Answer:
d. T1 = 7°C; T2 = 29°C
Explanation:
The temperature of a substance is a measure of the average kinetic energy of is particle: in fact, the two quantities are proportional to each other.
Moreover, heat is always transferred spontaneously from a hotter object (higher temperature) to a colder object (lower temperature).
In this problem, we have a container placed in contact with a bucket of hot water: heat flows from the hot water to the container, until the two are at the same temperature.
The amount of energy transferred between the two is proportional to the initial temperature difference between the container and the water:

Since in the two situations the amount of energy transferred is the same, then it means that the temperature difference between the two substances is the same in the two situations. So we can write:

And from the choices given, we see that the only option that satisfies this condition is

Answer:
6.574 g NaF into 300ml (0.25M HF) => Bfr with pH ~3.5
Explanation:
For buffer solution to have a pH-value of 3.5 the hydronium ion concentration [H⁺] must be 3.16 x 10⁻⁴M ( => [H⁺] = 10^-pH = 10⁻³°⁵ =3.16 x 10⁻⁴M).
Addition of NaF to 300ml of 0.25M HF gives a buffer solution. To determine mass of NaF needed use common ion analysis for HF/NaF and calculate molarity of NaF, then moles in 300ml the x formula wt => mass needed for 3.5 pH.
HF ⇄ H⁺ + F⁻; Ka = 6.6 x 10⁻⁴
Ka = [H⁺][F⁻]/[HF] = 6.6 x 10⁻⁴ = (3.16 x 10⁻⁴)[F⁻]/0.25 => [F⁻] = (6.6 x 10⁻⁴)(0.25)/(3.16x10⁻⁴) = 5.218M in F⁻ needed ( = NaF needed).
For the 300ml buffer solution, moles of NaF needed = Molarity x Volume(L)
= (5.218M)(0.300L) = 0.157 mole NaF needed x 42 g/mole = 6.574 g NaF needed.
Check using the Henderson - Hasselbalch Equation...
pH = pKa + log ([Base]/[Acid]); pKa (HF) = 3.18
Molarity of NaF = (6.572g/42g/mole)/(0.300 L soln) = 0.572M in NaF = 0.572M in F⁻.
pH = 3.18 + log ([0.572]/[0.25]) ≅ 3.5.
One can also back calculate through the Henderson -Hasselbalch Equation to determine base concentration, moles NaF then grams NaF.
The ratio of the areas of the signals in the h NMR spectrum of pentan-3-ol is 6: 4: 1: 1. The correct option is A.
<h3>What is a NMR spectrum?</h3>
Nuclear magnetic resonance spectroscopy is a spectroscopy that shows the detailed structure and chemical environment of a chemical element.
Pentan-3-ol contain 12 hydrogen atoms. In H-NMR spectra, hydrogen atoms have same environment gives a signal.
There are 4 different kinds of signals due of the 4 different environment experienced by these 12 hydrogens.
Thus, the ratio of the areas of the signals in the h NMR spectrum of pentan-3-ol is 6: 4: 1: 1. The correct option is A.
Learn more about NMR spectrum
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B. Gain information about the current state of the environment and ways to improve it.
Hey there
the answers are
A.
and
D. He could spray the fire extinguisher in the opposite direction of the space station
hope this helps