Given: C= 81.70% = 81.70g
H = 18.29% = 18.29g
<span>The number of moles is given by: n= Given mass (m)/Molar Mass (M)
</span>M of C = 12 g/mol
M of H = 1 g/mol
Thus, the number of moles of carbon = 81.70g / 12gmol= 6.83moles
and, the number of moles of hydrogen = 18.29/1g/mol = 18.29 moles
The ration of C moles with hydrogen :
H:C = 18.29moles/6.83moles= 2.67 ≈3
Thus, the empirical formula is C3H8
Answer:
Mass released = 8.6 g
Explanation:
Given data:
Initial number of moles nitrogen= 0.950 mol
Initial volume = 25.5 L
Final mass of nitrogen released = ?
Final volume = 17.3 L
Solution:
Formula:
V₁/n₁ = V₂/n₂
25.5 L / 0.950 mol = 17.3 L/n₂
n₂ = 17.3 L× 0.950 mol/25.5 L
n₂ = 16.435 L.mol /25.5 L
n₂ = 0.644 mol
Initial mass of nitrogen:
Mass = number of moles × molar mass
Mass = 0.950 mol × 28 g/mol
Mass = 26.6 g
Final mass of nitrogen:
Mass = number of moles × molar mass
Mass = 0.644 mol × 28 g/mol
Mass = 18.0 g
Mass released = initial mass - final mass
Mass released = 26.6 g - 18.0 g
Mass released = 8.6 g
Oxidation number of an atom is the charge that atom would have if the compound is composed of ions. In neutral substances that contains atoms of one element the oxidation number of an atom is zero. Thus atoms in O2, Ni2, and aluminium all have oxidation number of zero.
In this case, Ni2, the oxidation number of Ni atom is zero,
for NiO4-, assuming oxidation number of Ni is x
(x ×1) + (-2 × 4) = -1
x = + 7
Therefore, the oxidation number goes from 0 to +7
Answer:
1000000000000
Explanation:
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Answer:
1)The proximity of the positively charged phosphorous and negatively charged carbon stabilizes the charges.
2) Inductive effects and resonance stabilize the negative charge
Explanation:
both atoms have full octets of electrons( I.e Carbon and say phosphorus). The result can be viewed as a structure in which two adjacent atoms are connected by both a covalent and an ionic bond; normally written X+–Y−. Ylides are thus 1,2-dipolar compounds, and a subclass of zwitterions
