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Alinara [238K]
3 years ago
6

ose you has aniline as your unknown. What kind of solid derivative would be suitable to synthesize for aniline

Chemistry
1 answer:
gayaneshka [121]3 years ago
4 0

Answer:

Amides are suitable to synthesize aniline

Explanation:

Amide are basically used to synthesize aniline.

The chemical formula of Aniline is C6H5NH2 while that of Amide is RCONH2

Amide are solids that can be easily converted to amines.

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If an object has a density of 0.55 g/mL, what is its density in cg/L?
otez555 [7]

if              1 g is equal to 100 cg

then  0.55 g are equal to X cg

X = (0.55 × 100 ) / 1 = 55 cg

The density of the object is 55 cg/L.

7 0
3 years ago
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Hello, <br><br> So I was wondering if this is correct... Is it?
Sonja [21]
Looks correct but the second to last I would of put abiotic and biotic factors but I don’t know what’s right for you
4 0
3 years ago
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2, 3-diethyl–4-octene  what is structure of this pls help​
pantera1 [17]

Answer:

(CH2CH3).

!

CH-CH-CH-CH=CH-CH2-CH2-CH3

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(CH2CH3)

2, 3-diethyl–4-octene

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4 0
3 years ago
Which conversion factor is needed to solve the following problem?
77julia77 [94]
It is going to be  <span>Molar Volume
</span><span>3H2 + N2 --> 2NH3
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5 0
3 years ago
(Yield Problem)
alex41 [277]

Answer:

Percent Yield Fe  =  82.5%

Explanation:

The actual yield is the value produced after an experiment is conducted. The theoretical yield is the value calculated using the balanced chemical equation and atomic/molar masses.

To find the percent yield of iron (Fe), you need to (1) convert grams Al to moles Al (via atomic mass), then (2) convert moles Al to moles Fe (via mole-to-mole ratio from equation coefficients), then (3) convert moles Fe to grams Fe (via atomic mass), and then (4) calculate the percent yield. It is important to arrange the ratios in a way that allows for the cancellation of units. The final answer should have 3 sig figs to reflect the sig figs of the given values.

Atomic Mass (Mg): 24.305 g/mol

Atomic Mass (Fe): 55.845 g/mol

3 Mg + 2 FeCl₃ -----> 2 Fe + 3 MgCl₂

20.5 g Mg           1 mole              2 moles Fe            55.845 g
-----------------  x  -----------------  x  ----------------------  x  -----------------  =  
                           24.305 g           3 moles Mg             1 mole

=  31.4 g Fe

                                     Actual Yield
Percent Yield  =  ----------------------------------  x  100%
                                 Theoretical Yield

                               25.9 g Fe
Percent Yield  =  --------------------  x  100%
                               31.4 g Fe

Percent Yield  =  82.5%

5 0
2 years ago
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