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VashaNatasha [74]

3 years ago

9

The theoretical yield of 1,2-epoxycyclohexane is _______________ grams, when starting with 3.0 grams of trans-2-bromocyclohexano

l. (Enter the number using 3 significant figures, i.e. 1.22)

1 answer:

Amanda [17]3 years ago

7 0

Answer:

1.64g

Explanation:

The reaction scheme is given as;

2-bromocyclohexanol --> 1,2-epoxycyclohexane + HBr

From the reaction above,

1 mol of 2-bromocyclohexanol produces 1 mol of 1,2-epoxycyclohexane

3.0 grams of trans-2-bromocyclohexanol.

Molar mass = 179.05 g/mol

Number of moles = mass / molar mass = 3 / 179.05 = 0.016755 mol

This means 0.016755 mol of 1,2-epoxycyclohexane would be produced.

Molar mass = 98.143 g/mol

Theoretical yield = Number of moles * Molar mass

Theoretical yield = 0.016755 * 98.143 ≈ 1.64g

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Glucose is used by intestinal cells and red blood cells, while the rest reaches the liver, adipose tissue and muscle cells, where it is absorbed and stored as glycogen.

(it is saved to be used later)

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3 years ago

Choose all the correct information concerning chemical reactions: Involves small amounts of energy. Some of the mass is converte

suter [353]

<span>A chemical reactions :
Involves small amounts of energy.
</span><span> Only electrons are involved.
</span><span>The mass remains the same.</span>

8 0

3 years ago

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The stock concentration of NaOH is 0.5M. 0.5M = 0.500mol/1.0L.

valentinak56 [21]

Answer:a) 0.1 mole. b) 4g. c) 2% d) 196 mL

Explanation: in 200mL , 0.1mole

mw NaOH = 40g/mol —> 4g in 0.1 mole

4g in 200mL so 2g in 100mL

density NaOH = 1g/mL so if 4g in 200 mL, 4mL , 196 mL water

5 0

3 years ago

A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce

spin [16.1K]

Explanation:

The given data is as follows.

Volume of lake = $15 \times 10^{6} m^{3}$ = $15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}$

Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = $5.6 \times 15 \times 10^{9} mg$

= $84 \times 10^{9}$ mg

= $84 \times 10^{3}$ kg

Flow rate of river is 50 $m^{3} sec^{-1}$

Volume of water in 1 day = $50 \times 10^{3} \times 86400 liter$

= $432 \times 10^{7}$ liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are $2.9792 \times 10^{10} mg$ or $2.9792 \times 10^{4} kg$

Flow rate of sewage = $0.7 m^{3} sec^{-1}$

Volume of sewage water in 1 day = $6048 \times 10^{4}$ liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = $1.8144 \times 10^{10} mg$ or $1.8144 \times 10^{4}kg$

Therefore, total concentration of lake after 1 day = $\frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l$

= 6.8078 mg/l

$k_{D}$ = 0.2 per day

$L_{o}$ = 6.8078

Hence, $L_{liquid}$ = $L_{o}(1 - e^{-k_{D}t}$

$L_{liquid}$ = $6.8078 (1 - e^{-0.2 \times 1})$

= 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

= 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

5 0

4 years ago

Solid sodium azide (NaN3) produces solid sodium and nitrogen gas. How many grams of sodium azide are needed to yield a volume of

ch4aika [34]

Answer:

52.008 grams of sodium azide are needed to yield a volume of 26.5 L of nitrogen gas at a temperature of 295 K and a pressure of 1.10 atmospheres.

Explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P*V = n*R*T

In this case, the balanced reaction is:

2 NaN₃ → 2 Na + 3 N₂

You know the following about N₂:

P= 1.10 atm
V= 26.5 L
n=?

$R=0.082057 \frac{atm*L}{mol*K}$

T= 295 K

Replacing in the equation for ideal gas:

1.10 atm* 26.5 L= n* $0.082057 \frac{atm*L}{mol*K}$ *295 K

Solving:

$n=\frac{1.10 atm*26.5 L}{0.082057 \frac{atm*L}{mol*K} *295K}$

n= 1.2 moles

Now, the following rule of three can be applied: if 3 moles of N₂ are produced by stoichiometry of the reaction from 2 moles of NaN₃, 1.2 moles of N₂ are produced from how many moles of NaN₃?

$moles of NaN_{3}=\frac{1.2 molesofN_{2} *2 molesofNaN_{3} }{3 molesofN_{2} }$

moles of NaN₃= 0.8

Since the molar mass of sodium azide is 65.01 g / mol, then one last rule of three applies: if 1 mol has 65.01 grams of NaN₃, 0.8 mol how much mass does it have?

$mass of NaN_{3} =\frac{0.8 mol*65.01 grams}{1 mol}$

mass of NaN₃=52.008 grams

<u><em>52.008 grams of sodium azide are needed to yield a volume of 26.5 L of nitrogen gas at a temperature of 295 K and a pressure of 1.10 atmospheres.</em></u>

6 0

4 years ago

Read 2 more answers