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3 years ago

Where could a virologist find hemagglutinin?

1 answer:

vladimir2022 [97]3 years ago

6 0

A virologist could find hemagglutinin, "Hemagglutinin (HA) is a protein located at the virus most external layer, the envelope. It recognizes one sugar of our cell membrane, sialic acid, which it is responsible for the virus recognition and binding to our respiratory system cells."

I hope this helps!

- <em>DaMomsMom</em>

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3 years ago

Be sure to answer all parts.

Aleks04 [339]

Answer:

pH = 3.215

Explanation:

From the given information;

Using the equation for the dilution of a stock solution:

Since moles of C5H5N = moles of HNO3

Then:

$M_{C_5H_5N}\times V_{C_5H_5N}= M_{HNO_3}\times V_{HNO_3}$

$0.0750 M \times 2.65 L = 0.407 M \times V_{HNO_3}$

$V_{HNO_3}= \dfrac{0.0750 M \times 2.65 L}{0.407 M}$

$V_{HNO_3}=488.33 \ mL$

The reaction between C5H5NH and H2O is as follows:

$C_5H_5N^+H + H_2O$ ⇄ $C_5H_5N + H_3O^+$

$Molarity \ of \ C_5H_5N^+H = \dfrac{moles}{addition \ of\ the\ total\ volume}$

$Molarity \ of \ C_5H_5N^+H = \dfrac{0.0750 \ M \times 2.65 \ L}{2.65 \ L + 0.48833 \ L}$

$Molarity \ of \ C_5H_5N^+H = \dfrac{0.19875 \ ML}{3.13833 L }$

$= \ 0.06333\ M$

Now, the next step is to draw out the I.C.E table.

$C_5H_5N^+H + H_2O$ ⇄ $C_5H_5N + H_3O^+$

I 0.06333 0 0

C - x x x

E 0.06333 -x x x

$K_a = \dfrac{10^{-14}}{1.7 \times 10^{-19}} \\ \\ K_a = 5.8824 \times 10^{-6}$

$K_a = \dfrac{[C_5H_5N][H_3O^+] }{[C_5H_5N^+H] } \\ \\ 5.8824 \times 10^{-6} = \dfrac{x^2}{0.06333 - x}$

Assuming x < 0.06333

$x^2 = 5.8824 \times 10^{-6} \times 0.06333$

Then

$x^2 = 3.72532392 \times 10^{-7} \\ \\ x= \sqrt{3.72532392 \times 10^{-7}} \\ \\ x = 6.1035 \times 10^{-4} \ M$

$[H_3O^+] = x = 6.1035 \times 10^{-4} \ M \\ \\ pH = -log (6.1035 \times 10^{-4}) \\ \\ \mathbf{\\ pH = 3.215}$

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3 years ago

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Tropopause is the boundary between the troposphere and stratosphere.

Explanation:

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3 years ago

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Where could a virologist find hemagglutinin?​

Where could a virologist find hemagglutinin?