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Alborosie
3 years ago
7

When 1.00 g of boron is burned in o2(g) to form b2o3(s), enough heat is generated to raise the temperature of 733 g of water fro

m 18.0 ∘c to 38.0 ∘c?
Chemistry
1 answer:
Bas_tet [7]3 years ago
7 0
<span>Answer: For this problem, you would need to know the specific heat of water, that is, the amount of energy required to raise the temperature of 1 g of water by 1 degree C. The formula is q = c X m X delta T, where q is the specific heat of water, m is the mass and delta T is the change in temperature. If we look up the specific heat of water, we find it is 4.184 J/(g X degree C). The temperature of the water went up 20 degrees. 4.184 x 713 x 20.0 = 59700 J to 3 significant digits, or 59.7 kJ. Now, that is the energy to form B2O3 from 1 gram of boron. If we want kJ/mole, we need to do a little more work. To find the number of moles of Boron contained in 1 gram, we need to know the gram atomic mass of Boron, which is 10.811. Dividing 1 gram of boron by 10.811 gives us .0925 moles of boron. Since it takes 2 moles of boron to make 1 mole B2O3, we would divide the number of moles of boron by two to get the number of moles of B2O3. .0925/2 = .0462 moles...so you would divide the energy in KJ by the number of moles to get KJ/mole. 59.7/.0462 = 1290 KJ/mole.</span>
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Answer: Option (d) is the correct answer.

Explanation:

Steps involved for the given reaction will be as follows.

Step 1: 2NO \Leftrightarrow N_{2}O_{2}    (fast)

Rate expression for step 1 is as follows.

               Rate = k [NO]^{2}

Step 2: N_{2}O_{2} + H_{2} \rightarrow N_{2}O + H_{2}O

This step 2 is a slow step. Hence, it is a rate determining step.

Step 3. N_{2}O + H_{2} \rightarrow N_{2} + H_{2}O    (fast)

Here, N_{2}O_{2} is intermediate in nature.

All the steps are bimolecular and it is a second order reaction. Also, there is no catalyst present in this reaction.

Thus, we can conclude that the statement step 1 is the rate determining step, concerning this mechanism is not directly supported by the information provided.

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Whitepunk [10]

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ahrayia [7]

Answer:

See below.  

Step-by-step explanation:

Ethers react with HI at high temperature to produce an alky halide and an alcohol.

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If there is excess HI, the alcohol formed in Step 2 is also converted to an alkyl iodide:

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The ethanol reacts with excess HI in an Sₙ2 reaction to form ethyl iodide (c).

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