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1 year ago

Calculate the theoretical yield for the bromination of both stilbenes

Chemistry

1 answer:

podryga [215]1 year ago

4 0

Answer:

cinnamic acid - 150 mg

cis-stilbene - 100 μL

trans- stilbene - 100 mg

pyridinium tribromide - 200-385 mg

For this data:

moles of cinnamic acid = 0.150 g/148.16 g/mol = 0.001 mols

Theoretical mass of dibromoproduct formed = 0.001 mol x 307.97 g/mol = 0.312 g

cis-stilbene (100 ul = 0.1 ml)

moles of cis-stilbene = 0.1 ml x 1.01 g/mol/180.25 g/mol = 0.00056 mols

Theoretical mass of dibromoproduct formed = 0.00056 mol x 340.05 g/mol = 0.19 g

trans-stilbene

moles of tran-stilbene = 0.1 g/180.25 g/mol = 0.00055 mols

Theoretical mass of dibromoproduct formed = 0.00055 mol x 340.05 g/mol = 0.19 g

Explanation:

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What is a food web made up of

brilliants [131]

Answer:

A food web is made up of all the food chains in a single ecosystem. Each living thing in an ecosystem belongs to many food chains. A food chain is a path that energy takes through a certain ecosystem. Trophic Levels. Organisms in food webs are grouped into categories called trophic levels.

Explanation:

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7 0

2 years ago

Read 2 more answers

Rasheed calculates that his chemical reaction should produce 4 moles of product, but when he does the experiment, he gets only 3

Brut [27]

Answer:

The answer to your question is 75%

Explanation:

Data

Theoretical production = 4 moles

Experimental production = 3 moles

Percent yield = ?

Formula

$Percent yield = \frac{Experimental production}{Theoretical production} x 100$

Substitution

$Percent yield = \frac{3}{4} x 100$

Result

Percent yield = 75 %

8 0

2 years ago

There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.