Answer is: the compound is B₂O₃.
ω(O) = 68.94% ÷ 100%.
ω(O) = 0.6894; percentage of oxygen in the compound.
ω(X) = 31.06% ÷ 100%.
ω(X) = 0.3106; percentage of unknown element in the compound.
If we take 69.7 grams of the compound:
M(compound) = 69.7 g/mol.
n(compound) = 69.7 g ÷ 69.7 g/mol.
n(compound) = 1 mol.
n(O) = (69.7 g · 0.6894) ÷ 16 g/mol.
n(O) = 3 mol.
M(compound) = n(O) · M(O) + n(X) · M(X).
n(X) = 1 mol ⇒ M(X) = 21.7 g/mol; there is no element with this molecular weight.
n(X) = 2 mol ⇒ M(X) = 10.85 g/mol; this element is boron (B).
The average atomic mass of the imaginary element : 47.255 amu
<h3>Further explanation </h3>
The elements in nature have several types of isotopes
Isotopes are elements that have the same Atomic Number (Proton)
Atomic mass is the average atomic mass of all its isotopes
Mass atom X = mass isotope 1 . % + mass isotope 2.% ..
isotope E-47 47.011 amu, 87.34%
isotope E-48 48.008 amu, 6.895
isotope E-49 50.009 amu, 5.77%
The average atomic mass :
Answer:
They are called corpuscles
Explanation:
Answer is: <span>requirement of valine is 0,0222 mol for a100 kilograms adult.
</span>m(C₅H₁₁NO₂) = 2,60 g.
M(C₅H₁₁NO₂) = 5 · 12 + 11 · 1 + 1 · 14 + 2 · 16 · g/mol.
M(C₅H₁₁NO₂) = 117 g/mol.
n(C₅H₁₁NO₂) = m(C₅H₁₁NO₂) ÷ M(C₅H₁₁NO₂).
n(C₅H₁₁NO₂) = 2,60 g ÷ 117 g/mol.
n(C₅H₁₁NO₂) = 0,0222 mol.
n - amount of substance.
m - mass of substance.
M - molar mass of substance.
