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motikmotik
3 years ago
9

Please help! I don’t know the answer!

Chemistry
1 answer:
kow [346]3 years ago
3 0
The answer is 4 moles!
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Rocks and sand are nonliving .all organisms are
solong [7]
Hey there!

Rocks and sand are nonliving. All organisms are living.
Living organisms have five characteristics. Living organisms respond to a stimulus, need energy, grow, reproduce, and get rid of wastes. All living organisms consist of cells. 

Hope this helps! Have a Brainly day! :D
5 0
3 years ago
List out the effects seen inplants due to lack of phosphorus.​
valkas [14]

Answer:

Plants normally turn dark green and look stunted (both leaves and stalks). Older leaves are first affected and may become violet. Sometimes the brown tips of the leaves remain fragile and their maturity seems to be delayed.

Explanation:

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4 0
3 years ago
A combustion analysis of 5.214 g of a compound yields 5.34 g co 2 ​ , 1.09 g h 2 ​ o, and 1.70 g n 2 ​ . if the molar mass of th
nekit [7.7K]
Answer is: C₃H₃N₃O₃.
Chemical reaction: CₓHₓNₓOₓ + O₂ → aCO₂ + x/2H₂ + x/2N₂.
m(CₐHₓNₓ) = 5,214 g.
m(CO₂) = 5,34 g.
m(H₂) = 1,09 g.
m(N₂) = 1,70 g.
n(CO₂) = n(C) =  5,34 g ÷ 44 g/mol = 0,121 mol.
n(H₂O) = 1,09 g ÷18 g/mol = 0,06 mol.
n(H) = 2 · 0,0605 mol = 0,121 mol.
n(N₂) = 1,7 g ÷ 28 g/mol = 0,0607 mol.
n(N) = 0,0607 mol · 2 = 0,121 mol.
n(C) : n(H) : n(N) = 0,121 mol : 0,121 mol : 0,121 mol /: 0,121
n(C) : n(H) : n(N) = 1 : 1 : 1.
M(CHN) = 27 g/mol.
m(O₂) = 8,13 g - 5,214 g = 2,914 g.
n(O₂) = 2,914 g ÷ 32 g/mol = 0,09 mol.
n(CₓHₓNₓOₓ) = 5,214 g ÷ 129,1 g/mol = 0,0404 mol.
n(CₓHₓNₓOₓ) : n(CO₂) = 1 : 3.


3 0
4 years ago
Read 2 more answers
Dbv-qyag-dsc.join this now on meet​
Rasek [7]

Answer:

Don't post any question if isn't related to the topic or to your homework or assignment.

Explanation:

7 0
3 years ago
Draw the most stable resonance structure for the intermediate in the electrophilic aromatic bromination of aniline, anisole, and
ASHA 777 [7]

Answer:

Here's what I get

Explanation:

(a) Intermediates

The three structures below represent one contributor to the resonance-stabilized intermediate, in which the lone pair electrons on the heteroatom are participating (the + charge on the heteroatoms do not show up very well).

(b) Relative Stabilities

The relative stabilities decrease in the order shown.

N is more basic than O, so NH₂ is the best electron donating group (EDG) and will best stabilize the positive charge in the ring. However, the lone pair electrons on the N in acetanilide are also involved in resonance with the carbonyl group, so they are not as available for stabilization of the ring.

(c) Relative reactivities

The relative reactivities would be

C₆H₅-NH₂ >  C₆H₅-OCH₃ > C₆H₅-NHCOCH₃

4 0
3 years ago
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