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3 years ago

Consider the following numbered processes: 1. A → 2B 2. B → C + D 3. E → 2D ΔH for the process A → 2C + E is

Chemistry

1 answer:

aivan3 [116]3 years ago

6 0

Answer:

ΔH = ΔH₁ + ΔH₂ - ΔH₃

Explanation:

Given that:

1. A → 2B

2. B → C + D

3. E → 2D

Assuming from the corresponding ΔH for process 1, 2 and 3 are ΔH₁, ΔH₂, ΔH₃ respectively.

To estimate the ΔH for the process A → 2C + E

We multiply 2 with equation 2 where (B → C + D)

2B → 2C + 2D ⇒ 2ΔH₂

Also, let's switch equation (3), such that we have,

2D → E -ΔH₃

The summation of all the equation result into :

A → 2C + E

where; ΔH = ΔH₁ + ΔH₂ - ΔH₃

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Anton [14]

Answer
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3 years ago

Nitration of 1,4−dimethylbenzene (p−xylene) gives a single product having the molecular formula c8h9no2 in high yield. draw this

GarryVolchara [31]

Structure is in document below.
The mononitration of p-xylene can be easily carried out at 30 degrees C.
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8 0

3 years ago

Suppose the formation of iodine proceeds by the following mechanism:

NARA [144]

Answer:

Overall reaction

H2(g) + 2ICI(g) -----> I2(g) +2HCl(g)

Overall Rate = k1[H2] [ICl]

Explanation:

Overall reaction

H2(g) + 2ICI(g) -----> I2(g) +2HCl(g)

The overall reaction is the sum of the two two reactions shown in the question. After the two reactions are summed up properly, this overall reaction equation his obtained.

Since K1<<K2 it means that step 1 is slower than step 2. Recall that the rate if reaction depends on the slowest step of the reaction. Hence

Overall Rate = k1[H2] [ICl]

3 0

3 years ago

Conversion of gaseous nitrogen to liquid nitrogen is an

miss Akunina [59]

Answer:

true

Explanation:

you can't change it back, it's chemically changed

8 0

3 years ago

In the laboratory a student determines the specific heat of a metal. She heats 19.6 grams of zinc to 98.37°C and then drops it i

fgiga [73]

Answer:

The specific heat of zinc is 0.375 J/g°C

Explanation:

Step 1: Data given

Mass of zinc = 19.6 grams

Mass of water = 82.9 grams

Initial temperature of zinc = 98.37 °C

Initial temperature of water = 24.16 °C

Final temperature of water (and zinc) = 25.70 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculate Specific heat of zinc

Heat lost by zinc = heat won by water

Q=m*c*ΔT

Qzinc = -Qwater

m(zinc)*C(zinc)*ΔT(zinc) = -m(water)*C(water)*ΔT(water)

⇒ with mass of zinc = 19.6 grams

⇒ with C(zinc) = TO BE DETERMINED

⇒ with ΔT(zinc) = T2 -T1 = 25.70 - 98.37 = -72.67°C

⇒ with mass of water = 82.9 grams

⇒ with C(water) = 4.184 J/g°C

⇒ with ΔT(water) = T2 - T1 = 25.70 - 24.16 = 1.54

Qzinc = -Qwater

19.6g* C(zinc) * (-72.67°C) = - 82.9g* 4.184 J/g°C * 1.54 °C

-1424.332*C(zinc) = -534.155

C(zinc) = 0.375 J/g°C

The specific heat of zinc is 0.375 J/g°C

7 0

3 years ago