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nadya68 [22]
3 years ago
9

I begin the reaction with 0.45 g of beryllium. If my actual yield of beryllium chloride (mm = 79.91 g/mol) was 3.5 grams, what w

as my percent yield?
Please show work
Chemistry
1 answer:
trasher [3.6K]3 years ago
5 0

The percentage of yield was 777.78%

<u>Explanation:</u>

We have the equation,

Be [s]  +  2 HCl [aq]  →  BeCl 2(aq]  + H 2(g]  ↑ Be (s]  + 2 HCl [aq]  →  BeCl 2(aq]  + H 2(g] ↑

To find the percent yield we have the formula

Percentage of Yield= what you actually get/ what you should theoretically get  x 100

                                   =3.5 g/0.45 g 100

                                    = 777.78 %

The percentage of yield was 777.78%

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How many molecules of carbon dioxide are in 5.61 moles of carbon dioxide (CO2)?
gulaghasi [49]
<h3>Answer:</h3>

3.38 × 10²⁴ molecules CO₂

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

<u>Math</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>

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<u>Stoichiometry</u>

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 5.61 moles CO₂

[Solve] molecules CO₂

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                      \displaystyle 5.61 \ mooles \ CO_2(\frac{6.022 \cdot 10^{23} \ molecules \ CO_2}{1 \ mol \ CO_2})
  2. [DA] Multiply/Divide [Cancel out units]:                                                          \displaystyle 3.37834 \cdot 10^{24} \ molecules \ CO_2

<u>Step 4: Check</u>

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

3.37834 × 10²⁴ molecules CO₂ ≈ 3.38 × 10²⁴ molecules CO₂

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