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3 years ago

How to determine Km and Vm from a graph

Chemistry

1 answer:

ElenaW [278]3 years ago

5 0

Answer:

From the graph find the maximum velocity and half it i.e. Vmax/2. Draw a horizontal line from this point till you find the point on the graph that corresponds to it and read off the substrate concentration at that point. This will give the value of Km.

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Someone pls help me I will make you brain

wlad13 [49]

Answer:

It is true answer. Ozone = O3

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3 years ago

1. Alexa and her family travelled 5 hours south east on 1-45 to crystal

Gre4nikov [31]

Answer:

Average speed = 68 mph

Explanation:

Given that,

Total distance traveled by the family, d = 340 miles

The family traveled 5 hours southeast.

We need to find her average speed. The speed of an object is given by the total distance covered divided by time taken. So,

$v=\dfrac{d}{t}$

Put all the values,

$v=\dfrac{340}{5}\\\\v=68\ mph$

So, her average speed is equal to 68 mph.

5 0

3 years ago

51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b

Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

$a = \sqrt{8} r$

$a = \sqrt{8} \times 144 pm$

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}$

density d = 19.41 g/cm³

Similarly; For a body-centered cubic structure

$r = \dfrac{\sqrt{3}}{4}a$

where;

r = 144

$144 = \dfrac{\sqrt{3}}{4}a$

$a = \dfrac{144 \times 4}{\sqrt{3}}$

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V 3.68 × 10⁻²³ cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}$

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0

3 years ago

What type of cells are gametes?

ELEN [110]

Answer:

Reproductive cells(also known as sex cells) are gametes.

Explanation:

Have a great day :)

6 0

3 years ago

Read 2 more answers

What is the molarity of a 1.44 (m/v)% solution of an HCOOOH (Formic acid)?

AlexFokin [52]

This problem is providing the mass-volume percent of a formic acid solution so its molarity is required and found to be 0.313 M after the following calculations.

<h3>Molarity</h3>

In chemistry, units of concentration provide a measurable understanding of the relationship between the relative amounts of both solute and solvent. In the case of molarity, one must relate moles of solute and liters of solution as follows:

$M=\frac{mol\ solute}{Volume\ solution \ in \ L}$

In such a way, when given this mass-volume percent of 1.44% for the formic acid in the solution, one can assume there is 100 mL of solution and 1.44 g of solute (formic acid), which means one must convert the volume to liters and the mass to moles with:

$mol\ solute=\frac{1.44g}{46.03g/mol} =0.0313mol\\\\Volume\ in \liters: 100mL*\frac{1L}{1000mL}=0.100L$

Finally, we plug in these numbers in the equation for the calculation of molarity:

$M=\frac{0.0313mol}{0.100L}\\\\M=0.313M$

Learn more about molarity: brainly.com/question/10053901

8 0

2 years ago