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Pavlova-9 [17]
2 years ago
9

In details explain element

Chemistry
2 answers:
Mamont248 [21]2 years ago
7 0
Going off of what the other person said
An element can also be “each of more than one hundred substances that cannot be chemically interconverted or broken down into simpler substances and are primary constituents of matter. Each element is distinguished by its atomic number, i.e. the number of protons in the nuclei of its atoms.”
Dominik [7]2 years ago
6 0

Answer:

definition- a part or aspect of something abstract, especially one that is essential or characteristic

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Write equation to represent the Hydrogenation reactions​
garik1379 [7]

Explanation:

vegetable oils have long unsaturated carbon chains which can converted into vegetable ghee

4 0
2 years ago
What is the color of the indicator thymol blue in a solution that has a ph of 11?.
Leviafan [203]

Answer:

Red

Explanation:

Red color is evidenced when thymol blue indicator is in a solution having a pH of 11. A pH of 11 means that the solution is basic or alkaline. Therefore, the indicator turns red, indicating that the solution is alkaline.

When thymol blue indicator is in a acidic solution, the indicator remains blue.

8 0
2 years ago
What are some factors that affect weather?
Zarrin [17]
B is the right answer
6 0
2 years ago
Calculate the molarity (M) of a 600 ml solution of 0.54 moles of H2SO4.
icang [17]
Molarity (concentration) can be calculated by the equation:
Concentration = moles / volume in L = 0.54 mol / 0.6 L = 0.9 M
Hope this helps!
8 0
3 years ago
Phenolphthalein has a pKa of 9.7 and is colorless in its acid form and pink in its basic form. calculate [In-}/{HIn} for the fol
zvonat [6]

Answer:

1.58x10⁻⁵

2.51x10⁻⁸

0.0126

63.10

Explanation:

Phenolphthalein acts like a weak acid, so in aqueous solution, it has an acid form HIn, and the conjugate base In-, and the pH of it can be calculated by the Handerson-Halsebach equation:

pH = pKa + log[In-]/[HIn]

pKa = -logKa, and Ka is the equilibrium constant of the dissociation of the acid. [X] is the concentrantion of X. Thus,

i) pH = 4.9

4.9 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = - 4.8

[In-]/[HIn] = 10^{-4.8}

[In-]/[HIn] = 1.58x10⁻⁵

ii) pH = 2.1

2.1 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -7.6

[In-]/[HIn] = 10^{-7.6}

[In-]/[HIn] = 2.51x10⁻⁸

iii) pH = 7.8

7.8 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -1.9

[In-]/[HIn] = 10^{-1.9}

[In-]/[HIn] = 0.0126

iv) pH = 11.5

11.5 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = 1.8

[In-]/[HIn] = 10^{1.8}

[In-]/[HIn] = 63.10

6 0
3 years ago
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