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3 years ago

In what way does hemoglobin act as a buffer against changes in blood ph?

Chemistry

2 answers:

Nonamiya [84]3 years ago

6 0

Answer:

Excess acids in the red blood cells are removed by the hemoglobin

Explanation:

In what way does hemoglobin act as a buffer against changes in blood ph?

Excess acids in the red blood cells are removed by the hemoglobin . it is a good receptor of excess proton released by carbonic acid.

It helps to remove acids before it affects the PH of the blood. deoxygenated hemoglobin are better receptor of proton than the oxygenated one.

If hemoglobin does not act as a buffer, there will be changes in the blood's PH and can poison the blood.

Vesnalui [34]3 years ago

3 0

Answer:

Explanation:

Hemoglobin removes excess protons from the red blood cells so that they can be excreted through the kidneys.

Hemoglobin binds some of the excess protons released by carbonic acid.

Subsequent binding of oxygen is drastically reduced after the first one is bound.

Hemoglobin produces protons or hydroxide ions as needed to alter the blood pH.

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How many molecules are in the substance formula of 2C6H1206 (use coefficients)

FrozenT [24]

<u>Answer:</u> The number of formula units present in 2 moles of $C_6H_{12}O_6$ are $1.2044\times 10^{24}$

<u>Explanation:</u>

Formula units are defined as the number of molecules or atoms present in 1 mole of a compound or element respectively.

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of formula units

Here, 2 represents the number of moles of $C_6H_{12}O_6$

We are given:

Moles of $C_6H_{12}O_6$ (glucose) = 2 moles

Number of formula units of $C_6H_{12}O_6=(2\times 6.022\times 10^{23})=1.2044\times 10^{24}$

Hence, the number of formula units present in 2 moles of $C_6H_{12}O_6$ are $1.2044\times 10^{24}$

6 0

3 years ago

If 0.40 mol of H2 and 0.15 mol of O2 were to react as completely as possible to produce H2O what mass of reactant would remain?

liberstina [14]

Answer : The mass of reactant $H_2$ remain would be, 0.20 grams.

Solution : Given,

Moles of $H_2$ = 0.40 mol

Moles of $O_2$ = 0.15 mol

Molar mass of $H_2$ = 2 g/mole

First we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2H_2+O_2\rightarrow 2H_2O$

From the balanced reaction we conclude that

As, 1 mole of $O_2$ react with 2 mole of $H_2$

So, 0.15 moles of $O_2$ react with $0.15\times 2=0.30$ moles of $H_2$

From this we conclude that, $H_2$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

The moles of reactant $H_2$ remain = 0.40 - 0.30 = 0.10 mole

Now we have to calculate the mass of reactant $H_2$ remain.

$\text{ Mass of }H_2=\text{ Moles of }H_2\times \text{ Molar mass of }H_2$

$\text{ Mass of }H_2=(0.10moles)\times (2g/mole)=0.20g$

Therefore, the mass of reactant $H_2$ remain would be, 0.20 grams.

3 0

3 years ago

I don’t wanna fail if you know the answer pls help me :(

stealth61 [152]

Answer:

liquid and gas........................

7 0

3 years ago

What masses of KCl and H2O are needed to make 250 g of 5.00 % solution?

oksano4ka [1.4K]

Answer:

w/w means weight/weight

Therefore, in a 250 g of solution, 5% needs to be KCl,

i.e. 5% x 250 = 12.5 g of KCl is needed.

7 0

3 years ago

How many mass of chlorine molecule are found in 5 x 10^22 molecules of chlorine gas?

kompoz [17]

Answer;
Mass in gram/molecular mass=no.of molecule of Cl2 gas/Avogadro’s number(NA) - - - - - - - - -(1)
Here,
Molecular mass of cl2=35.5*2=71
No.of cl2 molecule=5*10^22
Avogadro’s number(NA)=6.032*10^23
Now,substituting these all value in equation one,we get,
Mass in gram =(no.of molecule)/NA)*molecular mass
Or,Mass in gram =(5*10^22/6.023*10^23)*71
Or,Mass in gram = 5.89gm.Ans

8 0

3 years ago