Answer:
This cannot be determined without knowing the actual mass of the objects.
Explanation:
its like trying to compare the letter A and letter B
C is the answer.
The temperature T<span> in degrees Celsius (°C) is equal to the temperature </span>T<span> in Kelvin (K) minus 273</span>°.
Gold has a heavy enough nucleus that its electrons must travel at speeds nearing the speed of light to prevent them from falling into the nucleus. This relativistic effect applies to those orbitals that have appreciable density at the nucleus, such as s and p orbitals. These relativistic electrons gain mass and as a consequence, their orbits contract. As these s and (to some degree) p orbits are contracted, the other electrons in d and f orbitals are better screened from the nucleus and their orbitals actually expand.
Since the 6s orbital with one electron is contracted, this electron is more tightly bound to the nucleus and less available for bonding with other atoms. The 4f and 5d orbitals expand, but can't be involved in bond formation since they are completely filled. This is why gold is relatively unreactive.
Hope it helps
Answer:
D is the answer to this problem
Answer:
See Explanation
Explanation:
The equation of the reaction;
KHSO4(aq) + KOH(aq) -------> K2SO4(aq) + H2O(l)
Number of moles of KHSO4 = 49.6 g/136.169 g/mol = 0.36 moles
Since the reaction is in a mole ratio of 1:1, 0.36 moles of K2SO4 is produced.
Number of moles of KOH = 25.3 g/56.1056 g/mol = 0.45 moles
Since the reaction is 1:1, 0.45 moles of K2SO4 is produced
Hence K2SO4 is the limiting reactant.
Mass of K2SO4 formed = 0.36 moles of K2SO4 * 174.26 g/mol = 62.7 g
So;
1 mole of KHSO4 reacts with 1 mole of KOH
0.36 moles of KHSO4 reacts with 0.36 * 1/1 = 0.36 moles of KOH
Amount of excess KOH = 0.45 moles - 0.36 moles = 0.09 moles
Mass of excess KOH = 0.09 moles * 56.1056 g/mol = 5 g of excess KOH
