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marusya05 [52]
3 years ago
15

Which type of radioactive decay can pass through the body?

Chemistry
1 answer:
Mkey [24]3 years ago
3 0
Q. Which type of radioactive decay can pass through the body?

A. Gamma Rays
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In 36 hours the amount of radioactive substance has reduced from 80 mg to 10 mg. What is the half-life of this isotope?
rusak2 [61]
Half-life refers to the time taken for half of the atoms in a radioactive substance to decay. Therefore in 1 half-life the 80 mg of substance will reduce to 40 mg. In another half-life it will reduce from 40 mg to 20 mg, then in a third half-life from 20 mg to 10 mg. Therefore the half-life is equal to the total time taken to decay from 80 mg to 10 mg divided by 3:
t(1/2) = 36 hours / 3 = 12 hours.
6 0
3 years ago
Separating techniques - Chromatography
vodomira [7]

The method that can be used to separate the mixture is chromatography.

<h3>What is chromatography?</h3>

"Chromatography" is obtained form a Greek word which literarily means, color writing. It is a method of separation which is common in separating a mixture of pigments.

To obtain the colors used, two solvents are mixed and the sample ink is dissolved in the solvents then spotted on a thin layer and put into a TLC chamber then the chromatogram is allowed to develop.

The various components of the pigment will appear on the chromatogram and can be identified using spectrophotometry. The Rf values of each component can also be used to identify it.'

Learn more about chromatography: brainly.com/question/26491567

7 0
2 years ago
What do plants need to live?
zimovet [89]
Sunlight, soil, water, leaves,
4 0
3 years ago
Read 2 more answers
Which heats up the fastest? treated glass, aluminum, stainless steel, or copper?
GrogVix [38]

Answer:

stainless steel

Explanation:

stainless steel would heat up fast because it is made of metal

8 0
2 years ago
An engine operates on a Carnot cycle that uses 1mole of an ideal gas as the
sattari [20]

Answer:

Step 1;

q = w = -0.52571 kJ, ΔS = 0.876 J/K

Step 2

q = 0, w = ΔU = -7.5 kJ, ΔH = -5.00574 kJ

Explanation:

The given parameters are;

P_i = 100 N·m

T_i = 327 K

P_f = 90 N·m

Step 1

For isothermal expansion, we have;

ΔU = ΔH = 0

w = n·R·T·ln(P_f/P_i) = 1 × 8.314 × 600.15 × ln(90/100) = -525.71

w ≈<em> -0.52571</em> kJ

At state 1, q = w = -0.52571 kJ

ΔS = -n·R·ln(P_f/P_i) = -1 × 8.314 × ln(90/100) ≈ 0.876

ΔS ≈ 0.876 J/K

Step 2

q = 0 for adiabatic process

ΔU = 25×(27 - 327) = -7,500

w = ΔU = <em>-7.5 kJ</em>

ΔH = ΔU + n·R·ΔT

ΔH = -7,500 + 8.3142 × 300 = -5,005.74

ΔH = ΔU = <em>-5.00574 kJ</em>

6 0
2 years ago
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