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2 years ago

What is the kinetic product obtained from the addition of 1 mole of bromine to 1,3-butadiene?

Chemistry

1 answer:

Sveta_85 [38]2 years ago

7 0

From the addition of 1 equivalent of HBr to 1,3-butadiene, 3-bromo-1-butene is obtained as a product.

What is moles?

Moles are defined as the smallest base unit that is equivalent to $6.02214 X 10^{23}$ units.

Moles are chemical entities, which are used to measure the mass or molecular mass of the compound, molecule, or atom. The moles are, for example can be explained as equal to the mass of the atom. For example, there is one mole in 12 grams of carboncarbon.

Thus the kinetic product formed is 3-bromo-1-butene.

Learn more about mole or mole fraction:

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if lavoisier heated 100 grams of powdered mercuric oxide to produce 93 grams of liquid mercury, how much oxygen would be release

frutty [35]

7 grams of Oxygen. The Law of conservation of mass says that mass can be neither created nor destroyed, so the total mass of the products (in this case the oxygen plus the mercury) must equal the total mass of the reactants (in this case, the mercuric oxide).

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3 years ago

State any two non-industrial uses of hydrogen gas (2mks)<br>

Brut [27]

Answer:

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2 years ago

In a Dam, If we doubled the depth of the dam the hydrostatic force will be ?

AnnZ [28]

Answer: Option (A) is the correct answer.

Explanation:

Force acting on a dam is as follows.

F = $\frac{1}{2}\rho g\omega H^{2}$ .......... (1)

Now, when we double the depth then it means H is increasing 2 times and then the above relation will be as follows.

F' = $\frac{1}{2}\rho g\omega (2)^{2}$

F' = $\frac{1}{2}\rho g\omega 4$ ........... (2)

Now, dividing equation (1) by equation (2) as follows.

$\frac{F}{F'}$ = $\frac{\frac{1}{2}\rho g\omega H^{2}}{\frac{1}{2}\rho g\omega 4}$

Cancelling the common terms we get the following.

$\frac{F}{F'}$ = $\frac{1}{4}$

4F = F'

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3 years ago

4. Calculate the number of carbon (C) atoms present in the 25 mL of sugar.

Nadusha1986 [10]

Answer: $8.38\times 10^{23}atoms$ of carbon

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

mass of sugar= $density\times volume=1.59g/ml\times 25ml=39.75g$

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{39.75g}{342.3g/mol}=0.116moles$

1 mole of sugar $C_{12}H_{22}O_{11})$ contains = $12\times 6.023\times 10^{23}=72.28\times 10^{23}atoms$ of carbon

0.116 mole of sugar $C_{12}H_{22}O_{11})$ contains = $\frac{72.28\times 10^{23}}{1}\times 0.116=8.38\times 10^{23}atoms$ of carbon

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3 years ago

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vazorg [7]

Answer:

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