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adoni [48]
3 years ago
10

A buret was improperly read to 1 decimal place giving a reading of 27.2 mL. If the actual volume is 27.26 mL, what is the percen

t error caused by this mistake?
Chemistry
1 answer:
Elenna [48]3 years ago
8 0
Using the significant figure it would be 27.3
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Predict the two most likely mechanisms for the reaction of 1-iodohexane with potassium tert-butoxide.
Rudiy27

Answer:

C) SN2 and E2

Explanation:

For this question, we have analyzed the <u>substrate</u> and the <u>base/nucleophile</u>. The substrate, in this case, is 1-iodohexane and the base/nucleophile is potassium tert-butoxide.

<u>Substrate</u>

<u />

In the 1-iodohexane the iodide "I" is bonded to a primary carbon (carbon 1). Therefore we will have a <u>primary substrate</u>. If we have a primary substrate an Sn1 can not take place. We can not have a <u>primary carbocation</u> due to this instability. So, we can disccard options A) and B).

<u>Base/nucleophile</u>

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In the potassium tert-butoxide we have an ionic compound. A positive charge is placed in the potassium atom a negative charge is placed in the oxygen of the ter-butoxide ion. So, we will have a <u>strong base</u> (a molecule with the ability to remove electrons) and a <u>strong nucleophile</u> (a molecule with ability to bond with an electrophile). With all this in mind, w<u>e can not have an E1 reaction</u>.

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Calculate the number of atoms in 0.25 moles of manganese?
White raven [17]
<h3>Answer:</h3>

1.5 × 10²³ atoms Mn

<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>Atomic Structure</u>

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

<u>Step 1: Define</u>

0.25 mol Mn

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 0.25 \ mol \ Mn(\frac{6.022 \cdot 10^{23} \ atoms \ Mn}{1 \ mol \ Mn})
  2. Multiply:                            \displaystyle 1.5055 \cdot 10^{23} \ atoms \ Mn

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

1.5055 × 10²³ atoms Mn ≈ 1.5 × 10²³ atoms Mn

3 0
3 years ago
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