Which activity performed by a medical professional is primarily based on an understanding of physics

Chemistry

1 answer:

scoundrel [369]3 years ago

3 0

MRI scanning, CAT scanning, radiotherapy, conventional X-ray imaging, ultrasound scans and ultrasound therapy.

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How much energy, in joules, is released when 17 kilograms of granite is cooled from 45 C to 21 C?

Lostsunrise [7]

Answer is: D. 327,992.8 J.
m(granite) = 17 kg = 17000g.
ΔT(granite) = 21°C - 45°C = -24°C (-24K).
cp(granite) = 0,804 J/g·°C, specific heat capacity of granite.
Q = m(granite) · ΔT(granite) · cp(granite).
Q = 17000 g ·(-24°C)· 0,804 J/g·K.
Q = -327990 J.
The granite lost 327990 joules of energy.
Q - amount of energy gained or lost.

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3 years ago

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14.Which of the following exhibit maximum paramagnetic behaviour – Fe3+ or

Naddik [55]

Answer:

Fe^3+

Explanation:

The electron configuration for Fe^3+ is; 1s2 2s2 2p6 3s2 3p6 3d5

The electron configuration for Ni^2+ is; 1s2 2s2 2p6 3s2 3p6 3d8

Now it is pertinent to recall that the 3d sublevel has a maximum occupancy of ten electrons. These ten electrons occupy a set of five degenerate orbitals. Having said that, it is clear that Ni^2+ ion will have two unpaired electrons while Fe^3+ will have five unpaired electrons.

Let us also not forget that paramagnetism has to do with the presence of unpaired electrons. That means that maximum paramagnetism refers to the presence of maximum number of unpaired electrons.

Since Fe^3+ has the greatest number of unpaired electrons among the duo, Fe^3+ will exhibit a maximum paramagnetic behavior.

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3 years ago

What type of reactive intermediate is formed upon irradiation of a solution of toluene (ph-ch3) and bromine?

timofeeve [1]

Answer: Benzylic radical . First Br radicals are produced which strip off a H- from the methyl group. The benzylic radical the reacts with Br2 to form benzylic bromide and another Br radical.

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3 years ago

Calculate the [H+] and pH of a 0.0040 M hydrazoic acid solution. Keep in mind that the Ka of hydrazoic acid is 2.20×10−5. Use th

Vera_Pavlovna [14]

Answer:

$[H^+]=0.000285$

$pH=3.55$

Explanation:

In this, we can with the ionization equation for the hydrazoic acid ( $HN_3$ ). So:

$HN_3~~H^+~+~N_3^-$

Now, due to the Ka constant value, we have to use the whole equilibrium because this is not a strong acid. So, we have to write the Ka expression:

$Ka=\frac{[H^+][N_3^-]}{[HN_3]}$

For each mol of $H^+$ produced we will have 1 mol of $N_3^-$ . So, we can use "X" for the unknown values and replace in the Ka equation:

$Ka=\frac{X*X}{[HN_3]}$

Additionally, we have to keep in mind that $HN_3$ is a reagent, this means that we will be consumed. We dont know how much acid would be consumed but we can express a subtraction from the initial value, so:

$Ka=\frac{X*X}{0.004-X}$