How many moles are 6.022(10^23) atoms of sodium

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

sergeinik [125]

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 $V_{max}$ .

5 0

3 years ago

A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0

2 years ago

Read 2 more answers

Counting atoms worksheet answer key i need help!!

katovenus [111]

Answer is in a pho $^{}$ to. I can only uplo $^{}$ ad it to a file host $^{}$ ing service. link below!

bit. $^{}$ ly/3a8Nt8n

6 0

3 years ago

Predict how many H1 NMR signals (individual resonances, not counting splitting) are expected for the compound.

Lyrx [107]

Answer:

3 H1 NMR signals

Explanation:

NB: kindly check the diagram of the chemical compound in the attached picture.

This particular Question is based on the part of chemistry which is known as spectroscopy. Spectroscopy is used in the Determination or in identifying chemical compounds. H'NMR works on the principle of nuclear magnetic resonance.

In order to solve this question, one has to count the number of hydrogen in unique location. The diagram in the attached show how hydrogen is been counted.

The numbers of signals is the number of different chemical environments in which hydrogen atoms are located.

NB: signals is also the same as peak in H'NMR.

Hence, the number of H1 NMR signals in this chemical compound is 3.

3 0

3 years ago

4. How many milligrams are in 5.25 x 10-13 kg?<br><br> the “-13” is an exponent

rusak2 [61]

5. 25 x 10⁻⁷mg

Explanation:

This is mass conversion from mg to kg;

The kg is a quantity of mass used to measure the amount of matter in a substance.

Given mass = 5.25 x 10⁻¹³kg

The kilo- is a prefix that denotes 10³

therefore;

1000g = 1kilogram

the milli- is a prefix that denotes 10⁻⁻³

1000mg = 1g

Now that we know this, we can convert:

5.25 x 10⁻¹³kg x $\frac{1000g}{1kg} x \frac{1000mg}{1g}$ = 5. 25 x 10⁻¹³ x 10⁶mg

= 5. 25 x 10⁻⁷mg

learn more:

Conversion brainly.com/question/1548911

#learnwithBrainly

8 0

3 years ago