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Virty [35]
1 year ago
13

Barium-131 is used in the detection of bone tumors. The half-life of barium-131 is approximately 12 days. How long will it take

for the radiation level of barium-131 to drop to 1/4 of its original level?.
Chemistry
1 answer:
Evgesh-ka [11]1 year ago
5 0

Barium-131's radiation level won't reach 1/4 of its initial level for 24 hours.

ln[A] t = -kt + ln[A] 0 is the integrated rate rule for the first-order reaction A's products.

A straight line is produced when the natural log of [A] is plotted as a function of time since this equation has the form y = mx + b.

How is the length of a half-life determined?

The amount of time needed for the reactant concentration to drop to half its initial value is known as the half-life of a reaction. A first-order reaction's half-life is a constant that is correlated with its rate constant:

t 1/2 = 0.693/k.

To know more about rate constant, visit:

brainly.com/question/20305871

#SPJ4

You might be interested in
Aqueous hydrochloric acid (HCl) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and liquid w
Naddika [18.5K]

Answer:

87.9 % is the percent yield of H₂O

Explanation:

This is the neutralization reaction. A base reacts with an acid to produce water and the correspondly ionic salt.

NaOH  +  HCl  → NaCl +  H₂O

As we have the mass of the two reactants, we must determine the limiting reactant.

Let's convert to moles, the mass of each reactant. (mass / molar mass)

21.1 g / 36.45 g/mol = 0.579 moles of HCl

46.3 g / 40g/mol = 1.15 moles of NaOH

Ratio is 1:1, so it is obviously that the limiting reactant is the HCl. For 1.15 moles of NaOH, i need the same amount of acid, but I only have 0.579 moles

Let's work with the products now. Ratio is 1:1 again, so If I have 0.579 moles of acid, I can produce 0.579 moles of H₂O.

How many grams are 0.579 moles of water? We should find it out as this

mol . molar mass = mass → 0.579 mol . 18 g/mol = 10.4 g

We were told that the production of water was 9.17 g, so let's determine the percent yield as this:

(Yield produced / Theoretical yield) . 100 =

(9.17 g / 10.4g ) . 100 = 87.9 %

6 0
4 years ago
What element is in group 13, period 4
Ludmilka [50]

Answer:

Gallium

Explanation:

6 0
3 years ago
Read 2 more answers
Identify each of the following sets of quantum numbers as allowed or not allowed in the hydrogen atom.
Westkost [7]

Answer:

See explanation.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to firstly recall the electron configuration of hydrogen:

1s^1

To realize that the principal quantum number is 1, the angular is 0 as well as the magnetic one; therefore we infer that all the given n's are not allowed, just l=0 is allowed as well as ml=0 yet the rest, are not allowed.

Best regards!

5 0
3 years ago
When excess dilute hydrochloric acid was added to sodium sulphite 960 of sulphuric (iv) oxide was produced. calculate the mass o
irakobra [83]

The mass of sodium sulfite that was used will be 1,890 grams.

<h3>Stoichiometric problems</h3>

First, the equation of the reaction:

NaSO_3 + 2HCl --- > NaCl_2 + H_2O + SO_2

The mole ratio of SO2 produced and sodium sulfite that reacted is 1:1.

Mole of 960 grams SO2 = 960/64 = 15 moles

Equivalent mole of sodium sulfite that reacted = 15 moles

Mass of 15 moles sodium sulfite = 15 x 126 = 1,890 grams

More on stoichiometric problems can be found here: brainly.com/question/14465605

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3 0
1 year ago
How much heat is needed to raise the temperature of 55.0 g sample of water by 65.0 oC.
sveticcg [70]

Answer: 14943.5 J

Explanation:

The quantity of heat energy (Q) required to raise the temperature of a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Given that,

Q = ?

Mass of water = 55.0g

C = 4.18 J/g°C

Φ = 65.0°C

Then, Q = MCΦ

Q = 55.0g x 4.18 J/g°C x 65.0°C

Q = 14943.5 J

Thus, 14943.5 joules of heat is needed to raise the temperature of water.

3 0
3 years ago
Read 2 more answers
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