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riadik2000 [5.3K]
3 years ago
8

How many atoms of aluminium are contained in 2.75 mol

Chemistry
1 answer:
horsena [70]3 years ago
4 0
<h3>Answer:</h3>

1.66 × 10²⁴ atoms Al

<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> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 2.75 mol Al

[Solve] atoms Al

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                     \displaystyle 2.75 \ mol \ Al(\frac{6.022 \cdot 10^{23} \ atoms \ Al}{1 \ mol \ Al})
  2. [DA] Multiply [Cancel out units]:                                                                       \displaystyle 1.65605 \cdot 10^{24} \ atoms \ Al

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.65605 × 10²⁴ atoms Al ≈ 1.66 × 10²⁴ atoms Al

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here, m = total mass = 475 + 125 = 600 g
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7 0
3 years ago
The system described by the reaction CO(g)+Cl2(g)⇌COCl2(g) is at equilibrium at a given temperature when PCO= 0.26 atm , PCl2= 0
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<u>Answer:</u> The new pressure of CO is 0.09 atm

<u>Explanation:</u>

For the given chemical reaction:

CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

The expression of K_p for above equation follows:

K_p=\frac{p_{COCl_2}}{p_{CO}\times p_{Cl_2}}             .......(1)

We are given:

p_{COCl_2}=0.66atm\\p_{CO}=0.26atm\\p_{Cl_2}=0.15atm

Putting values in above equation, we get:

K_p=\frac{0.66}{0.15\times 0.26}\\\\K_p=16.92

Addition pressure of chlorine added = 0.39 atm

Now, the equilibrium is re-established:

                    CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

Initial:            0.26     0.15             0.66

At eqllm:     0.26-x   0.54            0.66+x

Putting values in expression 1, we get:

16.92=\frac{(0.66+x)}{(0.26-x)\times 0.54}\\\\x=0.17

So, new pressure of CO = 0.26 - x = (0.26 - 0.17) = 0.09 atm

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4 0
3 years ago
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Temperature is an intensive property. Therefore temperature of combined sample of lead will be same as with individual pellets.

temperature of combined sample = 20^{0}\textrm{C}

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