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valentinak56 [21]
2 years ago
8

How many kilograms of the rock must be processed to obtain 2.0 kg of Pb ?

Chemistry
1 answer:
Pepsi [2]2 years ago
6 0
15 kilogram of the rock must be processed.
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When heating a solution to boiling on a hot plate, start by ___________. Then, turn the heat to ___________to start. If necessar
Aleonysh [2.5K]

Answer:

1. starting and stabilizing the stir function

2. a medium heat

3.turn up the heat setting

Explanation:

A chemical reaction can be defined as a chemical process which typically involves the transformation or rearrangement of the atomic, ionic or molecular structure of an element through the breakdown and formation of chemical bonds to produce a new compound or substance.

Some of the laboratory apparatus (equipment) used for conducting a chemical reaction are conical flask, Bunsen burner, beaker, tongs, crucible, round bottom flask etc.

When heating a solution to boiling on a hot plate, start by starting and stabilizing the stir function. Then, turn the heat to a medium heat to start. If necessary, turn up the heat setting after waiting about ten minutes without seeing boiling.

The safety precautions that must be taken when heating a solution to boiling on a hot plate;

I. A proper inspection of the round bottom flask for cracks, irregularities or any imperfection.

II. Ensure you avoid heating the flask while it is closed.

III. When suspending the flask on a hot plate, you should ensure that you use a clamp for stability.

5 0
3 years ago
What is the standard electrode potential for a galvanic cell constructed in the appropriate way from these two half-cells?
____ [38]

E

θ

Cell

=

+

2.115

l

V

Cathode

Mg

2

+

/

Mg

Anode

Ni

2

+

/

Ni

Explanation:

Look up the reduction potential for each cell in question on a table of standard electrode potential like this one from Chemistry LibreTexts. [1]

Mg

2

+

(

a

q

)

+

2

l

e

−

→

Mg

(

s

)

−

E

θ

=

−

2.372

l

V

Ni

2

+

(

a

q

)

+

2

l

e

−

→

Ni

(

s

)

−

E

θ

=

−

0.257

l

V

The standard reduction potential

E

θ

resembles the electrode's strength as an oxidizing agent and equivalently its tendency to get reduced. The reduction potential of a Platinum-Hydrogen Electrode under standard conditions (

298

l

K

,

1.00

l

kPa

) is defined as

0

l

V

for reference. [2]

A cell with a high reduction potential indicates a strong oxidizing agent- vice versa for a cell with low reduction potentials.

Two half cells connected with an external circuit and a salt bridge make a galvanic cell; the half-cell with the higher

E

θ

and thus higher likelihood to be reduced will experience reduction and act as the cathode, whereas the half-cell with a lower

E

θ

will experience oxidation and act the anode.

E

θ

(

Ni

2

+

/

Ni

)

>

E

θ

(

Mg

2

+

/

Mg

)

Therefore in this galvanic cell, the

Ni

2

+

/

Ni

half-cell will experience reduction and act as the cathode and the

Mg

2

+

/

Mg

the anode.

The standard cell potential of a galvanic cell equals the standard reduction potential of the cathode minus that of the anode. That is:

E

θ

cell

=

E

θ

(

Cathode

)

−

E

θ

(

Anode

)

E

θ

cell

=

−

0.257

−

(

−

2.372

)

E

θ

cell

=

+

2.115

Indicating that connecting the two cells will generate a potential difference of

+

2.115

l

V

across the two cells.

5 0
2 years ago
Read 2 more answers
A 0.150-kg sample of a metal alloy is heated at 540 Celsius an then plunged into a 0.400-kg of water at 10.0 Celsius, which is c
Zarrin [17]

Answer:

C_{alloy}=0.497\frac{J}{g\°C}

Explanation:

Hello there!

In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

Q_{allow}=-(Q_{water}+Q_{Al})

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})

Then, we solve for specific heat of the metallic alloy to obtain:

C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}

Thereby, we plug in the given data to obtain:

C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}

Regards!

3 0
2 years ago
What is the definition for floating
creativ13 [48]

Answer:

"1. buoyant or suspended in water or air.

2.not settled in a definite place; fluctuating or variable."

Explanation:

Hope this helps! :)

5 0
3 years ago
Read 2 more answers
If you have 0.045 L of 0.465 M potassium bromide. How many moles of potassium bromide are present?
Mekhanik [1.2K]
The answer to your question is letter A! Hope that helps
7 0
2 years ago
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