first order in A and second order overall.</em>

Part c: <em>Units of k is </em> $M^{-1}s^{-1}$ <em> where reaction is independent of the concentration of A and second order overall.</em>

Part d: <em>Units of k is </em> $M^{-3}s^{-1}$ <em> where reaction reaction is second order in both A and B.</em>

Explanation:

As the reaction is given as

$A+B \rightarrow C$

where as the rate is given as

$r=k[A]^x[B]^y$

where x is the order wrt A and y is the order wrt B.

Part a:

x=1 and y=2 now the reaction rate equation is given as

$r=k[A]^1[B]^2$

Now the units are given as

$r=k[A]^1[B]^2\\M/s =k[M]^1[M]^2\\M/s =k[M]^{1+2}\\M/s =k[M]^{3}\\M^{1-3}/s =k\\M^{-2}s^{-1} =k$

The units of k is $M^{-2}s^{-1}$

Part b:

x=1 and o=2

x+y=o

1+y=2

y=2-1

y=1

Now the reaction rate equation is given as

$r=k[A]^1[B]^1$

Now the units are given as

$r=k[A]^1[B]^1\\M/s =k[M]^1[M]^1\\M/s =k[M]^{1+1}\\M/s =k[M]^{2}\\M^{1-2}/s =k\\M^{-1}s^{-1} =k$

The units of k is $M^{-1}s^{-1}$

Part c:

x=0 and o=2

x+y=o

0+y=2

y=2

y=2

Now the reaction rate equation is given as

$r=k[A]^0[B]^2$

Now the units are given as

$r=k[B]^2\\M/s =k[M]^2\\M/s =k[M]^{2}\\M^{1-2}/s =k\\M^{-1}s^{-1} =k$

The units of k is $M^{-1}s^{-1}$

Part d:

x=2 and y=2

Now the reaction rate equation is given as

$r=k[A]^2[B]^2$

Now the units are given as

$r=k[A]^2[B]^2\\M/s =k[M]^2[M]^2\\M/s =k[M]^{2+2}\\M/s =k[M]^{4}\\M^{1-4}/s =k\\M^{-3}s^{-1} =k$

The units of k is $M^{-3}s^{-1}$

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3 years ago