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krok68 [10]
2 years ago
11

I would like some help por favor / please

Chemistry
1 answer:
CaHeK987 [17]2 years ago
7 0
What do you need help on
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Which water source becomes polluted as it travels over the earth?
never [62]

Run off

Explanation:

Run-off are the water sources that becomes polluted as they travel over the earth surface.

Surface run-off are water that moves on the earth surface. This water is a component of the water cycle which usually originate from precipitation.

  • As the water moves through different terrain it comes into contact with different pollutants.
  • Some of these run-off travels through mines, waste disposal sites among others.
  • This leads to the contamination of this form of water as it moves through these earth materials.
  • They are the most polluted water

Learn more:

Ocean water brainly.com/question/6760255

#learnwithBrainly

8 0
3 years ago
How many valence electrons configuration of oxygen?
Ganezh [65]
Oxygen had 6 valence electrons
3 0
3 years ago
Imagine a rock comprised entirely of an unstable isotope. After three half-lives, how much of the parent isotope remains
uysha [10]

Answer:  After three half-lives 1/8 (12.5%) of the original sample remains

7 0
2 years ago
There is a gas at 780 mm of Hg, in a volume of 5 liters and a temperature of 37 ​ C, the volume is changed to 5.5 liters and the
HACTEHA [7]

Answer:

32.8 C

Explanation:

- Use combined gas law formula and rearrange.

- Hope that helped! Please let me know if you need further explanation.

6 0
3 years ago
Be sure to answer all parts. A concentration cell consists of two Sn/Sn2+ half-cells. The electrolyte in compartment A is 0.23 M
kolbaska11 [484]

Answer:

Part A. The half-cell B is the cathode and the half-cell A is the anode

Part B. 0.017V

Explanation:

Part A

The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).

So, the half-cell B is the cathode and the half-cell A is the anode.

Part B

By the Nersnt equation:

E°cell = E° - (0.0592/n)*log[anode]/[cathode]

Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.

E°cell = 0 - (0.0592/2)*log(0.23/0.87)

E°cell = 0.017V

3 0
3 years ago
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