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3 years ago

What is the protein that helps us identify the blood type

Chemistry

1 answer:

Gekata [30.6K]3 years ago

7 0

Answer:

Antigens (more info below)

Explanation:

Antigens determine the blood type and may be either proteins or sugar molecule complexes (polysaccharides). Genes in the blood group antigen group provide instructions for the production of antigen proteins. Blood group antigen proteins serve a variety of functions within the red blood cell membrane.

hope I helped!

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Consider the cell pt | h2 (1 atm); h+ (? m) || hg2cl2(s); cl− (1 m) | hg 2 h+ + 2 e − → h2 e 0 = 0.00 v hg2cl2 + 2 e − → 2 hg +

Dmitry [639]

Answer:

1.39

Explanation:

[Hg2Cl2]= 1M

[H^+] = ????

E°cell= 0.35V

E= 0.268 V

Therefore E for the reaction must -0.082 V

n= 2 moles of electrons

From Nernst Equation:

E= E°cell- 0.0592/n log [Red]/[Ox]

0.0268= 0.35- 0.0592/2 log 1/[Ox]^2

-0.082= -0.0296 log 1/[Ox]^2

log 1/[Ox]^2= 0.082/0.0296

log 1/[Ox]^2= 2.77

1/[Ox]^2=Antilog (2.77)

[Ox]^2=1.698×10^-3

[Ox] = 0.0412 M

But pH= -log [H^+]= -log(0.0412)= 1.385

3 0

3 years ago

Read 2 more answers

Sometimes more than one egg is released at a time. What do you think would happen if two eggs were released and both were fertil

podryga [215]

Both eggs would be born but they would not be twins because twins are produced by one egg

7 0

3 years ago

I need the Formulas for:

andrey2020 [161]

Explanation:

mass = desity × volume

volume = mass/density...

i believe thats right...

btw u means Physics formulae only..lol

3 0

4 years ago

The stopcock connecting a 3.06 L bulb containing methane gas at a pressure of 9.61 atm, and a 6.65 L bulb containing oxygen gas

Contact [7]

Answer : The final pressure in the system is 4.22 atm.

Explanation :

First we have to calculate the moles of methane.

$PV=n_1RT$

where,

P = pressure of gas = 9.61 atm

V = volume of gas = 3.06 L

T = temperature of gas = T

$n_1$ = number of moles of methane gas = ?

R = gas constant

Now put all the given values in the ideal gas equation, we get:

$(9.61atm)\times (3.06L)=n_1\times RT$

$n_1=\frac{29.4}{RT}$

Now we have to calculate the moles of oxygen gas.

$PV=n_2RT$

where,

P = pressure of gas = 1.75 atm

V = volume of gas = 6.65 L

T = temperature of gas = T

$n_2$ = number of moles of oxygen gas = ?

R = gas constant

Now put all the given values in the ideal gas equation, we get:

$(1.75atm)\times (6.65L)=n_2\times RT$

$n_2=\frac{11.6}{RT}$

Now we have to determine the final pressure in the system after mixing the gases.

$P_{total}=(n_1+n_2)\times \frac{RT}{V_{total}}$

where,

$P_{total}$ = final pressure of gas = ?

$V_{total}$ = final volume of gas = (3.06 + 6.65)L = 9.71 L

T = temperature of gas = T

R = gas constant

Now put all the given values in the ideal gas equation, we get:

$P_{total}=(\frac{29.4}{RT}+\frac{11.6}{RT})\times \frac{RT}{9.71L}$

$P_{total}=4.22atm$

Therefore, the final pressure in the system is 4.22 atm.

4 0

3 years ago

What volume in mt, of 0.5a M1HCI solution is needed to neutralize 77 ml of 1.54 M NaOH solution?

Rainbow [258]

Answer:

237.2 mL.

Explanation:

We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.

(XMV) acid = (XMV) base.

where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.

M is the molarity of the acid or base.

V is the volume of the acid or base.

(XMV) HCl = (XMV) NaOH.

For HCl; X = 1, M = 0.5 M, V = ??? mL.

For NaOH, X = 1, M = 1.54 M, V = 77.0 mL.

∴ V of HCl = (XMV) NaOH / (XV) HCl = (1)(1.54 M)(77.0 mL) / (1)(0.5 M) = 237.2 mL.

8 0

3 years ago