The question says that when the ball is placed at the top of the hill it has a lot of potential energy. So I think the higher a ball is placed the more potential energy it will have
Here are the given:
<span>ΔHf° = –423 kJ/mol
</span> ΔHsub = 119 kJ/mol
IE = 469 kJ/mol
ΔHEA = –301 kJ/mol
BE = 161 kJ/mol
The lattice energy of the compound is solved using the formula:
U = <span>ΔHf° - </span>ΔHsub - BE - IE - ΔHEA
U = -423 - 119 - 161 - 469 - (-301)
U = -871 kJ/mol
Therefore, the lattice energy is 871 kJ/mol (released).
In a 0.20 M solution, a weak acid is 3.0% dissociated, the value of
H₃O⁺ =9.19 × 10⁻² OH⁻ = 1.09 × 10⁻¹³, pH =1.04 , pOH = 12.96
<h3>What is pH?</h3>
The term pH, which originally stood for "potential of hydrogen" (or "power of hydrogen"), is used in chemistry to describe how acidic or basic an aqueous solution is. Lower pH values are summarized for acidic solutions (solutions with higher H+ ion concentrations) than for basic or alkaline solutions.
The pH scale is inversely indicates to the concentration of hydrogen ions in the solution and is logarithmic.
⇒pH = -log(
)
Acidic solutions are those with a pH below 7, and basic solutions are those with a pH above 7, at a temperature of 25 °C (77 °F). At this temperature, solutions with a pH of 7 are neutral (e.g. pure water). The pH neutrality relies on temperature, falling below 7 if the temperature rises above 25 °C.
Lets find [H₃O⁺]
Because 12.5% of the weak acid dissociated, 12.5% of the concentration of the weak acid also produced H3O.
H₃O = HA × 0.125
= 0.735 M × 0.125
= 9.19 × 10⁻²
Lets find [OH⁻]
Using the Kw = 1.0 × 10⁻¹⁴
Kw = [H₃O⁺][OH⁻]
[OH⁻] = Kw / [H₃O⁺]
= 1.0 × 10⁻¹⁴ / 9.19 × 10⁻²
= 1.09 × 10⁻¹³
Lets find pH
pH = -log[H₃O⁺]
= -log(9.19 × 10⁻²)
= 1.04
Lets find pOH
Using pH +pOH = 14
pH +pOH = 14
pOH = 14 - pH
= 14 - 1.04
= 12.96
Learn more about pH
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Answer:
M(NaSO4)=23+32+16*4=119 g/moles
n=200/119=1.68
Explanation:
Answer:
question not clear can u rewrite