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3 years ago

………… is the Energy stored in chemical bonds.

Chemistry

2 answers:

sergiy2304 [10]3 years ago

7 0

Answer:

ANS is potential energy

Hope it helps

Katyanochek1 [597]3 years ago

3 0

Answer:Energy, potential energy, is stored in the covalent bonds holding atoms together in the form of molecules. This is often called chemical energy. ..

Explanation:This movement is a form of kinetic energy, and the more the molecules move the more kinetic energy they have. Molecules in solids don't move much, they just vibrate

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A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flo

atroni [7]

Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

Weight = $0.09 \times 16.04 + 0.91 \times 29$

= 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

$\frac{800}{27.8336}$

= 28.74 kmol/hr

Now, applying component balance as follows.

$0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}$

$F_{p}$ = 43.11 kmol/hr

$F_{A} + F_{B} = F_{p}$

$F_{B}$ = 43.11 - 28.74

= 14.37 kmol/hr

So, mass flow rate of pure (B), is $F_{B}$ = $14.37 \times 29$

= 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

Molecular weight of product stream = $Mol. weight \times 43.11 kmol/hr$

= $0.06 \times 16.04 + 0.94 \times 29$

= 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of $O_{2}$ into the product stream is as follows.

$0.21 \times 0.94 \times 43.11$

= $8.5099 kmol/hr \times 329 g/mol$

= 272.31 kg/hr

Therefore, calculate the mass % of $O_{2}$ into the stream as follows.

$\frac{272.31}{1216.67} \times 100$

= 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of $O_{2}$ in the product gas is 22.38%.

4 0

3 years ago

Using your knowledge of reactions and mechanisms, draw the mechanism for the formation of 2-methylpropene from 2-methyl-2-propan

BARSIC [14]

Answer:

Here's what I get

Explanation:

The reaction is an E1 elimination of an alcohol to form an alkene. It has three steps:

1. Protonation

The alcohol is protonated with aqueous sulfuric acid to convert it into a better leaving group.

2. Loss of the leaving group

A water molecule leaves in a unimolecular process to form a stable 3° carbocation.

3. Loss of an α-hydrogen

A water molecule removes an α-hydrogen, forming 2-methylpropene and regenerating the original hydronium ion.

4 0

3 years ago

A solution of rubbing alcohol is 78.5 % (v/v isopropanol in water. how many milliliters of isopropanol are in a 90.8 ml sample o

poizon [28]

.785 x 90.8mL = 71.3 mL

4 0

3 years ago

Which two of the following correctly describe a solution?

Mrac [35]

Answer:

C and A

Explanation:

i think it is because a solution is a mix of a solute and solvent, dont judge me if its wrong

8 0

2 years ago

(b) Draw the structure of the products when cyclopentene is reacted with:

oee [108]

Answer:

See explanation and image attached

Explanation:

When an alkene is reacted with cold, dilute and alkaline KMnO4, the product of the reaction is a 1,2-diol (vicinal diol). The products exhibit a “syn” stereochemistry with the addition of the two -OH groups to the same side of the alkene.

The reaction of cyclopentene with cold, dilute, alkaline KMnO4 is shown in the image attached.

6 0

3 years ago