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Ede4ka [16]
3 years ago
5

A chemist found a 60.00 gram sample containing copper and oxygen. Given the following percent composition, what is the mass of c

opper in the sample?
Copper- 57.08 %
Oxygen- 42.9296


choose an answer:
O 0.5708 g Cu
o 25.02 g Cu
34.25 g cu
63.5 g Cu
Chemistry
1 answer:
Strike441 [17]3 years ago
6 0
The answer should 34.25g of Cu.
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For the reaction, X + Y → A + B, ΔGo is –1324 kJ. Which one of the following statements is NOT valid concerning the reaction?
natima [27]

Answer:

c. The reaction will proceed rapidly from left to right.

Explanation:

The variation of the free Gibbs energy doesn't tell anything about the speed of reaction.

On the other hand, when ΔGo is negative: the reaction is spontaneous, thermodynamically favourable, and the products are more stable than the reactants

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3 years ago
If 5.0 kJ of energy is added to a 15.5-g sample of water at 10.°C, the water is Group of answer choices
BaLLatris [955]

Answer:

The water is completely vaporized at this stage.

Explanation:

The complete question is

If 5.0 kJ of energy is added to a 15.5-g sample of water at 10.°C, the water is

-boiling

-completely vaporized

-frozen solid

-decomposed

-still a liquid

Energy added = 50 kJ = 50000 J

mass of water = 15.5 g = 0.0155 kg

temperature of water = 10 °C

We know that the energy posses by a mass of water at a given temperature is given as

H = mcT

where H is the energy possessed by the mass of water

m is the mass of the water

c is the specific heat capacity of water = 4200 J/ kg- °C

T is the temperature of the water

substituting values, the energy of this amount of water is

H = 0.0155 x 4200 x 10 = 651 J

If 50 kJ is added to the water, the energy increases to

50000 J + 651 J = 50651 J

Temperature of this water at this stage will be gotten from

H = mcT

we solve for the new temperature

50651 = 0.0155 x 4200 x T

50651 = 65.1 x T

T = 50651/65.1 = 778.05 °C

This temperature is well over 100 °C, which is the vaporization temperature of water, but less than 3000 °C for its molecules to decompose.

3 0
3 years ago
Use the following half-reactions to write three spontaneous reactions, calculate E°cell for each reaction, and rank the oxidizin
Ainat [17]

Answer:

See explaination

Explanation:

1)

we know that

half cell with higher reduction potential is cathode

so

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

anode :

Cr(s) ---> Cr+3 + 3e-

so

overall reaction is

3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3

now

Eo cell = Eo cathode - Eo anode

so

EO cell = 1.77 + 0.74

Eo cell = 2.51 V

now

in this case

oxidizing agents are N20 and Cr+3

reducing agents are Cr and N2

higher the reduction potential , stronger the oxidizing agent

lower the reduction potential , stronger the reducing agent

so

oxidzing agents

N20 > Cr+3

reducing agents

Cr > N2

2)

cathode :

Au+ + e- --> Au

anode :

Cr ---> Cr+3 + 3e-

overall reaction

3Au+ + Cr ---> 3Au + Cr+3

Eo cell = 1.69 + 0.74

Eo cell = 2.43

now

oxidizing agents :

Au+ > Cr+3

reducing agents :

Cr > Au

3)

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

andoe :

Au ---> Au+ + e-

overall

2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20

Eo cell = 1.77 - 1.69

Eo cell = 0.08

oxidizing agents

N20 > Au+

reducing agents

Au > N2

8 0
3 years ago
Which of the following changes would improve the accuracy of the molar mass measurement carried out in this experiment:
aleksandrvk [35]
A. would not help and maybe make it harder to get accurate measurements.

b. would make more delta T which will improve its accuracy

c. will also increase delta T.
4 0
3 years ago
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