The number of C atoms in 0.524 moles of C is 3.15 atoms.
The number of
molecules in 9.87 moles
is 59.43 molecules.
The moles of Fe in 1.40 x
atoms of Fe is 0.23 x 
The moles of
in 2.30x
molecules of
is 3.81.
<h3>What are moles?</h3>
A mole is defined as 6.02214076 ×
of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.
A. The number of C atoms in 0.524 mole of C:
6.02214076 ×
x 0.524 mole
3.155601758 atoms =3.155 atoms
B. The number of
molecules in 9.87 moles of
:
6.02214076 ×
x 9.87
59.4385293 molecules= 59.43 molecules
C. The moles of Fe in 1.40 x
atoms of Fe:
1.40 x
÷ 6.02214076 × 
0.2324754694 x
moles.
0.23 x
moles.
D. The moles of
in 2.30x
molecules of
:
2.30x
÷ 6.02214076 × 
3.819239854 moles=3.81 moles
Learn more about moles here:
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Explanation:
The atoms of one element differs from the atoms of other elements in terms of the number of protons they contain. This is often taken as the atomic number of such an atom.
- The number of proton is the best indicator of the atom one is dealing with.
- Based on this number, elements are categorized into distinct columns and rows on the periodic table.
- The atomic number is the number of protons or positively charge particles in the atom.
II.
It is possible to change the identity of an atom. This is only possible by altering the atomic number of the atom.
Only nuclear reactions have this capability.
When an atom undergoes nuclear reaction that involves change in number of protons, transmutation occurs and a new atom forms.
Answer:
1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.
2) The pH of the solution after adding HCl is 12.6
Explanation:
10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.


There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.
NaOH + HCl ⇒ NaCl + H₂O
Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0
Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³
Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³
The concentration of NaOH is:
![[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M](https://tex.z-dn.net/?f=%5BNaOH%5D%3D%5Cfrac%7B1.0%20%5Ctimes%2010%5E%7B-3%7D%20mol%20%7D%7B25.0%20%5Ctimes%2010%5E%7B-3%7D%20L%7D%20%3D0.040M)
NaOH is a strong base so [OH⁻] = [NaOH].
Finally, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log 0.040 = 1.4
pH = 14 - pOH = 14 - 1.4 = 12.6