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disa [49]
2 years ago
15

AlBr3 can be used as a catalyst in the Friedel-Crafts alkylation reaction. The correct name for the compound represented by the

formula AlBr3 is
Chemistry
1 answer:
Andrew [12]2 years ago
7 0

The question is incomplete, the complete question is;

AlBr3 can be used as a catalyst in the Friedel-Crafts alkylation reaction. The correct name for the compound represented by the formula AlBr3 is —

aluminum bromide

monoaluminum tribromide

aluminide bromine

aluminum tribromide

Answer:

aluminum bromide

Explanation:

Having known that AlBr3 is an ionic compound and aluminium is the central atom here, we now have to ask ourselves if Aluminium exists in other stable oxidation states.

We must take cognizance of the fact that the oxidation number of the central atom in a compound becomes part of the name of that compound when other stable oxidation states for atoms of the same elements exists.

Since the +3 state is the only stable oxidation state for aluminium, the name of the compound is simply aluminium bromide.

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¿Se puede disolver acido sulfurico en NaCl ?
lord [1]

Answer:

no

Explanation:

3 0
3 years ago
Ethanol (C2H5OH) melts a - 144 oC and boils at 78 °C. The enthalpy of fusion of ethanol is 5.02 kj/mol, and its enthalpy of vapo
hammer [34]

<u>Answer:</u>

<u>For a:</u> The total heat required is 36621.5 J

<u>For b:</u> The total heat required is 58944.5 J

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the heat required at different temperature, we use the equation:

q=mc\Delta T         .........(1)

where,

q = heat absorbed

m = mass of substance

c = specific heat capacity of substance

\Delta T = change in temperature

To calculate the amount of heat required at same temperature, we use the equation:

q=m\times \Delta H      ........(2)

where,

q = heat absorbed

m = mass of substance

\Delta H = enthalpy of the reaction

The processes involved in the given problem are:

1.)C_2H_5OH(l)(35^oC)\rightarrow C_2H_5OH(l)(78^oC)\\2.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)

  • <u>For process 1:</u>

We are given:

Change in temperature remains the same.

m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=35^oC\\\Delta T=[T_2-T_1]=[78-35]^oC=43^oC=43K

Putting values in equation 1, we get:

q_1=42.0g\times 2.3J/g.K\times 43K\\\\q_1=4153.8J

  • <u>For process 2:</u>

We are given:

Conversion factor: 1 kJ = 1000 J

Molar mass of ethanol = 46 g/mol

m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{35.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g

Putting values in equation 2, we get:

q_2=42.0g\times 773.04J/g\\\\q_2=32467.7J

Total heat required = [q_1+q_2]

Total heat required = [4153.8J+32467.7J]=36621.5J

Hence, the total heat required is 36621.5 J

  • <u>For b:</u>

The processes involved in the given problem are:  

1.)C_2H_5OH(s)(-155^oC)\rightarrow C_2H_5OH(s)(-144^oC)\\2.)C_2H_5OH(s)(-144^oC)\rightarrow C_2H_5OH(l)(-144^oC)\\3.)C_2H_5OH(l)(-144^oC)\rightarrow C_2H_5OH(l)(78^oC)\\4.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)

  • <u>For process 1:</u>

We are given:

Change in temperature remains the same.

m=42.0g\\c_s=0.97J/g.K\\T_2=-144^oC\\T_1=-155^oC\\\Delta T=[T_2-T_1]=[-144-(-155)]^oC=11^oC=11K

Putting values in equation 1, we get:

q_1=42.0g\times 0.97J/g.K\times 11K\\\\q_1=448.14J

  • <u>For process 2:</u>

We are given:

m=42.0g\\\Delta H_{fusion}=5.02kJ/mol=\frac{5.02kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=109.13J/g

Putting values in equation 2, we get:

q_2=42.0g\times 109.13J/g\\\\q_2=4583.5J

  • <u>For process 3:</u>

We are given:

Change in temperature remains the same.

m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=-144^oC\\\Delta T=[T_2-T_1]=[78-(-144)]^oC=222^oC=222K

Putting values in equation 1, we get:

q_3=42.0g\times 2.3J/g.K\times 222K\\\\q_3=21445.2J

  • <u>For process 4:</u>

We are given:

m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{38.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g

Putting values in equation 2, we get:

q_4=42.0g\times 773.04J/g\\\\q_4=32467.7J

Total heat required = [q_1+q_2+q_3+q_4]

Total heat required = [448.14+4583.5+21445.2+32467.7]J=58944.5J

Hence, the total heat required is 58944.5 J

8 0
3 years ago
Nault 25000L 250 our = 25000L 15040 8. How many grams of CaCl2 are needed to make 150.0 mL of a 0.500 M CF solution? (Note: CaCl
yuradex [85]

Answer:

You need 8,324 g of CaCl₂ yo make this solution

Explanation:

Molarity is a way to express concentration in a solution, in units of moles of solute per liter of solution.

To know the grams of CaCl₂ it is necessary to know, first, the moles of this substance with the desired volume and concentration , thus:

0,1500 L × \frac{0,500 mol}{L} = 0,075 CaCl₂ moles

Now, with the molar mass of CaCl₂ you will obtain the necessary grams, thus:

0,075 CaCl₂ moles  × \frac{110,98 g}{mol} = 8,324 g of CaCl₂

So, you need <em>8,324 g of CaCl₂</em> to make 150,0 mL of a 0,500M solution

I hope it helps!

3 0
3 years ago
Which would best help will give brainlist first 1
son4ous [18]
It would definitely be hand lense
3 0
3 years ago
Read 2 more answers
Plz help thx brainliest btw :)
Marina86 [1]

Answer:

1. earth metal 2. halogen 3.The elements are arranged in seven horizontal rows, called periods or series, and 18 vertical columns, called groups. ... Elements in the periodic table are organized according to their properties.

Inventor: Dmitri Mendeleev please name most brainliest

3 0
2 years ago
Read 2 more answers
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