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Akimi4 [234]
3 years ago
7

Make a suggestion on how you will deal with the problem

Chemistry
2 answers:
Ilia_Sergeevich [38]3 years ago
8 0

Answer:

《<☆HOPE IT WILL HELP YOU 》>☆

Explanation:

Dealing with the problem by

Never lose your hope you're cofidance your power

Always take dlessing or suggeation from the elder or your parent

<em>•</em><em>●</em><em>○</em><em>Please</em><em> </em><em> </em><em>mark</em><em> </em><em>my ans</em><em> </em><em>as</em><em> </em><em>BRAIN</em><em> </em><em>LIST</em><em>○</em><em>●</em><em>•</em>

ollegr [7]3 years ago
5 0

Answer:

is it a research work?

Explanation:

if it is first u write about what is a climate what r due to climatic changes

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Tia learns that the temperature of rock in Earth's mantle varies depending on its location. Rock located close to
kifflom [539]

Answer:

Selling puffs and juuls right now. I have strawberry and mint i'm almost out of stalk hurry and buy one for just 1$ alone or you can get 2 juuls with a candy bag depending on how much candy or juuls you buy.

Explanation:

8 0
3 years ago
How does tension affect the speed of a wave in a rope
Ainat [17]
Hi sorry

I don’t know

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3 0
3 years ago
Using the solubility of Ca(IO3)2 that you determined in pure water, calculate the value of Ksp using only concentrations, that i
Aleks04 [339]

Answer:

4.6305 * 10^-6 mol^3.L^-3

Explanation:

Firstly, we write the value for the solubility of Ca(IO3)2 in pure water. This equals 0.0105mol/L.

We proceed to write the dissociation reaction equation for Ca(IO3)2

Ca(IO3)2(s) <——->Ca2+(aq) + 2IO3-(aq)

We set up an ICE table to calculate the Ksp. ICE stands for initial, change and equilibrium. Let the concentration of the Ca(IO3)2 be x. We write the values for the ICE table as follows:

Ca2+(aq). 2IO3-(aq)

I. 0. 0.

C. +x. +2x

E. x. 2x

The solubility product Ksp = [Ca2+][IO3-]^2

Ksp = x * (2x)^2

Ksp = 4x^3

Recall, the solubility value for Ca(IO3)2 in pure water is 0.0105mol/L

We substitute this value for x

Ksp = 4(0.0105)^2 = 4 * 0.000001157625 = 4.6305 * 10^-6

5 0
3 years ago
In addition to those in Table 14.3, other less stable nitrogen oxides exist. Draw a Lewis structure for each of the following:(c
Natalka [10]

Lewis structure for each of the following N₂O₃ with no N¬N bond is attached below.

Even though pi symmetry occupies the antibonding orbitals of NO, this is unimportant after the dimer forms. A sigma connection exists. The enthalpy of the newly formed sigma bond in the dimer is low because the loss of a particularly distinctive set of single-electron resonance forms that were available for no monomer offset the net gain in bond. When the whole free energy is taken into account, there is no gain because the entropic effects are on the order of 1030kJ/mol, and dimerization is entropically disfavored at G=17kJ/mol. Therefore, any little increase in enthalpy is cancelled out by the loss of entropy.

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brainly.com/question/17152685

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5 0
2 years ago
How is good at electron configurations in chemistry?
Paha777 [63]

Try to understand all the rules and laws like:

Aufbau Principle

Hund's rule

Pauli exclusion principle...

Then, you should understand the way in which you can fill the electrons in the orbitals!!

8 0
3 years ago
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