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3 years ago

4. (He] 2s^1 2p^4 is the correct electron configuration for: a) C b) O c) Si d) S

Chemistry

1 answer:

Aleksandr-060686 [28]3 years ago

8 0

Hey there!

[He]2s²2p⁴ is the electron configuration for O (oxygen).

He is the noble has before oxygen. It has 2 electrons.

The s orbital is filled, and there are 4 electrons in the p orbital, so 6 elements over to the right.

This makes a total of 8 electrons.

Hope this helps!

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Carbonic anhydrase of erythrocytes (Mr 30,000) has one of the highest turnover numbers known. It catalyzes the reversible hydrat

Afina-wow [57]

Explanation:

According to the given data, the turnover number can be calculated as follows.

Turnover number = $K_{cat} = \frac{V_{max}}{\text{Concentration of enzyme}}$

$V_{max} = \frac{\text{Moles of CO_{2} hydrolyzed}{second}$

Therefore, moles of $CO_{2}$ hydrolyzed is as follows.

Moles of $CO_{2}$ hydrolyzed = $\frac{Mass of CO_{2}}{Molar mass of CO_{2}}$

= $\frac{0.30}{44}$

= 0.00682 moles

Now, moles of $CO_{2}$ hydolyzed per second is calculated as follows.

Moles of $CO_{2}$ hydolyzed per second = $\frac{0.00682}{60}$

= $1.137 \times 10^{-4} moles/second = V_{max}$

And,

Moles of enzyme = $\frac{Mass}{\text{Molar mass}}$

= $\frac{10.0 \mu g}{30000}$

= $3.33 \times 10^{-10}$ moles

Therefore, the value of $K_{cat}$ is as follows.

$K_{cat} = \frac{1.137 \times 10^{-4} moles}{3.333 \times 10^{-10} moles}$

= $0.3411 \times 10^{6}$ per second

= $0.3411 \times 60 \times 10^{6}$ per minute

= $20.466 \times 10^{6}$ per minute

Thus, we can conclude that the turnover number ( $K_{cat}$ ) of carbonic anhydrase (in units of $min^{-1}$ ) is $20.466 \times 10^{6}$ per minute.

6 0

4 years ago

You are provided with 300.0 mL of a buffer solution consisting of 0.200 M H3BO3 and 0.250 M NaH2BO3.

My name is Ann [436]

Answer:

a. 9.34

b. 9.06

c. 6 mL

Explanation:

Part a.

The pH of a buffer solution is given by the Henderson-Hasselbach equation:

pH = pKa + log [A⁻] / [HA]

where pKa is the negative log of Ka for the weak acid H₃BO₃ and can be obtained from reference tables, [A⁻] and [HA] are the concentrations of the weak conjugate base H₂BO₃⁻ and and the weak acid H₃BO₃ respectively.

Proceeding with the calculations, we have

Ka H₃BO₃ = 5.80 x 10⁻¹⁰

pKa = - log (5.80 x 10⁻¹⁰) = 9.24

[H₂BO₃⁻ ] = 0.250 M

[H₃BO₃] = 0.200 M

pH = 9.24 + log (0.250/0.200) = 9.34

part b.

When 1.0 mL of 6.0 M HCl is added to the buffer , we know that it will react with the conjugate base in the buffer doing what buffers do: keeping the pH within a small range according to the capacity of the buffer:

H₂BO₃⁻ + H⁺ ⇒ H₃BO₃

So lets calculate the new concentrations of acid and conjugate base after reaction and apply the Henderson equation again:

Initial # of moles:

H₃BO₃ = 0.300 L x 0.200 mol/L = 0.06 mol

H₂BO₃⁻ = 0.200 L x 0.250 mol/L = 0.05 mol

mol HCl = 0.001 L x 6.0 mol/L = 0.006 mol

After reaction

H₃BO₃ = 0.06 mol + 0.006 mol = 0.066 mol

H₂BO₃⁻ = 0.05 mol - 0.006 mol = 0.044 mol

New pH

pH = 9.24 + log ( 0.044 / 0.66 ) = 9.06

Note: There is no need to calculate the new concentrations since we have a quotient in the expression where the volumes cancel each other.

Part c.

We will be using the Henderson-Hasselbach equationm again but now to calculate ratio [H₂BO₃⁻] / [HBO₃] that will give us a pH of 10.00. Thenwe will make use of the stoichiometry of the reaction to calculate the volume of NaOH required.

pH = pKa + log[H₂BO₃⁻]-[H₃BO₃]

10.00 = 9.24 + log [H₂BO₃⁻]-[H₃BO₃]

⇒[H₂BO₃⁻] / [H₃BO₃] = antilog (0.76) = 5.75

Initiall # moles:

mol H₃BO₃ = 0.06 mol

mol H₂BO₃ = 0.05 mol

after consumption of H₃BO₃ from the reaction with NaOH:

H₃BO₃ + NaOH ⇒ Na⁺ + H₂BO₃⁻ + H₂O

mol H₃BO₃ = 0.06 - x

mol H₂BO₃⁻ = 0.05+ x mol

Therefore we have the algebraic expression:

[H₂BO₃⁻] / [H₃BO₃] = mol H₂BO₃⁻ / mol HBO₃ = 5.75

( again volumes cancel each other)

0.05 + x / 0.06 - x = 5.75 ⇒ x = 0.044

SO 0.037 mol NaOH were required, and since we know Molarity = mol / V we can calculate the volume of 6.0 M NaOH added:

V = 0.044 mol / 6.0 mol/L = 0.0073 L

V = 7.3 mL

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