Answer:
2.52L
Explanation:
Given parameters:
T₁ = 400K
V₁ = 4L
T₂ = 252K
unknown
V₂ = ?
Solution:
To solve this problem, we are going to apply charle's law. The law states that the volume of a fixed mass of gas is directly proportional to temperature provided pressure is constant.
Mathematically,
Substitute and solve for V₂
V₂ = 2.52L
Answer:
see note under explanation
Explanation:
When describing system and surroundings the system is typically defined as the 'object of interest' being studied and surroundings 'everything else'. In thermodynamics heat flow is typically defined as endothermic or exothermic. However, one should realize that the terms endothermic and exothermic are in reference to the 'system' or object of interest being studied. For example if heat is transferred from a warm object to a cooler object it is imperative that the system be defined 1st. So, with that, assume the system is a warm metal cylinder being added into cooler water. When describing heat flow then the process is exothermic with respect to the metal cylinder (the system) but endothermic to the water and surroundings (everything else).
Answer:
a. 3; b. 5; c. 10; d. 12
Explanation:
pH is defined as the negative log of the hydronium concentration:
pH = -log[H₃O⁺] (hydronium concentration)
For problems a. and b., HCl and HNO₃ are strong acids. This means that all of the HCl and HNO₃ would ionize, producing hydronium (H₃O⁺) and the conjugate bases Cl⁻ and NO₃⁻ respectively. Further, since all of the strong acid ionizes, 1 x 10⁻³ M H₃O⁺ would be produced for a., and 1.0 x 10⁻⁵ M H₃O⁺ for b. Plugging in your calculator -log[1 x 10⁻³] and -log[1.0 x 10⁻⁵] would equal 3 and 5, respectively.
For problems c. and d. we are given a strong base rather than acid. In this case, we can calculate the pOH:
pOH = -log[OH⁻] (hydroxide concentration)
Strong bases similarly ionize to completion, producing [OH⁻] in the process; 1 x 10⁻⁴ M OH⁻ will be produced for c., and 1.0 x 10⁻² M OH⁻ produced for d. Taking the negative log of the hydroxide concentrations would yield a pOH of 4 for c. and a pOH of 2 for d.
Finally, to find the pH of c. and d., we can take the pOH and subtract it from 14, giving us 10 for c. and 12 for d.
(Subtracting from 14 is assuming we are at 25°C; 14, the sum of pH and pOH, changes at different temperatures.)
