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in an experiment 3.425g of lead oxide was reduced to form 3.105g of lead the empirical formula of the lead oxide is?

Answer:

Pb3O4

Explanation:

According to this question, 3.425g of lead oxide was reduced to form 3.105g of lead in an experiment. Since lead oxide contains both lead (Pb) and oxygen (O) element,

Mass of lead oxide = 3.425g

Mass of lead = 3.105g

Mass of oxygen = (3.425g - 3.105g) = 0.320g

Next, we convert each mass value to mole by dividing by respective molar mass

Pb = 3.105g ÷ 207.2 = 0.0149mol

O = 0.320g ÷ 16 = 0.02mol

Next, we divide each mole value by the smallest (0.0149)

Pb = 0.0149mol ÷ 0.0149mol = 1

O = 0.02mol ÷ 0.0149mol = 1.342

Multiply each ratio value by 3 to get:

Pb = 1 × 3 = 3

O = 1.342 × 3 = 4.026

The whole number ratio, approximately, of Pb and O is 3:4, hence, their empirical formula is Pb3O4.

4 0

3 years ago

A solution with a ph of 4 has _________ the concentration of h+ present compared to a solution with a ph of 5.

Marina CMI [18]

A solution with a pH of 4 has ten times the concentration of H⁺ present compared to a solution with a pH of 5.
pH is a numeric scale for the acidity or basicity of an aqueous solution. It is the negative of the base 10 logarithm of the molar concentration of hydrogen ions.
[H⁺] = 10∧-pH.
pH = 4 → [H⁺]₁ = 10⁻⁴ M = 0,0001 M.
pH = 5 → [H⁺]₂ = 10⁻⁵ M = 0,00001 M.
[H⁺]₁ / [H⁺]₂ = 0,0001 M / 0,00001 M.
[H⁺]₁ / [H⁺]₂ = 10.

7 0

3 years ago

What volume of water would be added to 16.5 ml of a 0.0813 m solution of sodium borate in order to get a 0.0200 m s?

raketka [301]

When diluting solutions from concentrated solutions the following formula can be used
c1v1 = c2v2
where c1 is concentration and v1 is volume of the concentrated solution
and c2 is concentration and v2 is volume of the diluted solution to be prepared
substituting these values
0.0813 M x 16.5 mL = 0.0200 M x V
V = 67.1 mL
the volume of the diluted solution prepared is 67.1 mL.
the volume of water that should be added to get a final volume of 67.1 mL is (67.1 - 16.5 ) = 50.6 mL
a volume fo 50.6 mL should be added

7 0

3 years ago

ENTER THE COEFFICIENTS DIRECTLY FROM THE BALANCED CHEMICAL EQUATION. do not reduce to lowest terms. for example: if 3 moles H2 r

dlinn [17]

Answer:

The coefficients are 2 for H₂O and 1 for Ca(OH)₂.

Explanation:

Let's consider the following reaction.

Ca(OH)₂(aq) + 2 HCl(aq) → CaCl₂(aq) + 2 H₂O(l)

According to the balanced equation, the molar ratio of H₂O to Ca(OH)₂ is 2:1. Using this conversion factor, we have the following proportion:

moles Ca(OH)₂. (2 mol H₂O ÷ 1 mol Ca(OH)₂) = moles H₂O

4 0

3 years ago

How many of the following statements about silver acetate, AgCH3COO, are true? i) More AgCH3CoO(S) will dissolve if the pH of th

shutvik [7]

Answer:

All three statements are true

Explanation:

Solubility equilibrium of silver acetate:

$AgCH_{3}COO\rightleftharpoons Ag^{+}+CH_{3}COO^{-}$

If pH is increased then concentration of $H^{+}$ increases in solution resulting removal of $CH_{3}COO^{-}$ by forming $CH_{3}COOH$ . Hence, according to Le-chatelier principle, equilibrium will shift towards right. So more $AgCH_{3}COO$ will dissolve
If $AgNO_{3}$ is added then concentration of $Ag^{+}$ increases in solution resulting shifting of equilibrium towards left in accordance with Le-chatelier principle. So less $AgCH_{3}COO$ will dissolve
Insoluble precipitate of AgOH is formed by adding NaOH in solution resulting removal of $Ag^{+}$ . So, more $AgCH_{3}COO$ will dissolve

Hence all three statements are true

8 0

3 years ago