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2 years ago

HELP PLEASE :D

Chemistry

1 answer:

ivanzaharov [21]2 years ago

8 0

$\bold{Heya~!}$

For the first drop down.

Your answer will be is :

<h3>C. Normal force. </h3>

And, the second drop down.

Your answer will be is :

<h3>B. The puck will move off in a different direction.</h3>

Hope It Helps ! ~

#LearnWithBrainly

$\underline{Answer~:}$

Jace ~

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A chemistry student needs 30.0 g of acetic acid for an experiment. He has available 1.2 kg of a 7.58% w/w solution of acetic aci

Tomtit [17]

Answer:

The mass of solution will be "395.78 g".

Explanation:

Given:

Mass of acetic acid,

= 30 g

Percentage,

= 7.58% w/w

Now,

⇒ % $=\frac{Mass \ of \ acetic \ acid}{Mass \ of \ solution}\times 100$

By putting the values, we get

⇒ $7.58=\frac{30}{Mass \ of \ solution}\times 100$

⇒ $Mass \ of \ solution=\frac{30\times 100}{7.58}$

⇒ $=\frac{3000}{7.58}$

⇒ $=395.78 \ g$

8 0

2 years ago

Choose the best option for the precursor to the tosylate intermediate Design a synthesis of cis-2-methylcyclopentyl acetate from

kumpel [21]

Answer:

To synthesize cis-2-methylcyclopentyl from methycyclopentanol, you need to replace the acetate hydroxyl group with acetate by inverting the configuration.

Explanation:

To understand the process, you need to understand the nucleophilic mechanism taking place in the process. This is the first stage of the process. Hydroxide is a poor leaving group, to it must be converted to a good leaving group. To effect the change, it is necessary to use p-toluenesuphate.

p-toluenesuphate is favored because this can be prepared by a reaction that alters none of the bonds attached to the stereogenic center.

The reaction of p-toluensulfonate with potassium acetate in acetic acid effects the conversion to give the final product: cis-2-methylcyclopentyl.

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2 years ago

How do you find the mole ratio of a compound?

tangare [24]

Simply divide the moles of a reactant by the number of moles of product

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2 years ago

When N,N-Dimethylaniline is treated with bromine, ortho and para products are observed. However, when N,N-Dimethylaniline is tre

lilavasa [31]

Answer:

See explanation below

Explanation:

To get a better understanding watch the picture attached.

In the case of the reaction with Bromine, the -N(CH₃)₂ is a strong ring activator, therefore, it promotes a electrophilic aromatic sustitution, so, in the mechanism of reaction, the lone pair of the Nitrogen, will move to the ring by resonance and activate the ortho and para positions. That's why the bromine wil go to the ortho and para positions, mostly the para position, because the -N(CH₃)₂ cause a steric hindrance in the ortho position.

In the case of the reaction with HNO₃/H₂SO₄, the acid transform the -N(CH₃)₂ in a protonated form, the anilinium ion, which is a deactivating of the ring, and also a strong electron withdrawing, so, the electrophile will go to the meta position instead.

Hope this helps.