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AlexFokin [52]
3 years ago
8

A sample of iron (III) chloride has a mass of 26.29g. How many moles would this be?

Chemistry
1 answer:
vekshin13 years ago
8 0
<h3>Answer:</h3>

0.1621 mol FeCl₃

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Compounds</u>

  • Determining compound formulas

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

26.29 g FeCl₃ (iron (III) chloride)

<u>Step 2: Identify Conversions</u>

Molar Mass of Fe - 55.85 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of FeCl₃ - 55.85 + 3(35.45) = 162.2 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 26.29 \ g \ FeCl_3(\frac{1 \ mol \ FeCl_3}{162.2 \ g \ FeCl_3})
  2. Multiply:                            \displaystyle 0.162084 \ moles \ FeCl_3

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 4 sig figs.</em>

0.162084 mol FeCl₃ ≈ 0.1621 mol FeCl₃

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Given the Henry’s law constant for O2(4.34*109Pa) at 25° C, calculate the molar concentration of oxygen in air-saturated and O2s
Flauer [41]

This is an incomplete question, here is a complete question.

The Henry's law constant for oxygen dissolved in water is 4.34 × 10⁹ g/L.Pa at 25⁰C.If the partial pressure of oxygen in air is 0.2 atm, under atmospheric conditions, calculate the molar concentration of oxygen in air-saturated and oxygen saturated water.

Answer : The molar concentration of oxygen is, 2.67\times 10^2mol/L

Explanation :

As we know that,

C_{O_2}=k_H\times p_{O_2}

where,

C_{O_2} = molar solubility of O_2 = ?

p_{O_2} = partial pressure of O_2 = 0.2 atm  = 1.97×10⁻⁶ Pa

k_H = Henry's law constant  = 4.34 × 10⁹ g/L.Pa

Now put all the given values in the above formula, we get:

C_{O_2}=(4.34\times 10^9g/L.Pa)\times (1.97\times 10^{-6}Pa)

C_{O_2}=8.55\times 10^3g/L

Now we have to molar concentration of oxygen.

Molar concentration of oxygen = \frac{8.55\times 10^3g/L}{32g/mol}=2.67\times 10^2mol/L

Therefore, the molar concentration of oxygen is, 2.67\times 10^2mol/L

8 0
3 years ago
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Answer:

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Explanation:

Distillation, or classical distillation, is the process of separating the components or substances from a liquid mixture by using selective <u>boiling and condensation.</u>

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3 years ago
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Answer:

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Explanation:

Given:

P = 15 MPa

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T = 350 °C

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= 18 g/mol

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R = 0.4615 kPa·m3/kg·K

Using ideal gas equation,

P × V = n × R × T

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V = (R × T)/P

V/M = (R × T)/P × m

= (0.4615 × 623)/1.5 × 10^4

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3 years ago
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As bromine is higher up in the periodic table than Iodine, it would have a smaller radius. Iodine would have a larger radius.

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1. MG

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4. NA

5. O

6. K

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