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rusak2 [61]
3 years ago
6

Answer the questions below:

Chemistry
2 answers:
Helga [31]3 years ago
8 0
Why do people always send those links like it’s helpful, i’ll help you!
vichka [17]3 years ago
8 0
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As noted in Assignment 1.4.1., alkylation of benzene with 1-chlorobutane in the presence of AlCl3 gives both butylbenzene and (1
Sunny_sXe [5.5K]

Answer:

Use the Bromotriflouride catalyst, BF₃

Explanation:

The BF₃ is most likely to yield less desired side products. The effect lies in the reaction mechanism.

BF₃ is a Lewis acid. Its role is to promote the ionization of the HF. This is achieved through the electrophilic mechanism. The reaction mechanism is as follows:

2 - methylpropene + H-F-BF₃ → H-F + H₃C + benzene

butylbenzene + F-BF₃ → tert-butylbenzene + H-F + BF₃ (regenerated catalyst)

6 0
3 years ago
Make the equations for the reactions between calcium and hydrochloric acid, name the products.
Goryan [66]

Answer:

Calcium carbonate reacts with hydrochloric acid to form carbon dioxide gas. 2HCl (aq) + CaCO 3(s) CaCl 2 (aq) + CO 2(g) + H 2 O (l).

3 0
3 years ago
20 mL of 80°C water is mixed with 20 mL of 0°C water in a perfect calorimeter. What is the final temperature?
dedylja [7]
To calculate for the final temperature, we need to remember that the heat rejected should be equal to the absorbed by the other system. We calculate as follows:

Q1 = Q2
(mCΔT)1 = (mCΔT)2

We can cancel m assuming the two systems are equal in mass. Also, we cancel C since they are the same system. This leaves us,

 (ΔT)1 = (ΔT)2
(T - 80) = (0 - T)
T = 40°C
8 0
3 years ago
What happens to the particles during condensation
Paraphin [41]

Answer:

molecus in a gas cool down

Explanation:

the molecules lose heat and energy do they slow down they move closer to other molecules and form a liquid

8 0
3 years ago
When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H
Debora [2.8K]

Answer:

Y=58.15\%

Explanation:

Hello,

For the given chemical reaction:

Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol  Al_2S_3

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3

Finally, we compute the percent yield with the obtained 2.10 g:

Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%

Best regards.

7 0
3 years ago
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