Answer:
The strength of an acid or alkali depends on the degree of dissociation of the acid or alkali in water. The degree of dissociation measures the percentage of acid molecules that ionise when dissolved in water. He could use universal indicators or litmus paper for this.
Explanation:
(See answer for the explanation)
Answer:
Its probably none of those.
Explanation:
White dwarf temperatures can exceed 100,000 Kelvin according to NASA (that's about 179,500 degrees Fahrenheit). Despite these sweltering temperatures, white dwarfs have a low luminosity as they're so small in size according to NMSU.
Answer:
A jump occurs when a core electron is removed.
Explanation:
A jump in ionization energy occurs when a core electron is removed. A large jump in the ionization energy easily be seen from the electronic configuration of an element.
For Beryllium, the electronic configuration of is 1s2 2s2.
There are two valence electrons in the outermost shell hence the ionization energy data for beryllium will show a sudden jump or increase in going from the second to the third ionization energy owing to the removal of a core electron
The electronic configuration for Nitrogen is 1s2 2s2 2p3. Five valence electrons are found in the outermost shell so the ionization energy data for nitrogen will show a sudden jump or increase in going from the fifth to sixth ionization energy because of the removal of a core electron
The electronic configuration of oxygen is 1s2 2s2 2p4. There are six valence electrons hence ionization energy for oxygen atom will show a sudden jump or increase in going from the sixth to the seventh ionization energy because of the removal of a core electron
The electronic configuration of Lithium is 1s2 2s1
There is one valence electron in its outermost shell so its ionization energy data will show a sudden jump or increase in going from the first to the second ionization energy because of the removal of a core electron.
Answer:
C
Explanation:
The higher the period the higher the activity of an element, therefore, since iodine is in period 6 and bromine is in period 5, the described reaction is not possible due to the fact that bromine is less active
B) 40%
The balanced equation indicates that for every 3 moles of H2 used, 2 moles of NH3 will be produced. So the reaction if it had 100% yield would produce (2.00 / 3) * 2 = 1.333333333 moles of NH3. But only 0.54 moles were produced. So the percent yield is 0.54 / 1.3333 = 0.405 = 40.5%. This is a close enough match to option "b" to be considered correct.
