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iren2701 [21]
3 years ago
12

5. पहले अठारह तत्वों के विभिन्न कक्षों में इलेक्ट्रॉन वितरण के नियम कोलिखिए।​

Chemistry
1 answer:
aniked [119]3 years ago
8 0

分門夌亥峉帕丑岶Explanation:

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All matter can be classified by their physical properties. Which substance can be classified as being soluble in water
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The substance that can be classified as being soluble in water would be salt. The solution is D. Salt specifically is composed of ions and the ionic compound dissociates within water.
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Explain the role of a primary consumer and a secondary consumer in a food web.
Usimov [2.4K]

Answer:

Explanation:

Secondary consumers are organisms that eat primary consumers for energy. Primary consumers are always herbivores, or organisms that only eat autotrophic plants.

Carnivores only eat other animals, and omnivores eat both plant and animal matter.

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3 years ago
Which of the changes gas to solid gas to liquid, liquid to solid liquid to gas,solid to liquid solid to gas will involve an outp
Verdich [7]

Answer:

Liquid to solid liquid to gas

Explanation:

Boiling a liquid result in a gas

7 0
2 years ago
I don't quite understand it
irina1246 [14]
So, you need to have same ammount of atoms on the left and on the right side of the equation. You need to count the ammount of attoms of every substance on the left, and make sure that on the right side the ammount is same. For example in the 1st one it’s 6Sn+2P4=2Sn3P4, so that you have 6atoms of Sn on the left and 6 atoms of Sn on the right, same with the P
6 0
3 years ago
Please help me
Wittaler [7]

Answer:

pH = 6.999

The solution is acidic.

Explanation:

HBr is a strong acid, a very strong one.

In water, this acid is totally dissociated.

HBr + H₂O  →  H₃O⁺  +  Br⁻

We can think pH, as - log 7.75×10⁻¹² but this is 11.1

acid pH can't never be higher than 7.

We apply the charge balance:

[H⁺] = [Br⁻] + [OH⁻]

All the protons come from the bromide and the OH⁻ that come from water.

We can also think [OH⁻] = Kw / [H⁺] so:

[H⁺] = [Br⁻] + Kw / [H⁺]

Now, our unknown is [H⁺]

[H⁺] =  7.75×10⁻¹² + 1×10⁻¹⁴ / [H⁺]

[H⁺] = (7.75×10⁻¹² [H⁺] + 1×10⁻¹⁴) /  [H⁺]

This is quadratic equation:  [H⁺]² - 7.75×10⁻¹² [H⁺] - 1×10⁻¹⁴

a = 1 ; b = - 7.75×10⁻¹² ; c = -1×10⁻¹⁴

(-b +- √(b² - 4ac) / (2a)

[H⁺] = 1.000038751×10⁻⁷

- log [H⁺] = pH → 6.999

A very strong acid as HBr, in this case, it is so diluted that its pH is almost neutral.

8 0
3 years ago
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