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anyanavicka [17]

3 years ago

14

8. Benzene is an aromatic hydrocarbon with molecular formula C6H6. a) Draw the ring structure of Benzene. [1] how will you prepa

re benzene from phenol

1 answer:

kvv77 [185]3 years ago

3 0

When phenol is heated with zinc dust benzene is obtained.

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I need help with this

Lelechka [254]

Answer:

The answer is its equal to the volume of its container.

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I hope this helps! :)

7 0

3 years ago

In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas to produce ga

Reptile [31]

Answer:

The rate at which ammonia is being produced is 0.41 kg/sec.

Explanation:

$N_2+3H_2\rightarrow 2NH_3$ Haber reaction

Volume of dinitrogen consumed in a second = 505 L

Temperature at which reaction is carried out,T= 172°C = 445.15 K

Pressure at which reaction is carried out, P = 0.88 atm

Let the moles of dinitrogen be n.

Using an Ideal gas equation:

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.88 atm\times 505 L}{0.0821 atm l/mol K\times 445.15 K}=12.1597 mol$

According to reaction , 1 mol of ditnitrogen gas produces 2 moles of ammonia.

Then 12.1597 mol of dinitrogen will produce :

$\frac{2}{1}\times 12.1597 mol=24.3194 mol$ of ammonia

Mass of 24.3194 moles of ammonia =24.3194 mol × 17 g/mol

=413.43 g=0.41343 kg ≈ 0.41 kg

505 L of dinitrogen are consumed in 1 second to produce 0.41 kg of ammonia in 1 second. So the rate at which ammonia is being produced is 0.41 kg/sec.

6 0

3 years ago

If you know your acid concentration is .650 M and you used 15.00 mL, what was the concentration of the base if you needed 35.5 m

Anvisha [2.4K]

Answer:

0.29

Explanation:

Since the name of the acid (and the equation) is not given, you must assume that it is a 1:1 ratio. Use equation: volume of acid x molarity of acid = volume of base x molarity of base (when the ratio is 1:1).

5 0

3 years ago

When 125 mL of 0.150 M Pb(NO3)2 is mixed with 145 mL of 0.200 M KBr, 4.92 g of PbBr2 is collected. Calculate the percent yield.

Semenov [28]

Answer:

Y = 92.5 %

Explanation:

Hello there!

In this case, since the reaction between lead (II) nitrate and potassium bromide is:

$Pb(NO_3)_2+2KBr\rightarrow PbBr_2+2KNO_3$

Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:

$0.125L*0.150\frac{molPb(NO_3)_2}{L} *\frac{1molPbBr_2}{1molPb(NO_3)_2} =0.01875molPbBr_2\\\\0.145L*0.200\frac{molKBr}{L} *\frac{1molPbBr_2}{2molKBr} =0.0145molPbBr_2$

Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:

$0.0145molPbBr_2*\frac{367.01gPbBr_2}{1molPbBr_2} =5.32gPbBr_2$

And the resulting percent yield:

$Y=\frac{4.92g}{5.32g} *100\%\\\\Y=92.5\%$

Regards!

4 0

3 years ago

In a titration experiment a student uses 1.4 m hbr solution and the indicator phenolphthalein to determine the concentration of

Likurg_2 [28]

The question is incomplete. Complete question is attached below:
...........................................................................................................................

Answer:
Given: conc. of HBr = 1.4 M
Volume of HBr = 15.4 mL
Volume of KOH = 22.10 mL

We know that, M1V1 = M2V2
(HBr) (KOH)

Therefore, M2 = M1V1/V2
= 1.4 X 15.4/22.10
= 0.9756 M

Concentration of KOH is 0.9756 M.

3 0

3 years ago

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