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3 years ago

What is the recommended way of preparing 100ml of 0.02M from 2M NaOH in the laboratory?

Chemistry

1 answer:

Zigmanuir [339]3 years ago

6 0

Answer:

The recommended way of preparing 100ml of 0.02 M from 2M NaOH in the laboratory is to obey the equation V1S1=V2S2.

Explanation:

According to the given question

V2=100ml

S2=0.02M

S1= 2M

Then V1=?

We all know that V1S1=V2S2

or V1=V2×S2÷S1

or V1= 100×0.02÷2

or V1 =1 ml

It can be concluded that 1ml of 2M NaOH solution and 99 ml distilled water should be mixed to prepare 100ml of 0.02 M NaOH solution.

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Why do two oxygen atoms form a double covalent bond?

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A lone oxygen atom has 6 electrons in its outer shell which is not very stable, whereas as full octet (8 outer shell electrons) is stable. In order to achieve this two oxygen atoms will share 4 electrons, each contributing 2 electrons. Since these electrons exist within the orbitals of both atoms, to oxygen atoms essentially achieve a full octet.

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3 years ago

Balance this equation and pick the answer with the coefficients that belong there.

geniusboy [140]

Answer:

2,3,6,1

Explanation:

The unbalanced reaction expression is given as:

AlBr₃ + K₂SO₄ → KBr + Al₂(SO₄)₃

We need to balanced this reaction equation. Our approach is a mathematical method where we assign variable a,b,c and d as the coefficients.

aAlBr₃ + bK₂SO₄ → cKBr + dAl₂(SO₄)₃

Conserving Al; a = 2d

Br: 3a = c

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S: b = 3d

O: 4b = 12d

Let a = 1, c = 3, d = $\frac{1}{2}$ b = $\frac{3}{2}$

Multiply through by 2 to give;

a = 2, b = 3, c = 6 and d = 1

2AlBr₃ + 3K₂SO₄ → 6KBr + Al₂(SO₄)₃

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2 years ago

Which scientist determined the charge of the electron?.

Alexus [3.1K]

Answer:

J.J. Thomson

Explanation:

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2 years ago

The following molecular equation represents the reaction that occurs when aqueous solutions of silver(I) nitrate and calcium chl

dsp73

Answer:

Ag+(aq) + Cl-(aq) —> AgCl(s)

Explanation:

2AgNO3(aq) + CaCl2(aq) —>2AgCl(s) + Ca(NO3)2(aq)

The balanced net ionic equation for the reaction above can be obtained as follow:

AgNO3(aq) and CaCl2(aq) will dissociate in solution as follow:

AgNO3(aq) —> Ag+(aq) + NO3-(aq)

CaCl2(aq) —> Ca2+(aq) + 2Cl-(aq)

AgNO3(aq) + CaCl2(aq) –>

2Ag+(aq) + 2NO3-(aq) + Ca2+(aq) + 2Cl-(aq) —> 2AgCl(s) + Ca2+(aq) + 2NO3-(aq)

Cancel out the spectator ions i.e Ca2+(aq) and 2NO3- to obtain the net ionic equation.

2Ag+(aq) + 2Cl-(aq) —> 2AgCl(s)

Divide through by 2

Ag+(aq) + Cl-(aq) —> AgCl(s)

The, the net ionic equation is

Ag+(aq) + Cl-(aq) —> AgCl(s)

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3 years ago

If you have 10 grams of Lithium Oxide what will the volume be? Show your work

gogolik [260]

Answer:

Volume occupied by 10 grams of Lithium Oxide at STP is 7.52 L

Explanation:

STP (Standard Temperature and Pressure) : It is used while performing calculation on gases and provide standards of measurements while performing experiment.

According to IUPAC (International Union of Pure And applied Chemistry) , standard temperature is 273.15 K and Pressure is 100 kPa (0.9869 atm). At STP condition 1 mole of substance occupies 22.4 L volume .

Molar mass of Lithium Oxide = 29.8 g/mol

$Li_{2}O$ = 2(6.9) + 15.99 = 29.8

Mass of 1 mole $Li_{2}O$ = 29.8 g

1 mole $Li_{2}O$ occupies, Volume = 22.4 L

29.8 g $Li_{2}O$ occupies, V = 22.4 L

1 g $Li_{2}O$ occupies ,V = $\frac{22.4}{29.8}$

1 g $Li_{2}O$ occupies ,V = 0.7516 L

10 g tex]Li_{2}O[/tex] occupies ,V = $0.7516 \times10$ L

V = 7.52 L

So, volume occupied by Lithium Oxide At STP is 7.52 L

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3 years ago

What is the recommended way of preparing 100ml of 0.02M from 2M NaOH in the laboratory?​

What is the recommended way of preparing 100ml of 0.02M from 2M NaOH in the laboratory?