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Kaylis [27]
3 years ago
14

Help me please I’ll give brainlest

Chemistry
1 answer:
drek231 [11]3 years ago
4 0

Answer:

50%

50%

0%

Explanation:

um- punnett squares are weird lol

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According to markovnikov's rule of the electrophilic addition to an alkene, the electrophile, usually a proton, is more likely t
stealth61 [152]

According to markovnikov's rule of the electrophilic addition to an alkene, the electrophile, usually a proton, is more likely to add to the less-substituted carbon in a double bond.

With additional substituents present in this configuration, the intermediate carbocation is stabilised by being located on the more-substituted carbon.

The nucleophile will then end up in a double bond on the more-substituted carbon in a reaction that follows Markovnikov's rule.The outcome of some addition reactions is described by Markovnikov's rule or Markownikoff's rule in organic chemistry. Vladimir Markovnikov, a Russian scientist, created the rule in 1870.

To learn more about Markovnikov's rule

brainly.com/question/14529644

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8 0
2 years ago
What is the mass of 7 moles of Nitrogen?
Igoryamba

Hey there!

Molar mass of nitrogen is 14.007 g.

We have 7 moles.

Multiply 7 by 14.007.

7 x 14.007 = 98.049

7 moles of nitrogen has a mass of 98.049 grams.

Hope this helps!

8 0
3 years ago
4 As a boulder rolls from the top of a hill to the bottom, how do its potential and kinetic energy change? *
user100 [1]

Answer:

c. Potential decreases and kinetic increases.

Explanation:

As the boulder is rolling down the hill, potential energy decreases as kinetic energy increases.

Please Brainliest me

4 0
3 years ago
Read 2 more answers
DO NOT COMMENT IF YOUR NOT GOING TO HELP ITS REALLY ANNOYING :/
blondinia [14]

Answer:

50 mL; 7

Explanation:

By looking at the graph, the boundary point where the solution turns from acidic to basic appears at approximately 50 mL.

Because this is a titration between a strong acid and strong base, the pH of the equivalence point is always 7 (aka neutral).  

8 0
3 years ago
A 1.00 l solution contains 3.50×10-4 m cu(no3)2 and 1.75×10-3 m ethylenediamine (en). the kf for cu(en)22+ is 1.00×1020. what is
Alik [6]
<span>Answer: A 1.00 L solution containing 3.00x10^-4 M Cu(NO3)2 and 2.40x10^-3 M ethylenediamine (en). contains 0.000300 moles of Cu(NO3)2 and 0.00240 moles of ethylenediamine by the formula Cu(en)2^2+ 0.000300 moles of Cu(NO3)2 reacts with twice as many moles of en = 0.000600 mol of en so, 0.00240 moles of ethylenediamine - 0.000600 mol of en reacted = 0.00180 mol en remains by the formula Cu(en)2^2+ 0.000300 moles of Cu(NO3)2 reacts to form an equal 0.000300 moles of Cu(en)2^2+ Kf for Cu(en)2^2+ is 1x10^20. so 1 Cu+2 & 2 en --> Cu(en)2^2+ Kf = [Cu(en)2^2+] / [Cu+2] [en]^2 1x10^20. = [0.000300] / [Cu+2] [0.00180 ]^2 [Cu+2] = [0.000300] / (1x10^20) (3.24 e-6) Cu+2 = 9.26 e-19 Molar since your Kf has only 1 sig fig, you might be expected to round that off to 9 X 10^-19 Molar Cu+2</span>
4 0
3 years ago
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