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2 years ago

A solution of pH 3 was mixed with a solution of pH 9. What is the final pH could be ?

Chemistry

1 answer:

chubhunter [2.5K]2 years ago

3 0

The solution of which contains of PH 3 called acidic medium which is more stronger acid called HCl when reacts to the stronger base base NaOH forms the acid base reaction .

According to acid base equation you consider ph5 as PH 9 if you understand it then tell me below

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Whats the voltage of CuCl2 + Zn -> ZnCl2 + Cu

gtnhenbr [62]

Answer:

Approximately $1.10\; {\rm V}$ under standard conditions.

Explanation:

Equation for the overall reaction:

${\rm CuCl_{2}}\, (aq) + {\rm Zn}\, (s) \to {\rm ZnCl_{2}} \, (aq) + {\rm Cu}\, (s)$ .

Write down the ionic equation for this reaction:

$\begin{aligned}& {\rm Cu^{2+}}\, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Zn}\, (s)\\ & \to {\rm Zn^{2+}} \, (aq) + 2\; {\rm Cl^{-}}\, (aq) + {\rm Cu}\, (s)\end{aligned}$ .

The net ionic equation for this reaction would be:

${\rm Cu^{2+}}\, (aq) + {\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + {\rm Cu}\, (s)$ .

In this reaction:

Zinc loses electrons and was oxidized (at the anode): ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ .
Copper gains electrons and was reduced (at the cathode): ${\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s)$ .

Look up the standard potentials for each half-reaction on a table of standard reduction potentials.

Notice that ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ is oxidation and is likely not on the table of standard reduction potentials. However, the reverse reaction, ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ , is reduction and is likely on the table.

$E(\text{anode}) = -0.7618\; {\rm V}$ for ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ , and
$E(\text{cathode}) = 0.3419\; {\rm V}$ for ${\rm Cu^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Cu} \, (s)$ .

The reduction potential of ${\rm Zn}\, (s) \to {\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}}$ would be $-E(\text{anode}) = -(-0.7618\; {\rm V}) = 0.7618\; {\rm V}$ , the opposite of the reverse reaction ${\rm Zn^{2+}}\, (aq) + 2\, {\rm e^{-}} \to {\rm Zn}\, (s)$ .

The standard potential of the overall reaction would be the sum of the standard potentials of the two half-reactions:

$\begin{aligned} E^{\circ} &= E^{\circ}(\text{cathode}) + (-E^{\circ}(\text{anode})) \\ &= 0.3419 - (-0.7618\; {\rm V}) \\ &\approx 1.10\; {\rm V}\end{aligned}$ .

7 0

1 year ago

Osmium metal, the densest element, has a density of 22.6 g/mL, while hydrogen, the least dense element, has a density of 8.99 e-

KengaRu [80]

Answer:

1. Volume of osmium = 0.044 mL

2. Volume of Hydrogen = 11123.47 mL

3. Osmium is 251390 times denser than hydrogen

Explanation:

From the question given above, the following data were obtained:

Density of osmium = 22.6 g/mL

Density of Hydrogen = 8.99×10¯⁵ g/mL

Mass of osmium = 1 g

Mass of Hydrogen = 1 g

1. Determination of the Volume of osmium.

Density of osmium = 22.6 g/mL

Mass of osmium = 1 g

Volume of osmium =?

Density = mass /volume

22.6 = 1 / volume

Cross multiply

22.6 × volume = 1

Divide both side by 22.6

Volume = 1 / 22.6

Volume of osmium = 0.044 mL

2. Determination of the Volume of Hydrogen.

Density of osmium = 8.99×10¯⁵ g/mL

Mass of osmium = 1 g

Volume of osmium =?

Density = mass /volume

8.99×10¯⁵ = 1 / volume

Cross multiply

8.99×10¯⁵ × volume = 1

Divide both side by 8.99×10¯⁵

Volume = 1 / 8.99×10¯⁵

Volume of Hydrogen = 11123.47 mL

3. Determination of the number of times osmium is denser than hydrogen.

Density of osmium (Dₒ) = 22.6 g/mL

Density of Hydrogen (Dₕ) = 8.99×10¯⁵ g/mL

Dₒ / Dₕ = 22.6 / 8.99×10¯⁵

Dₒ / Dₕ = 251390

Cross multiply

Dₒ = 251390 × Dₕ

Thus, osmium is 251390 times denser than hydrogen.

5 0

2 years ago

What is the next atomic orbital in the series below? 1s, 2s, 2p, 3s, 3p, ___

Angelina_Jolie [31]

D.4s this would be the answer because after eight electrons are added to the third orbital two have to be added to the fourth before more can be added to the third one. The third orbital can hold 18 electrons, but it first has to have eight, then two on the fourth orbital, then the rest of the ten can be added

7 0

2 years ago

Read 2 more answers

13) The velocity of an object dropped in air will continue to increase

melamori03 [73]

Answer:

Because due to the presence air resistance

7 0

2 years ago

Vanillin, a flavoring agent, is made up of carbon, hydrogen, and Oxygen atoms. When a sample of Vanillin weighing 2.500g burns i

kirza4 [7]

To determine the empirical formula of the compound given, we need to determine the ratio of each element in the compound. To do that, we need to know the amount of carbon, hydrogen and oxygen atoms in the compound. We determine these from the given amounts of the products after burning.

Given:
2.5 g vanillin
5.79 g CO2
1.18 g H2O

Solution:

mol C = 5.79 g CO2 ( 1 mol CO2 / 44.01 g CO2) ( 1 mol C / 1 mol CO2) = 0.132 mol C

mol H = 1.18 g H2O ( 1 mol g H2O / 18.02 g H2O) ( 2 mol H / 1 mol H2O) = 0.131 mol H

We convert them into grams,
0.132 mol C (12.01 g / 1 mol) = 1.59 g
0.131 mol H (1.01 g / 1 mol) = 0.132 g
mass of C and H in vanillin = 1.722 g
mass of O = 2.5 - 1.722 = 0.778 g O

mol O = 0.778 g O ( 1 mol O / 16 g ) = 0.049 mol O

Dividing the number of moles of each element with the smallest value of number of mole, we will have the empirical formula:

moles ratio
C 0.132 / 0.049 2.69 x 3 = 8
H 0.131 / 0.049 2.67 x 3 = 8
O 0.049 / 0.049 1 x 3 = 3

The empirical formula would be C8H8O3.

6 0

3 years ago