contrast copy and fill in this graphic organizer to contrast how different elements, isotopes, and ions are produced

Kaylis [27]

Answer:

Here

Explanation:

Chemists need the mole concept to bridge the gap between the microscopic world of atoms to the macroscopic world of humans. As you know, the molecular level consists of particles that are invisible to us.

7 0

3 years ago

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Consider the neutralization reaction 2 HNO 3 ( aq ) + Ba ( OH ) 2 ( aq ) ⟶ 2 H 2 O ( l ) + Ba ( NO 3 ) 2 ( aq ) A 0.120 L sample

k0ka [10]

Answer:

The concentration of the HNO3 solution is 0.150 M

Explanation:

Step 1: Data given

Volume of the unknown HNO3 sample = 0.120 L

Volume of the 0.200 M Ba(OH)2 = 45.1 mL

Step 2: The balanced equation

2HNO3 + Ba(OH)2 ⟶ Ba(NO3)2 + 2H2O

Step 3: Calculate moles Ba(OH)2

moles Ba(OH)2 = molarity * volume

moles Ba(OH)2 = 0.200 M * 0.0451 L

moles Ba(OH)2 = 0.00902 moles

Step 4: Calculate moles of HNO3

For 1 mole of Ba(OH)2 we need 2 moles of HNO3

For 0.00902 moles of Ba(OH)2 we need 2*0.00902 = 0.01804 moles

Step 5: Calculate molarity of HNO3

molarity = moles / volume

molarity = 0.01804 / 0.120 L

Molarity = 0.150 M HNO3

The concentration of the HNO3 solution is 0.150 M

6 0

3 years ago

The addition of 0.275 L of 1.62 M KCl to a solution containing Ag+ and Pb2+ ions is just enough to precipitate all of the ions

musickatia [10]

Answer:

The mass of PbCl₂ is 45.88 grams and the mass of AgCl is 16.48 grams.

Explanation:

As mentioned in the given question, the addition of 0.275 L of 1.62 M KCl is done in a solution that comprise Ag⁺ and Pb²⁺ ions so that all the ions get precipitated. Therefore, the moles of KCl present is,

Moles of KCl = 0.275 L × 1.62 M = 0.445 moles

Now the reaction will be,

Ag⁺ + Pb²⁺ + KCl ⇒ AgCl + PbCl₂ + 3K⁺

Now let us assume that the formation of x moles of AgCl and y moles of PbCl₂ is taking place.

Therefore, mass of AgCl will be x × molecular mass, which will be equal to x × 143.32 grams = 143.32 x grams

Now the mass of PbCl2 formed will be,

y × molecular mass = y × 278.1 grams = 278.1 y grams

Now the total precipitate will be,

62.37 grams = 143.32 x + 278.1 y -----------(i)

Now as AgCl and PbCl₂ requires 1:2 ratio of KCl, this shows that x moles of AgCl will require x moles of KCl and y mol of PbCl₂ will require 2*y moles of PbCl₂. Therefore,

x + 2y = total mass of KCl

x + 2y = 0.445 moles ------ (ii)

On solving equation (i) and (ii) we get,

x as 0.115 and y as 0.165

Now the mass of AgCl will be,

143.32 × 0.115 = 16.48 grams

The mass of PbCl₂ will be,

278.1 × 0.165 = 45.88 grams.

3 0

3 years ago

DO NOT ANSWER IF YOU DO NOT KNOW WHAT TO DO

Mice21 [21]

Answer:

im gunna save my answer so no one can take it:)

Explanation:

I tried my best

4 0

3 years ago

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Determine the pH of the resulting solution if 25 mL of 0.400 M strychnine (C21H22N2O2) is added to 50 mL of 0.200 M HCl? Assume

DIA [1.3K]

Answer:

pH = 4.56

Explanation:

The strychnine reacts with HCl as follows:

C₂₁H₂₂N₂O₂ + HCl ⇄ C₂₁H₂₂N₂O₂H⁺ + Cl⁻

For strychnine buffer:

pOH = 5.74 + log [C₂₁H₂₂N₂O₂H⁺] / [C₂₁H₂₂N₂O₂]

Initial moles of C₂₁H₂₂N₂O₂ are:

0.025L * (0.400 mol / L) = 0.01 moles C₂₁H₂₂N₂O₂

And of HCl are:

0.05L * (0.200 mol / L) = 0.01 moles HCl

That means after the reaction, you will have just 0.01 moles of C₂₁H₂₂N₂O₂H⁺ in 50mL + 25mL = 0.075L. And molarity is:

[C₂₁H₂₂N₂O₂H⁺] = 0.01 mol / 0.075L = 0.1333M

This conjugate acid, is in equilibrium with water as follows:

C₂₁H₂₂N₂O₂H⁺(aq) + H₂O(l) ⇄ C₂₁H₂₂N₂O₂ + H₃O⁺

Where Ka = Kw / Kb = 1x10⁻¹⁴ / 1.8x10⁻⁶ = 5.556x10⁻⁹

Ka is defined as:

Ka = 5.556x10⁻⁹ = [C₂₁H₂₂N₂O₂] [H₃O⁺] / [C₂₁H₂₂N₂O₂H⁺]

In equilibrium, concentrations are:

C₂₁H₂₂N₂O₂ = X

H₃O⁺ = X

C₂₁H₂₂N₂O₂H⁺ = 0.1333M - X

Replacing in Ka expression:

5.556x10⁻⁹ = [X] [X] / [0.1333M - X]

7.39x10⁻¹⁰ - 5.556x10⁻⁹X = X²

7.39x10⁻¹⁰ - 5.556x10⁻⁹X - X² = 0

Solving for X:

X = - 2.72x10⁻⁵M → False solution. There is no negative concentrations

X = 2.72x10⁻⁵M → Right solution.

As H₃O⁺ = X

H₃O⁺ = 2.72x10⁻⁵M

And pH = -log H₃O⁺

4 0

3 years ago