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OLga [1]
3 years ago
13

Read excerpt and answer 3 questions I’ll mark brainliest. Pls thank you

Chemistry
1 answer:
mario62 [17]3 years ago
6 0

Answer:

please also share the excerpt

Explanation:

Thank you

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__ Cu + ___ HNO₃ → ___ _Cu(NO₃)₂ + ___ NO₂ + ___ H₂O
ivolga24 [154]

Answer:

Cu + 4HNO3 --->   Cu(NO3)2 + 2NO2 + 2H2O.

Explanation:

Balancing:

Cu + 4HNO3 --->   Cu(NO3)2 + 2 NO2 + 2H2O.

3 0
3 years ago
Read 2 more answers
The neutron main responsibility is to
jarptica [38.1K]
Neutrons are responsible for combining with protons in the nucleus together to be held by the strong force. Too many neutrons in an atom will result in radioactive decay due to the neutrons' force overpowering the strong force.
5 0
3 years ago
Draw the major organic product in the reaction scheme below. Be sure to clearly show stereochemistry (if applicable).
Juliette [100K]

Answer:

see explanation below

Explanation:

You are missing the reaction scheme, but in picture 1, I found a question very similar to this, and after look into some other pages, I found the same scheme reaction, so I'm gonna work on this one, to show you how to solve it. Hopefully it will be the one you are asking.

According to the reaction scheme, in the first step we have NaNH2/NH3(l). This reactant is used to substract the most acidic hydrogen in the alkine there. In this case, it will substract the hydrogen from the carbon in the triple bond leaving something like this:

R: cyclopentane

R - C ≡ C (-)

Now, in the second step, this new product will experiment a SN2 reaction, and will attack to the CH3 - I forming another alkine as follow:

R - C ≡ C - CH3

Finally in the last step, Na in NH3 are reactants to promvove the hydrogenation of alkines. In this case, it will undergo hydrogenation in the triple bond and will form an alkene:

R - CH = CH - CH3

In picture 2, you have the reaction and mechanism.

6 0
3 years ago
How many milliliters of 0.0050 N KOH are required to neutralize 41 mL of 0.0050 M H2SO4?
Kipish [7]

Answer:

V KOH = 41 mL

Explanation:

for neutralization:

  • ( V×<em>C </em>)acid = ( V×<em>C </em>)base

∴ <em>C </em>H2SO4 = 0.0050 M = 0.0050 mol/L

∴ V H2SO4 = 41 mL = 0.041 L

∴ <em>C</em> KOH = 0.0050 N = 0.0050  eq-g/L

∴ E KOH = 1 eq-g/mol

⇒ <em>C</em> KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L

⇒ V KOH = ( V×<em>C </em>) acid / <em>C </em>KOH

⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)

⇒ V KOH = 0.041 L

4 0
3 years ago
Do centromeres divide at anaphase i or ii
ddd [48]
Anaphase 1 is when centromeres divide
7 0
3 years ago
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