Answer:sodium
Explanation: because its not hard to look on the PERIODIC TABLE
Answer:
See explanation
Explanation:
a) The magnitude of intermolecular forces in compounds affects the boiling points of the compound. Neon has London dispersion forces as the only intermolecular forces operating in the substance while HF has dipole dipole interaction and strong hydrogen bonds operating in the molecule hence HF exhibits a much higher boiling point than Ne though they have similar molecular masses.
b) The boiling points of the halogen halides are much higher than that of the noble gases because the halogen halides have much higher molecular masses and stronger intermolecular forces between molecules compared to the noble gases.
Also, the change in boiling point of the hydrogen halides is much more marked(decreases rapidly) due to decrease in the magnitude of hydrogen bonding from HF to HI. The boiling point of the noble gases increases rapidly down the group as the molecular mass of the gases increases.
Answer:
4) Titration
Explanation:
Titration is a common process used to determine the concentration of acids. It does this by adding a solution of base with a known concentration to the acid until it reaches neutralization.
Answer:
Helium
Explanation: The ionization energy decreases from the top to bottom in groups. And increases from left to the right across a period. Therefor Helium has tge largest first ionization energy, while francium has one of the lowest.
Answer:
1.58x10⁻⁵
2.51x10⁻⁸
0.0126
63.10
Explanation:
Phenolphthalein acts like a weak acid, so in aqueous solution, it has an acid form HIn, and the conjugate base In-, and the pH of it can be calculated by the Handerson-Halsebach equation:
pH = pKa + log[In-]/[HIn]
pKa = -logKa, and Ka is the equilibrium constant of the dissociation of the acid. [X] is the concentrantion of X. Thus,
i) pH = 4.9
4.9 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = - 4.8
[In-]/[HIn] = 
[In-]/[HIn] = 1.58x10⁻⁵
ii) pH = 2.1
2.1 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = -7.6
[In-]/[HIn] = 
[In-]/[HIn] = 2.51x10⁻⁸
iii) pH = 7.8
7.8 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = -1.9
[In-]/[HIn] = 
[In-]/[HIn] = 0.0126
iv) pH = 11.5
11.5 = 9.7 + log[In-]/[HIn]
log[In-]/[HIn] = 1.8
[In-]/[HIn] = 
[In-]/[HIn] = 63.10
