Answer:
a)15.5 mmHg is the partial pressure of water vapor.
b) 724.5 mmHg is the partial pressure of hydrogen gas.
Explanation:
Total pressure of gases = T = P = 740 mmHg
Vapor pressure of the water = 
Partial pressure of the hydrogen gas = 
Partial pressure of the water = 
15.5 mmHg is the partial pressure of water vapor.
Using Dalton's law of partial pressure:



724.5 mmHg is the partial pressure of hydrogen gas.
Answer:
7.35 moles of oxygen
Explanation:
First of all, for 1 mole of H₂CO₃ we have 3 moles of oxygen (can be deduced from the chemical formula of the acid), then the moles of oxygen in 2.45 mole of the compound, which are given in the question, from the carbonic acid will be:
If in 1 mole of H₂CO₃ we have 3 moles of oxygen
The in 2.45 moles of H₂CO₃ we have X moles of oxygen
X = ( 3 × 2.45 ) / 1 = 7.35 moles of oxygen
Answer:
a. withdraws electrons inductively
b. donates electrons by hyperconjugation
c. donates electrons by resonance
d. withdraws electrons inductively
Explanation:
a. The bromide ion is a highly electronegative ion (in the halide series). Electronegative substituents on acids increase the acidity by inductive electron withdrawal method. The higher the electronegativity of a substance, the greater the acidity. The halogens have this order of electronegativity:
F > Cl > Br>I
b. The carboxyl groups have a stabilization of the sigma and pi bonds. This is achieved through a special delocalization of electrons. Because of the delocalization, hyperconjugation is the result effect.
c. The NHCH₃ group has a highly electonegative nitrogen atom that pulls the electron cloud towards itself. In this case, it withdraws electrons inductively. As a result, it donates electrons by resonance.
d. The OCH₃ group has a highly electonegative oxygen atom. This oxygen atom withdraws electron cloud towards itself. As a result, it withdraws electrons inductively.
Yes it is correct for that answer
The given concentration of boric acid = 0.0500 M
Required volume of the solution = 2 L
Molarity is the moles of solute present per liter solution. So 0.0500 M boric acid has 0.0500 mol boric acid present in 1 L solution.
Calculating the moles of 0.0500 M boric acid present in 2 L solution:

Converting moles of boric acid to mass:

Therefore, 6.183 g boric acid when dissolved and made up to 2 L with distilled water gives 0.0500 M solution.