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Mars2501 [29]
3 years ago
11

The radioisotope phosphorus-32 is used in tracers for measuring phosphorus uptake by plants. The half-life of phosphorus-32 is 1

4.3 days. How much time is required for the activity of a sample of phosphorus-32 to fall to 7.34 percent of its original value
Chemistry
1 answer:
Harman [31]3 years ago
3 0

Answer:

54 days

Explanation:

We have to use the formula;

0.693/t1/2 =2.303/t log Ao/A

Where;

t1/2= half-life of phosphorus-32= 14.3 days

t= time taken for the activity to fall to 7.34% of its original value

Ao=initial activity of phosphorus-32

A= activity of phosphorus-32 after a time t

Note that;

A=0.0734Ao (the activity of the sample decreased to 7.34% of the activity of the original sample)

Substituting values;

0.693/14.3 = 2.303/t log Ao/0.0734Ao

0.693/14.3 = 2.303/t log 1/0.0734

0.693/14.3 = 2.6/t

0.048=2.6/t

t= 2.6/0.048

t= 54 days

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__HgO __Hg + __O2 what is the reaction
Gelneren [198K]

Answer:

2HgO <---- 2Hg + O2

Explanation:

8 0
3 years ago
A student dissolves 3.9g of aniline (C6H5NH2) in 200.mL of a solvent with a density of 1.05 g/mL . The student notices that the
tino4ka555 [31]

Answer:

2.1 × 10⁻¹ M

2.0 × 10⁻¹ m

Explanation:

Molarity

The molar mass of aniline (solute) is 93.13 g/mol. The moles corresponding to 3.9 g are:

3.9 g × (1 mol/93.13 g) = 0.042 mol

The volume of the solution is 200 mL (0.200 L). The molarity of aniline is:

M = 0.042 mol/0.200 L = 0.21 M = 2.1 × 10⁻¹ M

Molality

The moles of solute are 0.042 mol.

The density of the solvent is 1.05 g/mL. The mass corresponding to 200 mL is:

200 mL × 1.05 g/mL = 210 g = 0.210 kg

The molality of aniline is:

m = 0.042 mol/0.210 kg = 0.20 m = 2.0 × 10⁻¹ m

5 0
3 years ago
Which of the following is the correct sequence for the discovery of subatomic particles? a Electrons&gt; Protons &gt; Neutrons b
laila [671]

Answer:

C

Explanation;

Maybe is C

4 0
3 years ago
Read 2 more answers
Which of the following atoms would have a charge of +1?
erica [24]

Answer: An atom with 6 protons, 5 electrons, and 7 neutrons

Explanation: In this case, neutrons do not matter as they have a charge of 0, or no charge. A proton has a charge of +1 and an electron has a charge of -1. Since there are 6 protons, the total charge of the protons would be +6. Since there are 5 electrons the total charge of the electrons would be -5. +6 - 5 would result in a charge of +1. This means that this atom would have an overall charge of + 1. Basically, if there is one more proton than electron, then the overall charge of the atom will be +1 but if there is one more electron than proton, then the overall charge of the atom will be -1.

3 0
3 years ago
Thallium-207 decays exponentially with a half life of 4.5 minutes. if the initial amount of the isotope was 28 grams, how many g
Agata [3.3K]
An exponential decay law has the general form: A = Ao * e ^ (-kt) =>

A/Ao = e^(-kt)

Half-life time => A/Ao = 1/2, and t = 4.5 min

=> 1/2 = e^(-k*4.5) => ln(2) = 4.5k => k = ln(2) / 4.5 ≈ 0.154

Now replace the value of k, Ao = 28g  and t = 7 min to find how many grams of Thalium-207 will remain:

A = Ao e ^ (-kt) = 28 g * e ^( -0.154 * 7) = 9.5 g

Answer 9.5 g.
7 0
3 years ago
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