Physical change
Giddy Up!!!!!
Given that the volume and amount of water are kept constant,
P/T = constant
P₁/T₁ = P₂/T₂
Normal atmospheric pressure is 746 mmHg and normal boiling point of water is 100 °C.
746/100 = 589/T₂
T₂ = 79.0 °C
Carbon-14 is a radioactive isotope used to date organic material. Its consistent rate of decay allows the age of an object to be determined by the proportion of carbon-14 to other carbon isotopes. This process is called radiocarbon dating. Carbon-14 is also used as a radioactive tracer for medical tests.
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.)
Answer: 400K
Explanation:
Given that,
Original volume of balloon V1 = 3.0L
Original temperature of balloonT1 = 27°C
Convert the temperature in Celsius to Kelvin
(27°C + 273 = 300K)
New volume of balloon V2 = 4.0L
New temperature of balloon T2 = ?
Since volume and temperature are given while pressure is constant, apply the formula for Charle's law
V1/T1 = V2/T2
3.0L/300K = 4.0L/T2
To get the value of T2, cross multiply
3.0L x T2 = 4.0L x 300K
3.0LT2 = 1200LK
Divide both sides by 3.0L
3.0LT2/3.0L = 1200LK/3.0L
T2 = 400K
Thus, at a temperature of 400 Kelvin, the balloon would have a volume of 4.0L.
