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marishachu [46]
3 years ago
10

Calculate the total quantity of heat required to convert 25.0 g of liquid ccl4(l from 35.0°c to gaseous ccl4 at 76.8°c (the norm

al boiling point for ccl4. the specific heat of ccl4(l is its heat of fusion is and its heat of vaporization is
Chemistry
1 answer:
GREYUIT [131]3 years ago
6 0

The solution would be like this for this specific problem:

<span>25.0 g CCl4 x (1 mole CCl4 / 153.8 g CCl4) = 0.163 moles CCl4 

 </span>(29.82 kJ / mole)(0.163 moles) = 4.86 kJ 

total heat = 1.11 kJ + 4.86 kJ = 5.97 kJ 

<span>I hope this helps and if you have any further questions, please don’t hesitate to ask again. </span>

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Sunny_sXe [5.5K]

Answer:

2,67 L

Explanation:

8 0
2 years ago
Use the following half-reactions to write three spontaneous reactions, calculate E°cell for each reaction, and rank the oxidizin
Ainat [17]

Answer:

See explaination

Explanation:

1)

we know that

half cell with higher reduction potential is cathode

so

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

anode :

Cr(s) ---> Cr+3 + 3e-

so

overall reaction is

3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3

now

Eo cell = Eo cathode - Eo anode

so

EO cell = 1.77 + 0.74

Eo cell = 2.51 V

now

in this case

oxidizing agents are N20 and Cr+3

reducing agents are Cr and N2

higher the reduction potential , stronger the oxidizing agent

lower the reduction potential , stronger the reducing agent

so

oxidzing agents

N20 > Cr+3

reducing agents

Cr > N2

2)

cathode :

Au+ + e- --> Au

anode :

Cr ---> Cr+3 + 3e-

overall reaction

3Au+ + Cr ---> 3Au + Cr+3

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oxidizing agents :

Au+ > Cr+3

reducing agents :

Cr > Au

3)

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

andoe :

Au ---> Au+ + e-

overall

2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20

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Eo cell = 0.08

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reducing agents

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8 0
3 years ago
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3 0
2 years ago
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Answer:

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Explanation:

The name of the compound is derived from the name of the elements present  in it.

The rule followed while naming the compound are:

1. The first element (always the cation) is named as such .

2. The second element (The anion) end with "-ate ,  -ide ," etc

3. NO prefix is added while naming the first element.

For example : Bi2 can't be named as Dibismuth

Na2 = Can't be named as disodium

Hence the compound :

Bi2(CO3)3 contain two element : Bi and CO3. Here , Bi = cation (named as such) and CO3 = anion (named according to rules)

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5 0
3 years ago
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igomit [66]

Answer:

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The molecules attached are two methyl groups.  Since you have two methyl groups, the name will be dimethyl.  Start counting from the side closest to the double bond.  This gives you a 3, 5-dimethyl group.

5 0
3 years ago
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