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Ilia_Sergeevich [38]
3 years ago
6

Why do we use ammonium formate as a hydrogen surrogate instead of using hydrogen gas in this experiment?

Chemistry
1 answer:
NemiM [27]3 years ago
7 0

Answer:

Ammonium formate can also be used in palladium on carbon (Pd/C) reduction of functional groups. In the presence of Pd/C, ammonium formate decomposes to hydrogen, carbon dioxide, and ammonia.

Explanation:

This hydrogen gas is adsorbed onto the surface of the palladium metal, where it can react with various functional groups.

You might be interested in
Why can liquids change shape but solids cannot
Cerrena [4.2K]

Answer:

Explanation:

the forces between the molecules are stronger in solid than in liquids

5 0
3 years ago
How much energy (heat) is required to convert 52.0 g of ice at –10.0°C to steam at 100°C? Specific heat of ice 2.09 J/g • °C Spe
dexar [7]

Answer: The energy (heat) required to convert 52.0 g of ice at –10.0°C to steam at 100°C is 157.8 kJ

Explanation:

Using this formular, q = [mCpΔT] and = [nΔHfusion]

The energy that is needed in the different physical changes is thus:

The heat needed to raise the ice temperature from -10.0°C to 0°C is given as as:

q = [mCpΔT]

q = 52.0 x 2.09 x 10

q = 1.09 kJ

While from 0°C to 100°C is calculated as:

q = [mCpΔT]

q = 52.0 x 4.18 x 100

q = 21.74 kJ

And for fusion at 0°C is called Heat of fusion and would be given as:

q = n ΔHfusion

q = 52.0 / 18.02 x 6.02

q = 17.38 kJ

And that required for vaporization at 100°C is called Heat of vaporization and it's given as:

q = n ΔHvaporization

q = 52.0 / 18.02 x 40.7

q = 117.45 kJ

Add up all the energy gives 157.8 kJ

5 0
3 years ago
How much heat, in calories, is needed to raise the temperature of 125.0 g of Lead (c lead = 0.130 J/g°C) from 17.5°C to 41.1°C?
pav-90 [236]
95.6 cal
are needed.
Explanation:
Use the following equation:
q
=
m
c
Δ
T
,
where:
q
is heat energy,
m
is mass,
c
is specific heat capacity, and
Δ
T
is the change in temperature.
Δ
T
=
T
final
−
T
initial
Known
m
=
125 g
c
Pb
=
0.130
J
g
⋅
∘
C
T
initial
=
17.5
∘
C
T
final
=
42.1
∘
C
Δ
T
=
42.1
∘
C
−
17.5
∘
C
=
24.6
∘
C
Unknown
q
Solution
Plug the known values into the equation and solve.
q
=
(
125
g
)
×
(
0.130
J
g
⋅
∘
C
)
×
(
24.6
∘
C
)
=
400. J

(rounded to three significant figures)
Convert Joules to calories
1 J
=
0.2389 cal
to four significant figures.
400
.
J
×
0.2389
cal
1
J
=
95.6 cal

(rounded to three significant figures)
95.6 cal
are needed.
8 0
3 years ago
List out the moles for each element in the following compound ZnBr2
natulia [17]

Answer:

If the colors in a chromatography are able to dissolve and travel up a paper wick, what kind of chemical property do the colors have when mixed with rubbing alcohol?

(You may need to search "Chemical Properties")

4 0
2 years ago
According to cell theory _____.
bulgar [2K]

Answer:

Every single living thing are comprised of cells. Cells are the fundamental units of structure and capacity in living things. New cells are delivered from existing cells.

4 0
3 years ago
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