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Elden [556K]
3 years ago
15

Protein a binds to protein b to form a complex, ab: a + b  ab a cell contains an equilibrium mix of protein a at a concentratio

n of 1 µm, protein b at a concentration of 1µm, and the complex ab also at 1 µm.
a. What is the equilibrium constant, keq, for the reaction a + b  ab?
b. What would the equilibrium constant be if a, b, and ab were each present in equilibrium at a concentration of 1 nm?
c. At this lower concentration, about how many extra hydrogen bonds would be needed to hold a and b together tightly enough to form the same proportion of the ab complex? Note: free energy change is related to the equilibrium constant by the equation ∆go = -2.3 rt log k, where r is 1.98 x 10-3 kcal/k mole) and t is 310 k.
Chemistry
1 answer:
nasty-shy [4]3 years ago
3 0

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Special Structures in Plant Cells. Most organelles are common to both animal and plant cells. However, plant cells also have features that animal cells do not have: a cell wall, a large central vacuole, and plastids such as chloroplasts.

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3 years ago
When nitrogen dioxide (NO2) from car exhaust combines with water in the air, it forms nitric acid (HNO3), which causes acid rain
yKpoI14uk [10]

<u>Answer:</u>

<u>For a:</u> The number of molecules of nitrogen dioxide is 4.52\times 10^{23}

<u>For b:</u> The mass of nitric acid formed is 54.81 grams

<u>For c:</u> The mass of nitric acid formed is 206 grams

<u>Explanation:</u>

The given chemical reaction follows:

3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq.)+NO(g)

  • <u>For a:</u>

By Stoichiometry of the reaction:

1 mole of water reacts with 3 moles of nitrogen dioxide

So, 0.250 moles of water will react with \frac{3}{1}\times 0.250=0.75mol of nitrogen dioxide

According to mole concept:

1 mole of a compound contains 6.022\times 10^{23} number of molecules.

So, 0.75 moles of nitrogen dioxide will contain 0.75\times 6.022\times 10^{23}=4.52\times 10^{23} number of molecules

Hence, the number of molecules of nitrogen dioxide is 4.52\times 10^{23}

  • <u>For b:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of nitrogen dioxide = 60.0 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

\text{Moles of nitrogen dioxide}=\frac{60.0g}{46g/mol}=1.304mol

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide produces 2 mole of nitric acid

So, 1.304 moles of nitrogen dioxide will produce = \frac{2}{3}\times 1.304=0.870 moles of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 0.870 moles

Putting values in equation 1, we get:

0.870mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(0.870mol\times 63g/mol)=54.81g

Hence, the mass of nitric acid formed is 54.81 grams

  • <u>For c:</u>
  • <u>For nitrogen dioxide:</u>

Given mass of nitrogen dioxide = 225 g

Molar mass of nitrogen dioxide = 46 g/mol

Putting values in equation 1, we get:

\text{Moles of nitrogen dioxide}=\frac{225g}{46g/mol}=4.90mol

  • <u>For water:</u>

Given mass of water = 55.2 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

\text{Moles of water}=\frac{55.2g}{18g/mol}=3.06mol

By Stoichiometry of the reaction:

3 moles of nitrogen dioxide reacts with 1 mole of water

So, 4.90 moles of nitrogen dioxide will react with = \frac{1}{3}\times 4.90=1.63mol of water

As, given amount of water is more than the required amount. So, it is considered as an excess reagent.

Thus, nitrogen dioxide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 mole of nitrogen dioxide produces 2 moles of nitric acid

So, 4.90 moles of nitrogen dioxide will produce \frac{2}{3}\times 4.90=3.27mol of nitric acid

Now, calculating the mass of nitric acid from equation 1, we get:

Molar mass of nitric acid = 63 g/mol

Moles of nitric acid = 3.27 moles

Putting values in equation 1, we get:

3.27mol=\frac{\text{Mass of nitric acid}}{63g/mol}\\\\\text{Mass of nitric acid}=(3.27mol\times 63g/mol)=206g

Hence, the mass of nitric acid formed is 206 grams

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3 years ago
State the period and group to which aluminum belongs to in the periodic table ​
FinnZ [79.3K]

Explanation:

aluminium belongs to group 13 and period 3

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A student throws a 2kg ball at 4m/s. What is the ball’s momentum?
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The answer is 8 m/s2
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3 years ago
Nitrogen dioxide reacts with water to form nitric acid and nitrogen monoxide according to the equation: 3NO2(g)+H2O(l)→2HNO3(l)+
Vesnalui [34]

Answer:

2 mol NO2

Explanation:

                                      3NO2(g)+H2O(l)→2HNO3(l)+NO(g)

from reaction                3 mol        1 mol

given                           11 mol          3 mol

for 3 mol NO2  -----   1 mol H2O

for x mol NO2  -----   3 mol H2O

3:x = 1:3

x = 3 *3/1 = 9 mol NO2

So, for 3 mol H2O are needed only 9 mol NO2.

But we have 11 mol NO2. So, NO2 is in excess, and

11 mol NO2 - 9 mol NO2 = 2 mol NO2 will be left after reaction.

3 0
3 years ago
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