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kotegsom [21]
2 years ago
10

Define Evaporation.Thanks​

Chemistry
2 answers:
emmainna [20.7K]2 years ago
6 0

Evaporation happens when a liquid turns into a gas. It can be easily visualized when rain puddles “disappear” on a hot day or when wet clothes dry in the sun.

Kay [80]2 years ago
5 0

Answer:

evapouration is a type of vaporization that occur on the surface of liquid as it change into the gas phase.

hope it helps.

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What is this please helppp​
svetoff [14.1K]

Answer:

it looks like deoxyribose...im not sure tho

Explanatio

7 0
2 years ago
In the formula for the compound XCl4, the X could represent<br> (1) C (3) Mg<br> (2) Η (4) Zn
Anvisha [2.4K]
Correct Answer: option 1 i.e. C

Reason: 
The the compound of interest i.e.  XCl4, since there are 4 Cl atoms bonded to X. This signifies that the valency of X is 4.

There atomic number of C is 6. It's electronic configuration is giving by 1s2 2s2 2p2. Thus, there are 4 electrons in valence shell of C. This signifies that valency of C is 4. Hence the compound present in present case is CCl4.
7 0
3 years ago
Read 2 more answers
Vitamin c is known chemically by the name ascorbic acid determine the empirical formula of ascorbic acid if it is composed of 40
kow [346]

Answer:

=C_3H_4O_3

Explanation:

When percentage composition is given, and asked for the empirical formula, it is simplest to  assume 100 g of material. Thus,

Mass C = 40.92 g.  Moles C = 40.92 g x 1 mole/12 g = 3.41 moles C

Mass H = 4.58 g.  Moles H = 4.58 g x 1 mole/1.0 g = 4.58 moles H

Mass O = 54.50 g.  Moles O = 54.50 g x 1 mole/16 g = 3.41 moles O

Now, we want to get the moles into whole numbers, so we begin by dividing all by the smallest, i.e. divide all values by 3.41.

Moles C = 3.41/3.41 = 1

Moles H = 4.58/3.41 = 1.34

Moles O = 3.41/3.41 = 1

Now, in order to get 1.34 to be a whole number we multiply it (and all others) by 3

Moles C = 1x3 = 3

Moles H = 1.34x3 = 4

Moles O = 1x3 = 3

Empirical Formula =C_3H_4O_3

3 0
3 years ago
What is The pH and pOH of a 1.00x10-3M solution of CH3COOH (Ka=1.75x10-5) ?
zaharov [31]

Answer:

<h2>pH = 3.9</h2><h2>pOH = 10.1</h2>

Explanation:

Since CH _ 3COOH is a weak acid to find the pH of CH _ 3COOH we use the formula

pH =  -  \frac{1}{2}   log(Ka)  -  \frac{1}{2}  log(c)

where

Ka is the acid dissociation constant

c is the concentration

From the question

Ka of CH _ 3COOH = 1.75 × 10^-5

c = 1.00 × 10-³M

Substitute the values into the above formula and solve for the pH

That's

pH =  \frac{1}{2} ( -  log(1.75 \times {10 }^{ - 5} -  log(1.00 \times  {10}^{ - 3} )  )  \\   =  \frac{1}{2} (4.757  + 3) \\  =  \frac{1}{2}  \times 7.757) \\  = 3.8785 \:  \:  \:  \:  \:  \:  \:  \:

We have the answer as

<h3>pH = 3.9</h3>

To find the pOH we use the formula

pH + pOH = 14

pOH = 14 - pH

pOH = 14 - 3.9

We have the answer as

<h3>pOH = 10.1</h3>

Hope this helps you

4 0
3 years ago
If a buffer solution is 0.220 M in a weak acid ( Ka=7.4×10−5) and 0.540 M in its conjugate base, what is the pH?
valkas [14]

Answer: the pH of the solution is 4.52

Explanation:

Consider the weak acid as Ha, it is dissociated as expressed below

HA     H⁺  +  A⁻

the Henderson -Haselbach equation can be expressed as;

pH = pKa + log( [A⁻] / [HA])

the weak acid is dissociated into H⁺ and A⁻ ions in the solution.

now the conjugate base of the weak acid HA is

HA(aq) {weak acid}     H⁺(aq)  +  A⁻(aq) {conjugate base}

so now we calculate the value of Kₐ as well as pH value by substituting the values of the concentrations into the equation;

pKₐ = -logKₐ

pKₐ = -log ( 7.4×10⁻⁵ )

pKₐ = 4.13

now thw pH is

pH = pKₐ  + log( [A⁻] / [HA])

pH = 4.13 + log( [0.540] / [0.220])

pH = 4.13 + 0.3899

pH = 4.5199 = 4.52

Therefore the pH of the solution is 4.52

6 0
3 years ago
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