Answer:
B. Decreasing the pressure applied to the gas molecules
Explanation:
According to Boyle's Law, the pressure of the gas is inversely proportional to the volume of the gas. So, the option B is correctly implied to it.
Other values such as Temperature, Number of molecules are inversely proportional to the volume of the gas.
Size of the gas molecules is negligible as compared to volume.
Answer:
Chemical reactivity increases down a group and decreases from left to right of a period.
Explanation:
The higher the ionization energy is, the lower the reactivity is. Since the ionization energy is highest in the top right corner of the periodic table, we can assume that the most reactive elements are in the opposite bottom left corner. This is because the electrons that react are farther away from the nucleus thus experience less attraction to the nucleus (called nuclear shielding). Therefore their electrons are more easily removed than elements that don't ecperience nuclear shielding.
Complete Question:
Ions to calculate the p-values: Na⁺, Cl⁻, and NH₄⁺
Answer:
pNa = 0.307
pCl = 0.093
pNH₄ = 0.503
Explanation:
The p-value is calculated by the antilog of the concentration of the substance of interest. For example, pH = -log[H⁺]. Thus, first, let's find the ions concentration.
Both substances are salts that solubilize completely, thus, by the solution reactions:
NaCl → Na⁺ + Cl⁻
NH₄Cl → NH₄⁺ + Cl⁻
So, for both reactions the stoichiometry is 1:1:1 and the concentration of the ions is equal to the concentration of the salts.
[Na⁺] = 0.493 M
[Cl⁻] = 0.493 + 0.314 = 0.807 M
[NH₄⁺] = 0.314 M
The p-values are:
pNa = -log[Na⁺] = -log(0.493) = 0.307
pCl = -log[Cl⁻] = -log(0.807) = 0.093
pNH₄ = -log[NH₄⁺] = -log(0.314) = 0.503
Molarity = moles / liter
a) M = 2/4 = 0.5 M
b) Moles = 4/(30 + 16 + 1)
= 0.085
M = 0.085 / 2 = 0.0425 M
c) Moles = 5.85 / (23 + 35.5)
= 0.1
M = 0.1 / 0.4
= 0.25 M
for it to be balanced in this case would be " <em>4</em> C6H6 + <em>6</em> CI2 = <em>3</em> C6H5CI + <em>9</em> HCI" therefore it's be a <u>Double Replacement</u>
