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3 years ago

What is a magnetic field

Chemistry

1 answer:

babymother [125]3 years ago

5 0

a region around a magnetic material or a moving electric charge within which the force of magnetism acts.

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Why is it important to measure the mass of an atom?

victus00 [196]

Because without the mass, you can't find the molar mass and other formulas in Chemistry.

8 0

3 years ago

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Two elevators carry five passengers to the fifth floor. However, the elevators do not do the same work. Choose the best factor f

Kryger [21]

Answer:

B the weight of the passengers

8 0

3 years ago

How do Cilia and flagella help cells?

otez555 [7]

Answer:

They are protection from the cell!

Explanation:

They move liquid around the cells and have the same internal structure as each other. They are also found in most microorganisms. Hope this helps :)

3 0

3 years ago

How many rings does an alkane have if its formula is c10h16?

stepladder [879]

If it is saturated compound then we can calculate the double bond equivalent which will be equal to the number of rings in the compound

the double bond equivalent can be calculated using following formula

$DBE = (C + 1 -\frac{H}{2} - \frac{X}{2} +\frac{N}{2} )$

Where

H = number of Hydrogen atoms

C = number of carbon atoms

X= number of halogen atoms

N = number of nitrogen atoms

DBE = (10 + 1 - 16 / 2 ) = 3

Hence there are three rings in the compound

3 0

3 years ago

Calculate the equilibrium concentration of H 3 O H3O in a 0.20 M M solution of oxalic acid. Express your answer to two significa

Black_prince [1.1K]

Answer: The equilibrium concentration of $H_3O^+$ ion is $8.3064\times 10^{-2}M$

Explanation:

We are given:

Molarity of oxalic acid solution = 0.20 M

Oxalic acid $(H_2C_2O_4)$ is a weak acid and will dissociate 2 hydrogen ions.

The chemical equation for the first dissociation of oxalic acid follows:

$H_2C_2O_4(aq.)+H_2O\rightleftharpoons H_3O^+(aq.)+HC_2O_4^-(aq.)$

Initial: 0.20

At eqllm: 0.20-x x x

The expression of first equilibrium constant equation follows:

$Ka_1=\frac{[H_3O^+][HC_2O_4^{-}]}{[H_2C_2O_4]}$

We know that:

$Ka_1\text{ for }H_2C_2O_4=0.059$

Putting values in above equation, we get:

$0.059=\frac{x\times x}{(0.20-x)}\\\\x=-0.142,0.083$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of hydronium ion = x = 0.083 M

The chemical equation for the second dissociation of oxalic acid:

$HC_2O_4^-(aq.)+H_2O\rightarrow H_3O^+(aq.)+C_2O_4^{2-}(aq.)$

Initial: 0.083

At eqllm: 0.083-y 0.083+y y

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H_3O^+][C_2O_4^{2-}]}{[HC_2O_4^-]}$

We know that:

$Ka_2\text{ for }H_2C_2O_4=6.4\times 10^{-5}$

Putting values in above equation, we get:

$6.4\times 10^{-5}=\frac{(0.083+y)\times y}{(0.083-y)}\\\\y=-0.083,0.0000639$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of hydronium ion = y = 0.0000639 M

Total concentration of hydronium ion = [x + y] = [0.083 + 0.0000639] = 0.0830639 M

Hence, the equilibrium concentration of $H_3O^+$ ion is $8.3064\times 10^{-2}M$

7 0

3 years ago