We are told that KOH is being used to completely neutral H₂SO₄ according to the following reaction:
KOH + H₂SO₄ → H₂O + KHSO₄
If KOH can completely neutralize H₂SO₄, then there must be an equal amount of moles of each as they are in a 1:1 ratio:
0.025 L x 0.150 mol/L = .00375 mol KOH
0.00375 mol KOH x 1 mole H₂SO₄/1 mole KOH = 0.00375 mol H₂SO₄
We are told we have 15 mL of H₂SO₄ initially, so now we can find the original concentration:
0.00375 mol / 0.015 L = 0.25 mol/L
The concentration of H₂SO₄ being neutralized is 0.25 M.
Answer:
c. HF can participate in hydrogen bonding.
Explanation:
<u>The boiling points of substances often reflect the strength of the </u><u>intermolecular forces</u><u> operating among the molecules.</u>
If it takes more energy to separate molecules of HF than of the rest of the hydrogen halides because HF molecules are held together by stronger intermolecular forces, then the boiling point of HF will be higher than that of all the hydrogen halides.
A particularly strong type of intermolecular attraction is called the hydrogen bond, <em>which is a special type of dipole-dipole interaction between the hydrogen atom in a polar bond</em>, such as N-H, O-H, or F-H, and an electronegative O, N, or F atom.
The ideal gas law may be written as

where
p = pressure
ρ =density
T = temperature
M = molar mass
R = 8.314 J/(mol-K)
For the given problem,
ρ = 0.09 g/L = 0.09 kg/m³
T = 26°C = 26+273 K = 299 K
M = 1.008 g/mol = 1.008 x 10⁻³ kg/mol
Therefore

Note that 1 atm = 101325 Pa
Therefore
p = 2.2195 x 10⁵ Pa
= 221.95 kPa
= (2.295 x 10⁵)/101325 atm
= 2.19 atm
Answer:
2.2195 x 10⁵ Pa (or 221.95 kPa or 2.19 atm)
The mass of hydrogen atoms that is measured at 54 u given the relationship is 89.64×10¯²⁴ g
<h3>Conversion scale </h3>
1 u = 1.66×10¯²⁴ g
<h3>How to determine the mass of hydrogen atoms </h3>
- Mass of Hydrogen (u) = 54 u
- Mass of Hydrogen (g) =?
1 u = 1.66×10¯²⁴ g
Therefore
54 u = 54 × 1.66×10¯²⁴ g
54 u = 89.64×10¯²⁴ g
Thus, the mass of the hydrogen atoms measured at 54 u is 89.64×10¯²⁴ g
Learn more about conversion:
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