The total number of atoms in 7.10g of chlorine is 1.204 × 10²³atoms.
HOW TO CALCULATE NUMBER OF ATOMS:
- The number of atoms in a substance can be calculated by multiplying the number of moles in that substance by Avogadro's number as follows:
- no. of atoms = no. of moles × 6.02 × 10²³ mol-¹
- The number of moles in 7.10g of Cl is calculated as follows:
no. of moles = mass ÷ molar mass
no. of moles = 7.10g ÷ 35.5g/mol
no. of moles = 0.2mol
no of atoms = 0.2mol × 6.02 × 10²³
no. of atoms = 1.204 × 10²³atoms.
- Therefore, the total number of atoms in 7.10g of chlorine is 1.204 × 10²³atoms.
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Answer : The mass of sample is, 267.5 grams.
Explanation :
Density : It is defined as the mass of a substance contained per unit volume.
Formula used :

Given:
Volume of Pb = 
Density of Pb = 
Now put all the given values in the above formula, we get the mass of Pb.


Therefore, the mass of sample is, 267.5 grams.
C. crop rotation
Explanation:
Crop rotation is a farming practice that is commonly used to maintain a healthy soil.
This practices helps to conserve and appropriate a soil for the best use.
During crop rotation, different crop species are planted during a growing season. The patterns are designed to cultivate both crops that replenish and deplete the nutrients in soil.
Through this the fertility of the soil is sustained all year round.
This helps to avoid the use of pesticides on crops.
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Answer: a) Anode: 
Cathode: 
b) Anode : Cr
Cathode : Au
c) 
d) 
Explanation: -
a) The element Cr with negative reduction potential will lose electrons undergo oxidation and thus act as anode.The element Au with positive reduction potential will gain electrons undergo reduction and thus acts as cathode.
At cathode: 
At anode: 
b) At cathode which is a positive terminal, reduction occurs which is gain of electrons.
At anode which is a negative terminal, oxidation occurs which is loss of electrons.
Gold acts as cathode ad Chromium acts as anode.
c) Overall balanced equation:
At cathode:
(1)
At anode:
(2)
Adding (1) and (2)

d)
= -0.74 V
= 1.40 V

Using Nernst equation :
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Au^{3+}]}{[Cr^{3+}]^}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BAu%5E%7B3%2B%7D%5D%7D%7B%5BCr%5E%7B3%2B%7D%5D%5E%7D)
where,
n = number of electrons in oxidation-reduction reaction = 3
= standard electrode potential = 2.14 V
![E_{cell}=2.14-\frac{0.0592}{3}\log \frac{[1.0}{[1.0]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D2.14-%5Cfrac%7B0.0592%7D%7B3%7D%5Clog%20%5Cfrac%7B%5B1.0%7D%7B%5B1.0%5D%7D)

Thus the standard potential for an electrochemical cell with the cell reaction is 2.14 V.