Answer:
Explanation:
To solve this problem, we must understand the relationship between mass of a substance and the number of atoms.
Atoms are the smallest indivisible particles of any matter. A substance can be made up of several number of atoms in their space.
The mass of any substance is a function of the amount of atoms its contains.
The mass of a substance is related in chemistry to the amount of atoms its contains using the parameter called the number of moles.
A mole is the amount of substance that contains the Avogadro's number of particles. This number is 6.02 x 10²³ particles. The particles here can be protons, neutrons, electrons, atoms e.t.c.
Now,
Number of moles = 
Molar mass of copper = 63.6g/mole
Number of moles = 
= 0.03mole
Since 1 mole of a substance contains 6.02 x 10²³atoms
0.03 mole of copper will contain 0.03 x 6.02 x 10²³atoms
= 1.89 x 10²² atoms
He needs to add 1.89 x 10²² atoms to make 2g of the sample.
O2 is the limited reactant
Answer:
Option-C (27.36% Na, 1.20% H, 14.30% C, and 57.14% O)
Explanation:
<em>Percent Composition</em> is defined as the <u><em>%age by mass of each element present in a compound</em></u>. Therefore, it is a relative amount of each element present in a compound.
Calculating Percent Composition of NaHCO₃:
1: Calculating Molar Masses of all elements present in NaHCO₃:
a) Na = 22.99 g/mol
b) H = 1.01 g/mol
c) C = 12.01 g/mol
d) O₃ = 16.0 × 3 = 48 g/mol
2: Calculating Molecular Mass of NaHCO₃:
Na = 22.99 g/mol
H = 1.01 g/mol
C = 12.01 g/mol
O₃ = 48 g/mol
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Total 84.01 g/mol
3: Divide each element's molar mass by molar mass of NaHCO₃ and multiply it by 100:
For Na:
= 22.99 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 27.36 %
For H:
= 1.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 1.20 %
For C:
= 12.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 14.29 % ≈ 14.30 %
For O:
= 48.0 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100
= 57.13 % ≈ 57.14 %
C, The atomic mass. This could also cause certain elements (i.e. Uranium, Plutonium) to radioactively decay in process called nuclear fission.
From the calculation, the standard free energy of the system is -359kJ.
<h3>What is the standard free-energy?</h3>
The standard free-energy is the energy present in the system. We have to first obtain the cell potential using the formula;
Ereduction - E oxidation = 0.96 V - 0.34 V = 0.62 V
Using the formula;
ΔG = -nFEcell
ΔG =-(6 * 96500 * 0.62)
ΔG =-359kJ
Learn more about free energy:brainly.com/question/15319033
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