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3 years ago

Define valence electrons.

Chemistry

1 answer:

vodka [1.7K]3 years ago

8 0

Answer:

Electrons on the outermost shell of an atom. They are responsible for the chemical properties of an atom.

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Answer:

yes, albert is better grffffrr#fffffrttt.

Explanation:

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3 years ago

Read 2 more answers

Read the two statements given below. Statement 1: All living matter at the smallest level is made of cells Statement 2: New anim

9966 [12]

Answer:

Statement 1: All living matter at the smallest level is made of cells

Explamation:

All living things are made of cells; the cell itself is the smallest fundamental unit of structure and function in living organisms.

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3 years ago

In the symbol 3p4

kompoz [17]

Answer:

a. the 3 represents the principal energy level

Explanation:

3 is the principal energy level. The p is the sublevel. 4 is the possible occupying electron.

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3 years ago

Will a precipitate of magnesium fluoride form when 300. mL of 1.1 × 10 –3 M MgCl 2 are added to 500. mL of 1.2 × 10 –3 M NaF? [K

Tju [1.3M]

Answer:

No precipitate is formed.

Explanation:

Hello,

In this case, given the dissociation reaction of magnesium fluoride:

$MgF_2(s)\rightleftharpoons Mg^{2+}+2F^-$

And the undergoing chemical reaction:

$MgCl_2+2NaF\rightarrow MgF_2+2NaCl$

We need to compute the yielded moles of magnesium fluoride, but first we need to identify the limiting reactant for which we compute the available moles of magnesium chloride:

$n_{MgCl_2}=0.3L*1.1x10^{-3}mol/L=3.3x10^{-4}molMgCl_2$

Next, the moles of magnesium chloride consumed by the sodium fluoride:

$n_{MgCl_2}^{consumed}=0.5L*1.2x10^{-3}molNaF/L*\frac{1molCaCl_2}{2molNaF} =3x10^{-4}molMgCl_2$

Thus, less moles are consumed by the NaF, for which the moles of formed magnesium fluoride are:

$n_{MgF_2}=3x10^{-4}molMgCl_2*\frac{1molMgF_2}{1molMgCl_2}=3x10^{-4}molMgF_2$

Next, since the magnesium fluoride to magnesium and fluoride ions is in a 1:1 and 1:2 molar ratio, the concentrations of such ions are:

$[Mg^{2+}]=\frac{3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =3.75x10^{-4}M$

$[F^-]=\frac{2*3x10^{-4}molMg^{+2}}{(0.3+0.5)L} =7.5x10^{-4}M$

Thereby, the reaction quotient is:

$Q=(3.75x10^{-4})(7.5x10^{-4})^2=2.11x10^{-10}$

In such a way, since Q<Ksp we say that the ions tend to be formed, so no precipitate is formed.

Regards.

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3 years ago

Scientists utilize models for a variety of different purposes, but each type of scientific model has limitations. What might be

Nezavi [6.7K]

They are based in current knowledge

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4 years ago