At the start of the _________, about 10,000 years ago, humans began planting crops and domesticating animals.

Consider the nitration by electrophilic aromatic substitution of salicylamide to iodosalicylamide. Reaction scheme illustrating

kvv77 [185]

<u>Answer:</u> The percent yield of the reaction is 68.68%.

<u>Explanation:</u>

To calculate the mass of salicylamide, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of salicylamide = 1.06 g/mL

Volume of salicylamide = 3.65 mL

Putting values in above equation, we get:

$1.06g/mL=\frac{\text{Mass of salicylamide}}{3.65mL}\\\\\text{Mass of salicylamide}=(1.06g/mL\times 3.65mL)=3.869g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of salicylamide = 3.869 g

Molar mass of salicylamide = 137.14 g/mol

Putting values in equation 1, we get:

$\text{Moles of salicylamide}=\frac{3.869g}{137.14g/mol}=0.0295mol$

The chemical equation for the conversion of salicylamide to iodo-salicylamide follows:

$\text{salicylamide }+NaI+NaOCl+EtOH\rightarrow \text{iodo-salicylamide }$

By Stoichiometry of the reaction:

1 mole of salicylamide produces 1 mole of iodo-salicylamide

So, 0.0295 moles of salicylamide will produce = $\frac{1}{1}\times 0.0295=0.0295moles$ of iodo-salicylamide

Now, calculating the mass of iodo-salicylamide from equation 1, we get:

Molar mass of iodo-salicylamide = 263 g/mol

Moles of iodo-salicylamide = 0.0295 moles

Putting values in equation 1, we get:

$0.0295mol=\frac{\text{Mass of iodo-salicylamide}}{263g/mol}\\\\\text{Mass of iodo-salicylamide}=(0.0295mol\times 263g/mol)=7.76g$

To calculate the percentage yield of iodo-salicylamide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of iodo-salicylamide = 5.33 g

Theoretical yield of iodo-salicylamide = 7.76 g

Putting values in above equation, we get:

$\%\text{ yield of iodo-salicylamide}=\frac{5.33g}{7.76g}\times 100\\\\\% \text{yield of iodo-salicylamide}=68.68\%$

Hence, the percent yield of the reaction is 68.68%.

4 0

4 years ago

POINTSSS!!!!!!

Lelu [443]

1 mole CO₂----------- 6.02x10²³ molecules
4 moles CO₂ ---------- ??

4 x ( 6.02x10²³) / 1 =

= 2.41 x 10²⁴ / 1 => 2.41 x 10²⁴ molecules of CO₂

Answer C

4 0

3 years ago

How many moles are in 8.63 x 103 atoms of Li?

Ira Lisetskai [31]

<h3>Answer:</h3>

1.43 × 10⁻²⁰ mol Li

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<h3>Explanation:</h3>

<u>Step 1: Define</u>

8.63 × 10³ atoms Li

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

Avogadro's Number

<u>Step 3: Convert</u>

Set up: $\displaystyle 8.63 \cdot 10^3 \ atoms \ Li(\frac{1 \ mol \ Li}{6.022 \cdot 10^{23} \ atoms \ Li})$
Multiply/Divide: $\displaystyle 1.43355 \cdot 10^{-20} \ moles \ Li$

<u>Step 4: Check</u>

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

1.43355 × 10⁻²⁰ mol Li ≈ 1.43 × 10⁻²⁰ mol Li

4 0

3 years ago

How do you solve for the Atomic Mass in chemistry?

loris [4]

Https://www.google.com/search?q=how+to+solve+fir+atomic+mass+in+chemisty&ie=UTF-8&oe=UTF-8&hl=en-us&client=safari#kpvalbx=1
Here is the link to a great video that explains your question nicely, hope this helps.

6 0

4 years ago

Read 2 more answers

What are the coefficients when the chemical equation below is balanced?

timofeeve [1]

Answer:

C is correct

Explanation:

4 0

3 years ago