the amount of heat produced from the combustion of 24.3 g benzene (c6h6) is ΔH = -976.5 kJ
There are two moles of benzene involved in the process (C6H6). Since the heat of this reaction is -6278 kJ, the burning of 2 moles of benzene will result in a heat loss of 6278 kJ. This reaction is exothermic.
Enthalpy, or the value of H, is a unit of measurement for heat that relies on the amount of matter present (number of moles).
Thus, 24.3 g of benzene contains:
n = mass/molar mass, where n = 24.3/78.11, and n = 0.311 moles.
2 moles = 6278 kJ
0.311 moles =x
By the straightforward direct three rule:
2x = -1953.08 x = -976.5 kJ
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First, we need to calculate moles of hydrazoic acid NH3:
moles NH3 = molarity * volume
= 0.15 m * 0.025 L
= 0.00375 moles
moles NaOH = molarity * volume
= 0.15 m * 0.015 L
= 0.00225 moles
after that we shoul get the total volume = 0.025L + 0.015L
= 0.04 L
So we can get the concentration of NH3 & NaOH by:
∴[NH3] = moles NH3 / total volume
= 0.00375 moles / 0.04 L
= 0.09375 M
∴[NaOH] = moles NaOH / total volume
= 0.00225 moles / 0.04 L
= 0.05625 M
then, when we have the value of Ka of NH3 so we can get the Pka value from:
Pka = -㏒Ka
= - ㏒ 1.9 x10^-5
= 4.7
finally, by using H-H equation we can get PH:
PH = Pka + ㏒[salt/ basic]
PH = 4.7 +㏒[0.05625/0.09375]
∴ PH = 4.48
To explain your first paragraph which includes your thesis. The second paragrph supports your first pragraph
When u add the solution to the chemical's
