The answer is 64.907 amu.
The atomic mass of an element is the average of the atomic masses of its isotopes. The relative abundance of isotopes must be taken into consideration, therefore:
atomic mass of copper = atomic mass of isotope 1 * abundance 1 + atomic mass of isotope 2 * abundance 2
We know:
atomic mass of copper = 63.546 amu
The atomic mass of isotope 1 is: 62.939 amu
The abundance of isotope 1 is: 69.17% = 0.6917
The atomic mass of isotope 1 is: x
The abundance of isotope 2: 100% - 69.17% = 30.83% = 0.3083
Thus:
63.546 amu = 62.939 amu * 0.6917 + x * 0.3083
63.546 <span>amu = 43.535 amu + 0.3083x
</span>⇒ 63.546 amu - 43.535 amu = 0.3083x
⇒ 20.011 amu = 0.3083x
⇒ x = 20.011 amu ÷ 0.3083 = 64.907 amu
Answer:
e. 3.08 x 10⁻² mol of ions.
Explanation:
- Every 1.0 mole of any compound contains Avogadro's number of molecules (6.022 x 10²³).
- We can get the no. of moles of NiCl₂ using cross multiplication:
1.0 mol NiCl₂ contains → 6.022 x 10²³ molecules.
??? mol NiCl₂ contains → 6.188 x 10²¹ molecules.
∴ The no. of moles of NiCl₂ = (1.0 mol)(6.188 x 10²¹ molecules)/(6.022 x 10²³ molecules) = 1.028 x 10⁻² mol.
- NiCl₂ is ionized according to the equation:
NiCl₂ → Ni²⁺ + 2Cl⁻.
Which means that every 1.0 mol of NiCl₂ is ionized to produce 3.0 moles (1.0 mol of Ni²⁺ and 2 moles of Cl⁻).
<em>∴ The total moles of ions are released</em> = 3 x 1.028 x 10⁻² mol = <em>3.083 x 10⁻² mol of ions.</em>
<span>2C2H6 + 7O2 = 4CO2 + 6H2O
</span>
According to the equation of the reaction of ethane combustion, ethane and carbon dioxide have following stoichiometric ratio:
n(C2H6) : n(CO2) = 1 : 2
n(CO2) = 2 x n(C2H6)
n(CO2) = 2 x 5.2 = 10.4 mole of CO2 is formed
The percent of O in Cr₂O₃ : 31.58%
<h3>Further explanation</h3>
Given
Cr = 52.00 amu, O = 16.00 amu
Required
The percent of O
Solution
MW Cr₂O₃ = 2 x Ar Cr + 3 x Ar O
MW Cr₂O₃ = 2.52+3.16
MW Cr₂O₃ =152 amu
