Weather... weather is the obvious answer
Answer: 1.4x10-3 g N2O4
Explanation: First convert molecules of N2O4 to moles using Avogadro's Number. Then convert moles to mass using the molar mass of N2O4.
9.2x10^18 molecules N2O4 x 1 mole N2O4 / 6.022x10²³ molecules N2O4
= 1.53x10-5 moles N2O4
1.53x10-5 moles N2O4 x 92 g N2O4/ 1 mole N2O4
= 1.4x10-3 g N2O4
An isotope of nitrogen containing 7 neutrons would be nitrogen-7
Given buffer:
potassium hydrogen tartrate/dipotassium tartrate (KHC4H4O6/K2C4H4O6 )
[KHC4H4O6] = 0.0451 M
[K2C4H4O6] = 0.028 M
Ka1 = 9.2 *10^-4
Ka2 = 4.31*10^-5
Based on Henderson-Hasselbalch equation;
pH = pKa + log [conjugate base]/[acid]
where pka = -logKa
In this case we will use the ka corresponding to the deprotonation of the second proton i.e. ka2
pH = -log Ka2 + log [K2C4H4O6]/[KHC4H4O6]
= -log (4.31*10^-5) + log [0.0451]/[0.028]
pH = 4.15
Answer:
Option B. A
Explanation:
From the question given above, the following data were obtained:
C(s) + 2H₂ (g) —> CH₄ (g). ΔH = –74.9 kJ
From the reaction above, we can see that the enthalpy change (ΔH) is negative (i.e –74.9 KJ) which implies that the heat content of the reactants is greater than the heat content of the products. Thus, the reaction is exothermic reaction.
For an exothermic reaction, the energy profile diagram is drawn in such a way that the heat content of reactants is higher than the heat content of products because the enthalpy change
(ΔH) is always negative.
Thus, diagram A (i.e option B) gives the correct answer to the question.
