Saturated fat, milk, cheese, and meat.
Answer:
Option 3. The catalyst does not affect the enthalpy change (
) of a reaction.
Explanation:
As its name suggests, the enthalpy change of a reaction (
) is the difference between the enthalpy of the products and the reactants.
On the other hand, a catalyst speeds up a reaction because it provides an alternative reaction pathway from the reactants to the products.
In effect, a catalyst reduces the activation energy of the reaction in both directions. The reactants and products of the reaction won't change. As a result, the difference in their enthalpies won't change, either. That's the same as saying that the enthalpy change
of the reaction would stay the same.
Refer to an energy profile diagram. Enthalpy change of the reaction
measures the difference between the two horizontal sections. Indeed, the catalyst lowered the height of the peak. However, that did not change the height of each horizontal section or the difference between them. Hence, the enthalpy change of the reaction stayed the same.
<span><span>V. C. Wynne-Edwards and Konrad Lorenz were
the first authors to raise the concept of
altruism</span><span>. This was later
reiterated by </span>David Sloan Wilson, E. O. Wilson and Elliott Sober<span> in 1994 whose works referred to social organisms such as ants</span></span>.
<span>4NH</span>₃<span> + 6NO → 5N</span>₂<span> + 6H</span>₂<span>O
mol of NO = </span>
=
= 0.93 mol
Based on the balance equation mole ratio of NH₃ : NO is 4 : 6
= 2 : 3
If mol of NO = 0.93 mol
then mol of NH₃ =
= 0.62 mol
Mass of ammonia = mol × molar mass
= 0.62 mol × 17.03 g/mol
= 10.54 g
Therefore B is the best answer
Answer:
NO3-
Explanation:
Given the reaction equation;
Au(s) + 3NO3-(aq) + 6H+(aq)→Au3+(aq) + 3NO2(g) + 3H2O (l).
We can consider the oxidation states of species on the left and right hand sides of the reaction equation;
Au is in zero oxidation state on the left hand side and an oxidation state of +3 on the righthand side.
NO3- is in oxidation state of +5 on the righthand side and NO2 is in + 4 oxidation state.
H+ is in + 1 oxidation state on both the left and right hand sides of the reaction equation.
Since reduction has to do with a decrease in oxidation number, it follows that NO3- was reduced in the reaction.