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bogdanovich [222]
2 years ago
13

Suppose you have samples of three unknown solids. Explain how you could use their properties to

Chemistry
1 answer:
alexandr402 [8]2 years ago
7 0

Answer:https://es.khanacademy.org/science/ap-chemistry-beta/x2eef969c74e0d802:intermolecular-forces-and-properties/x2eef969c74e0d802:properties-of-solids/v/ionic-solids

Explanation:

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One single covalent bond, hope this helps!
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The student who answers the riddle will get the prize ? underline pronoun<br>.​
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the student <u>who</u> answers the riddle will get the prize

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Give the name of reaction in the<br> preparation of alkene from alcohol<br> compounds.
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3 0
3 years ago
Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin
STALIN [3.7K]

Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of KI and AgNO_3.

\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}

Molar mass of KI = 166 g/mole

\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole

and,

\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

KI+AgNO_3\rightarrow KNO_3+AgI

From the balanced reaction we conclude that

As, 1 mole of KI react with 1 mole of AgNO_3

So, 0.0178 mole of KI react with 0.0178 mole of AgNO_3

From this we conclude that, AgNO_3 is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgI

From the reaction, we conclude that

As, 1 mole of KI react to give 1 mole of AgI

So, 0.0178 moles of KI react to give 0.0178 moles of AgI

Thus,

Moles of AgI = Moles of I^- anion = Moles of Ag^+ cation = 0.0178 moles

Now we have to calculate the molarity of iodide anion in the solution.

\text{Concentration of }AgNO_3=\frac{\text{Moles of }AgNO_3}{\text{Volume of solution}}

\text{Concentration of }AgNO_3=\frac{0.0178mol}{0.350L}=0.0508M

Therefore, the final molarity of iodide anion in the solution is 0.0508 M.

3 0
3 years ago
I NEED HELP ASAP! PLEASE BE GENUINE
love history [14]

1. The molar mass of the unknown gas obtained is 0.096 g/mol

2. The pressure of the oxygen gas in the tank is 1.524 atm

<h3>Graham's law of diffusion </h3>

This states that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass i.e

R ∝ 1/ √M

R₁/R₂ = √(M₂/M₁)

<h3>1. How to determine the molar mass of the gas </h3>
  • Rate of unknown gas (R₁) = 11.1 mins
  • Rate of H₂ (R₂) = 2.42 mins
  • Molar mass of H₂ (M₂) = 2.02 g/mol
  • Molar mass of unknown gas (M₁) =?

R₁/R₂ = √(M₂/M₁)

11.1 / 2.42 = √(2.02 / M₁)

Square both side

(11.1 / 2.42)² = 2.02 / M₁

Cross multiply

(11.1 / 2.42)² × M₁ = 2.02

Divide both side by (11.1 / 2.42)²

M₁ = 2.02 / (11.1 / 2.42)²

M₁ = 0.096 g/mol

<h3>2. How to determine the pressure of O₂</h3>

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

  • Volume (V) = 438 L
  • Mass of O₂ = 0.885 kg = 885 g
  • Molar mass of O₂ = 32 g/mol
  • Mole of of O₂ (n) = 885 / 32 = 27.65625 moles
  • Temperature (T) = 21 °C = 21 + 273 = 294 K
  • Gas constant (R) = 0.0821 atm.L/Kmol
  • Pressure (P) =?

The pressure of the gas can be obtained by using the ideal gas equation as illustrated below:

PV = nRT

Divide both side by V

P = nRT / V

P = (27.65625 × 0.0821 × 294) / 438

P = 1.524 atm

Learn more about Graham's law of diffusion:

brainly.com/question/14004529

Learn more about ideal gas equation:

brainly.com/question/4147359

6 0
2 years ago
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