Answer:
+VE
Explanation:
If we look at the reaction profile pictured in the question, we can easily identify A as the enthalpy of the reaction. The enthalpy of reaction (ΔHrxn) is usually defined as the difference between the total enthalpy (heat content) of the products of a reaction and the total enthalpy (heat content) of the reactants in that reaction.
Looking at the figure, we can see that the enthalpy of products is greater than the enthalpy of reactants, hence ∆Hrxn is positive as stated in the answer above.
There are:
3.41 moles of C
4.54 moles of H
3.40 moles of O.
Why?
To solve the problem, the first thing that we need to do is to write the chemical formula of the ascorbic acid.
Now, we know that there are 100 grams of the compound, so, the masses of each element will represent the percent in the compound.
We have that:
To know the percent of each element, we need to to the following:
So, we know that for the 100 grams of the compound, there are:
40.92 grams of C
4.58 grams of H
54.49 grams of O
We know the molecular masses of each element:
Now, to calculate the number of moles of each element, we need to divide the mass of each element by the molecular mass of each element:
Hence, we have that there are 3.41 moles of C, 4.54 moles of H, and 3.40 moles of O.
Have a nice day!
Answer:
N₂ + 3H₂ → 2NH₃ ΔH = - 92.2KJ
Explanation:
Let's write out the chemical equation between Nitrogen and Hydrogen to Form Ammonia.
Nitrogen + Hydrogen = Ammonia
N₂ + H₂ → NH₃
A Thermochemical Equation is a balanced stoichiometric chemical equation that includes the enthalpy change, ΔH.
The balanced stoichiometric chemical equation is given as;
N₂ + 3H₂ → 2NH₃
92.2 kJ of energy are evolved for each mole of N2(g) that reacts. And from the equation, 1 mole of N2 reacts.
The enthalpy change, ΔH = - 92.2KJ. The negative sign is because heat is being evolved.
The balanced thermochemical equation;
N₂ + 3H₂ → 2NH₃ ΔH = - 92.2KJ
Answer:
Element 2
Explanation:
If we look at the model stated for element 1, it is clear that element 1 must be a noble gas. It has eight electrons in its outermost shell this implies that it has already attained a complete octet of electrons and is reluctant towards chemical reaction.
The second element belongs to group 16 since it has six electrons on its outermost shell. It is certainly more reactive than element 1 which is a noble gas.
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
