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1 year ago

Why does the reactivity of nonmetals increase moving left to right on the periodic table?

Chemistry

1 answer:

zhuklara [117]1 year ago

4 0

Answer: This is because the number of shell increases .

Explanation: On moving from left to write on the periodic table the reactivity of non metals increases because number of shells increases and the force with which the nucleus hold electrons decreases.

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Determine the mole fraction of the solute if you have 3.4 grams KCl in 8.1 grams H2O

Yakvenalex [24]

.42; 3.4 divided by 8.1 is about .42

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3 years ago

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What is the abbreviated electron configuration of cobalt?

drek231 [11]

It's Co just look at the periodic table

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Aqueous acetic acid is neutralized by aqueous barium hydroxide. express your answer as a balanced molecular equation. identify a

Studentka2010 [4]

The balanced molecular equation for the given reaction is:

$2CH_3COOH(aq) \ + \ Ba(OH)_2(aq) \rightarrow \ Ba(CH_3COO)_2(aq) + 2H_2O(l).$

<h3>FURTHER EXPLANATION</h3>

To check that the equation is balanced count how many atoms are present for each element in the reactant and product side. If they are the same before and after reaction for all elements, then the reaction is deemed balanced.

The atom counting for the equation is shown below:

$2CH_3COOH(aq) \ + \ Ba(OH)_2(aq) \rightarrow \ Ba(CH_3COO)_2(aq) + 2H_2O(l).$

Reactants → Products

C: (2 x 2) =4 C: (2 x 2) = 4

H: (2 x 4) + (1 x 2) =10 H: (3 x 2) + (2 X 2) = 10

O: (2 x 2) + (1 x 2) = 6 O: (2 x 2) + (2 x 1) = 6

Ba: 1 Ba: 1

Since the number of atoms of each element are similar in the reactants and products, the equation is balanced.

To determine the state of the substances, consider their solubility.

The reactants are both aqueous (aq) as indicated in the problem.

The first product, $Ba(CH_3COO)_2$ is aqueous (aq) because based on the solubility rule, it will dissolve in water. Acetates are generally soluble.

The other product is water which will be liquid (l) since it is the solvent used to dissolve the substances.

<h3>LEARN MORE</h3>

Solubility Rules brainly.com/question/12984314
Net Ionic Equation brainly.com/question/12980075

Keywords: acid-base, neutralization, balancing equations, molecular equation

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3 years ago

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The second-order diffraction for a gold crystal is at an angle of 22.20o for X-rays of 154 pm. What is the spacing between the c

Alenkinab [10]

Answer: The spacing between the crystal planes is $4.07\times 10^{-10}m$

Explanation:

To calculate the spacing between the crystal planes, we use the equation given by Bragg, which is:

$n\lambda =2d\sin \theta$

where,

n = order of diffraction = 2

$\lambda$ = wavelength of the light = $154pm=1.54\times 10^{-10}m$ (Conversion factor: $1m=10^{12}pm$ )

d = spacing between the crystal planes = ?

$\theta$ = angle of diffraction = 22.20°

Putting values in above equation, we get:

$2\times 1.54\times 10^{-10}=2d\sin (22.20)\\\\d=\frac{2\times 1.54\times 10^{-10}}{2\times \sin (22.20)}\\\\d=4.07\times 10^{-10}m$

Hence, the spacing between the crystal planes is $4.07\times 10^{-10}m$

4 0

3 years ago

When the following aqueous solutions are mixed together, a precipitate forms. Balance the net ionic equation in standard form fo

rewona [7]

Answer:

For (a): The balanced net ionic equation is $2Ag^{+}(aq)+S^{2-}(aq)\rightarrow Ag_2S(s)$ and the sum of coefficients is 4

For (b): The balanced net ionic equation is $Pb^{2+}(aq)+2Cl^{-}(aq)\rightarrow PbCl_2(s)$ and the sum of coefficients is 4

For (c): The balanced net ionic equation is $Ca^{2+}(aq)+CO_3^{2-}(aq)\rightarrow CaCO_3(s)$ and the sum of coefficients is

For (d): The balanced net ionic equation is $Ba^{2+}(aq)+2OH^{-}(aq)\rightarrow Ba(OH)_2(s)$ and the sum of coefficients is 4

For (e): The balanced net ionic equation is $Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s)$ and the sum of coefficients is 3

Explanation:

Net ionic equation is defined as the equations in which spectator ions are not included.

Spectator ions are the ones that are present equally on the reactant and product sides. They do not participate in the reaction.

For (a): Sodium sulfide and silver nitrate

The balanced molecular equation is:

$Na_2S(aq)+2AgNO_3(aq)\rightarrow 2NaNO_3(aq)+Ag_2S(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+S^{2-}(aq)+2Ag^+(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+Ag_2S(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$2Ag^{+}(aq)+S^{2-}(aq)\rightarrow Ag_2S(s)$

Sum of the coefficients = [2 + 1 + 1] = 4

For (b): Lead(II) nitrate and sodium chloride

The balanced molecular equation is:

$2NaCl(aq)+Pb(NO_3)_2(aq)\rightarrow 2NaNO_3(aq)+PbCl_2(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+2Cl^{-}(aq)+Pb^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+PbCl_2(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Pb^{2+}(aq)+2Cl^{-}(aq)\rightarrow PbCl_2(s)$

Sum of the coefficients = [2 + 1 + 1] = 4

For (c): Calcium nitrate and potassium carbonate

The balanced molecular equation is:

$K_2CO_3(aq)+Ca(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+CaCO_3(s)$

The complete ionic equation follows:

$2K^{+}(aq)+CO_3^{2-}(aq)+Ca^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2K^+(aq)+2NO_3^-(aq)+CaCO_3(s)$

As potassium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Ca^{2+}(aq)+CO_3^{2-}(aq)\rightarrow CaCO_3(s)$

Sum of the coefficients = [1 + 1 + 1] = 3

For (d): Barium nitrate and sodium hydroxide

The balanced molecular equation is:

$2NaOH(aq)+Ba(NO_3)_2(aq)\rightarrow 2NaNO_3(aq)+Ba(OH)_2(s)$

The complete ionic equation follows:

$2Na^{+}(aq)+2OH^{-}(aq)+Ba^{2+}(aq)+2NO_3^{-}(aq)\rightarrow 2Na^+(aq)+2NO_3^-(aq)+Ba(OH)_2(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions

The net ionic equation follows:

$Ba^{2+}(aq)+2OH^{-}(aq)\rightarrow Ba(OH)_2(s)$

Sum of the coefficients = [2 + 1 + 1] = 4

For (e): Silver nitrate and sodium chloride

The balanced molecular equation is:

$NaCl(aq)+AgNO_3(aq)\rightarrow NaNO_3(aq)+AgCl(s)$

The complete ionic equation follows:

$Na^{+}(aq)+Cl^{-}(aq)+Ag^{+}(aq)+NO_3^{-}(aq)\rightarrow Na^+(aq)+NO_3^-(aq)+AgCl(s)$

As sodium and nitrate ions are present on both sides of the reaction. Thus, they are considered spectator ions.

The net ionic equation follows:

$Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s)$

Sum of the coefficients = [1 + 1 + 1] = 3

8 0

3 years ago