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Novosadov [1.4K]
3 years ago
15

Molecules that influence protein activity by binding to sites that are distinct from the active site are called ______________ _

______________.
Chemistry
1 answer:
Rudiy273 years ago
6 0
<span>Answer: allosteric effectors</span>
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(3 pt) The cost of a space mission is roughly $60 million. What type of obstacle to space exploration does this represent? A. fu
erastova [34]
Im pretty sure its A funding because thats alot of money
7 0
3 years ago
Read 2 more answers
Calculate the pH of the following simple solutions:
IceJOKER [234]

Answer :

(1) pH = 1.27

(2) pH = 13.35

(3) The given solution is not a buffer.

Explanation :

<u>(1) 53.1 mM HCl</u>

Concentration of HCl = 53.1mM=53.1\times 10^{-3}M

As HCl is a strong acid. So, it dissociates completely to give hydrogen ion and chloride ion.

So, Concentration of hydrogen ion= 53.1\times 10^{-3}M

pH : It is defined as the negative logarithm of hydrogen ion concentration.

pH=-\log [H^+]

pH=-\log (53.1\times 10^{-3})

pH=1.27

<u>(2) 0.223 M KOH</u>

Concentration of KOH = 0.223 M

As KOH is a strong base. So, it dissociates completely to give hydroxide ion and potassium ion.

So, Concentration of hydroxide ion= 0.223 M

Now we have to calculate the pOH.

pOH=-\log [OH^-]

pOH=-\log (0.223)

pOH=0.65

Now we have to calculate the pH.

pH+pOH=14\\\\pH=14-pOH\\\\pH=14-0.65=13.35

<u>(3) 53.1 mM HCl + 0.223 M KOH</u>

Buffer : It is defined as a solution that maintain the pH of the solution by adding the small amount of acid or a base.

It is not a buffer because HCl is a strong acid and KOH is a strong base. Both dissociates completely.

As we know that the pH of strong acid and strong base solution is always 7.

So, the given solution is not a buffer.

5 0
3 years ago
Write the lewis structure for mgi2. draw the lewis dot structure for mgi2. include all lone pairs of electrons.
viktelen [127]
Lewis Structure is drawn in following steps,

1) Calculate Number of Valence Electrons:
    
# of Valence electrons in Mg  =  2
# of Valence electrons in I      =  7
# of Valence electrons in I      =  7
                                               ---------
Total Valence electrons          =  16

2) Draw Mg as a central atom surround it by two atoms of Iodine.

3) Connect each Iodine atom to Mg, and subtract two electrons per bond. In this case we will subtract 4 electrons from total valence electrons. i.e.

Total Valence electrons           16
- Four electrons                    -   4
                                              ----------
                                                  12

4) Now start adding the remaining 12 electrons on more electronegative atoms i.e. Iodine.

The final lewis structure formed is as follow, 

4 0
3 years ago
20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the
kirill115 [55]

Answer:

Choice A: approximately 0.12\; \rm M.

Explanation:

Note that the unit of concentration, \rm M, typically refers to moles per liter (that is: 1\; \rm M = 1\; \rm mol\cdot L^{-1}.)

On the other hand, the volume of the two solutions in this question are apparently given in \rm cm^3, which is the same as \rm mL (that is: 1\; \rm cm^{3} = 1\; \rm mL.) Convert the unit of volume to liters:

  • V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L.
  • V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L.

Calculate the number of moles of \rm H_2SO_4 formula units in that 0.02\; \rm L of the 0.09\; \rm M solution:

\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}.

Note that \rm H_2SO_4 (sulfuric acid) is a diprotic acid. When one mole of \rm H_2SO_4 completely dissolves in water, two moles of \rm H^{+} ions will be released.

On the other hand, \rm NaOH (sodium hydroxide) is a monoprotic base. When one mole of \rm NaOH formula units completely dissolve in water, only one mole of \rm OH^{-} ions will be released.

\rm H^{+} ions and \rm OH^{-} ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid \rm H_2SO_4 dissolves in water completely, it will take two moles of \rm OH^{-} to neutralize that two moles of \rm H^{+} produced. On the other hand, two moles formula units of the monoprotic base \rm NaOH will be required to produce that two moles of \rm OH^{-}. Therefore, \rm NaOH and \rm H_2SO_4 formula units would neutralize each other at a two-to-one ratio.

\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O.

\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2.

Previous calculations show that 0.0018\; \rm mol of \rm H_2SO_4 was produced. Calculate the number of moles of \rm NaOH formula units required to neutralize that

\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}.

Calculate the concentration of a 0.03\; \rm L solution that contains exactly 0.0036\; \rm mol of \rm NaOH formula units:

\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}.

3 0
3 years ago
How many atoms are present in 3 moles of chromium?
Murljashka [212]

Answer:

1.80 x 10^24 atoms

Explanation:

3moles × 6.022×10^23 atoms/mole

3 0
3 years ago
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