Write the molecular formula for a compound with the possible elements C, H, N and O that exhibits a molecular ion at M+ = 84.095

1 answer:

8 0

Use the rule of 13 - the formula of a compound is a multiple n of 13 (the molar mass of CH) plus a remainder r. For O, add O and remove CH₄ and for N, add N and remove one CH₂.
84 ÷ 13 = 6r6, the integer equals the number of carbon atoms and the integer plus the remainder equals the number of hydrogen atoms, so possible formula is C₆H₁₂.
M(C₆H₁₂) = 6 · 12,00 + 12 · 1,008 = 84,096.
Add one oxygen atom to molecular formula: C₅H₈O.
M(C₅H₈O) = 5 · 12,00 + 8 · 1,008 + 15,995 = 84,059, not.
Add one nitrogen atom to molecular formula: C₅H₁₀N.
M(C₅H₁₀N) = 5 · 12,00 + 10 · 1,008 + 14,003 = 84,083, not.
If we add nitrogen and oxygen mass will be lower than 84,095, so molecular formula is C₆H₁₂.

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Read 2 more answers

The value of ka for nitrous acid (hno2) at 25 ∘c is 4.5×10−4. what is the value of δg at 25 ∘c when [h+] = 5.9×10−2m , [no2-] =

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To get the value of ΔG we need to get first the value of ΔG°:

when ΔG° = - R*T*㏑K

when R is constant in KJ = 0.00831 KJ

T is the temperature in Kelvin = 25+273 = 298 K

and K is the equilibrium constant = 4.5 x 10^-4

so by substitution:

∴ ΔG° = - 0.00831 * 298 K * ㏑4.5 x 10^-4

= -19 KJ

then, we can now get the value of ΔG when:

ΔG = ΔG° - RT*㏑[HNO2]/[H+][NO2]

when ΔG° = -19 KJ

and R is constant in KJ = 0.00831

and T is the temperature in Kelvin = 298 K

and [HNO2] = 0.21 m & [H+] = 5.9 x 10^-2 & [NO2-] = 6.3 x 10^-4 m

so, by substitution:

ΔG = -19 KJ - 0.00831 * 298K* ㏑(0.21/5.9x10^-2*6.3 x10^-4 )

= -40

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