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anzhelika [568]
3 years ago
8

The atomic number is the same as the number of ________.

Chemistry
1 answer:
Zanzabum3 years ago
5 0

Answer:

number of electrons

Explanation:

(happy to help)

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A student prepared a stock solution by dissolving 10.0 g of KOH in enough water to make 150. mL of solution. She then took 15.0
yanalaym [24]

Answer: The concentration of KOH for the final solution is 0.275 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

Molarity=\frac{n\times 1000}{V_s}

where,

n = moles of solute

 V_s = volume of solution in ml = 150 ml

moles of solute =\frac{\text {given mass}}{\text {molar mass}}=\frac{10.0g}{56g/mol}=0.178moles

Now put all the given values in the formula of molality, we get

Molality=\frac{0.178\times 1000}{150}=1.19M

According to the dilution law,

C_1V_1=C_2V_2

where,

C_1 = molarity of stock solution = 1.19 M

V_1 = volume of stock solution = 15.0 ml

C_2 = molarity of diluted solution = ?

V_2 = volume of diluted solution = 65.0 ml

Putting in the values we get:

1.19\times 15.0=M_2\times 65.0

M_1=0.275M

Therefore, the concentration of KOH for the final solution is 0.275 M

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3 years ago
John Dalton thought that atoms
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Answer:

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2 years ago
Penicillinase, also known as β‑lactamase, is a bacterial enzyme that hydrolyzes and inactivates the antibiotic penicillin. Penic
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Explanation:

The given data is as follows.

        V_{max} = 6.8 \times 10^{-10} \mu mol/min

          K_{m} = 5.2 \times 10^{-6} M

Now, according to Michaelis-Menten kinetics,

              V_{o} = V_{max} \times [\frac{S}{(S + Km)}]

where, S = substrate concentration = 10.4 \times 10^{-6} M

Now, putting the given values into the above formula as follows.

        V_{o} = V_{max} \times [\frac{S}{(S + Km)}]

        V_{o} = 6.8 \times 10^{-10} \mu mol/min \times [\frac{10.4 \times 10^{-6} M}{(10.4 \times 10^{-6}M + 5.2 \times 10^{-6} M)}]

            V_{o} = 6.8 \times 10^{-10} \mu mol/min \times 0.667

                              = 4.5 \times 10^{-10} \mu mol/min

This means that V_{o} would approache V_{max}.

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I will take a stab at it, but there are not equations, did you forget them?
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