Answer:
It's explained below.
Explanation:
An everyday situation is when we raise an object.
Now, when we raise an object, energy is transferred to the Earth object system and thus the gravitational field energy of the system will increase.
Now, this energy is usually released when the object falls. The mechanism of this release is known as gravitational force.
In the same manner, two magnetic and electrically charged objects that are interacting at a distance will exert forces on each other and this can lead to transfer of energy between the interacting objects.
<u>Answer:</u> The balanced chemical equation is written below.
<u>Explanation:</u>
Galvanization is defined as the process in which a protective layer of zinc is applied to iron or steel to prevent the metal from rusting.
Zinc prevents the oxidation of iron and acts as a reducing agent in the process.
The half reaction for the process follows:
<u>Oxidation half reaction:</u> 
<u>Reduction half reaction:</u> 
Net chemical equation: 
Hence, the balanced chemical equation is written above.
Answer:
The answer to your question is 122.4 g of O₂
Explanation:
Data
mass of O₂ = ?
moles of H₂O = 7.65
Process
1.- Write the balanced chemical reaction
2H₂O ⇒ 2H₂ + O₂
2.- Convert the moles of H₂O to grams
molar mass of H₂O = 2 + 16 = 18 g
18 g of H₂O ---------------- 1 mol
x ----------------- 7.65 moles
x = (7.65 x 18) / 1
x = 137.7 g H₂O
3.- Calculate the grams of O₂
36 g of H₂O -------------------- 32 g of O₂
137.7 g of H₂O ------------------- x
x = (32 x 137.7) / 36
x = 122.4 g of O₂
<span>The answer is
101.1032 g/mol</span>
Answer:
The concentration of chloride ion is 
Explanation:
We know that 1 ppm is equal to 1 mg/L.
So, the
content 100 ppm suggests the presence of 100 mg of
in 1 L of solution.
The molar mass of
is equal to the molar mass of Cl atom as the mass of the excess electron in
is negligible as compared to the mass of Cl atom.
So, the molar mass of
is 35.453 g/mol.
Number of moles = (Mass)/(Molar mass)
Hence, the number of moles (N) of
present in 100 mg (0.100 g) of
is calculated as shown below:

So, there is
of
present in 1 L of solution.