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Mariulka [41]
3 years ago
9

A radioactive isotope of potassium (k) has a half-life of 20 minutes. if a 43 gram sample of this isotope is allowed to decay fo

r 80 minutes, how many grams of the radioactive isotope will remain?
Chemistry
1 answer:
Ksenya-84 [330]3 years ago
8 0

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 43 g

m (final mass after time T) = ? (in g)

x (number of periods elapsed) = ?

P (Half-life) = 20 minutes

T (Elapsed time for sample reduction) = 80 minutes

Let's find the number of periods elapsed (x), let us see:

T = x*P

80 = x*20

80 = 20\:x

20\:x = 80

x = \dfrac{80}{20}

\boxed{x = 4}

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

m =  \dfrac{m_o}{2^x}

m =  \dfrac{43}{2^{4}}

m = \dfrac{43}{16}

\boxed{\boxed{m = 2.6875\:g}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

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4 0
3 years ago
What is the limiting reactant when 4 mol P4 and 4 mol S8 react.
In-s [12.5K]

Taking into account the reaction stoichiometry, P₄ will be the limiting reagent.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

8 P₄ + 3 S₈ → 8 P₄S₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • P₄: 8 moles
  • S₈: 3 moles
  • P₄S₃: 8 moles

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 3 moles of S₈ reacts with 8 moles of P₄, 4 moles of S₈ reacts with how many moles of P₄?

moles of P_{4} =\frac{4 moles of S_{8} x8 moles of P_{4} }{3 moles of S_{8}}

<u><em>moles of P₄= 10.667 moles</em></u>

But 10.667 moles of P₄ are not available, 4 moles are available. Since you have less amount of moles than you need to react with 4 moles of S₈, P₄ will be the limiting reagent.

Learn more about the reaction stoichiometry:

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6 0
1 year ago
Estimate the ph of the resulting solution prepared by mixing 1.0 mole of solid disodium phosphate (na2hpo4) and 1.25 mole of hyd
GenaCL600 [577]

The HCl added = 1.25 moles

and the moles of Na2HPO4 = 1 mole

Now when acid is added in the given solution of Na2HPO4

One mole of H+ will react with one mole of Na2HPO4 to given one mole of NaH2PO4

Na2HPO4  + H+  ---> NaH2PO4

Now this one mole formed NaH2PO4 will further react with 0.25 moles of H+ left to form 0.25 moles of  H3PO4 and 0.75 moles of NaH2PO4 will remain in the solution

So this will result into formation of a buffer of phosphoric acid and NaH2PO4

NaH2PO4  + H+  ---> H3PO4

pKa of H3PO4 = 2.1

so pH = pKa + log [salt] / [acid] = 2.1 + log [0.75 / 0.25] = 2.58

so the pH will be in between 2.1 to 7.2

7 0
3 years ago
A sample of an unknown metal has a mass of 120.4 g. As the sample cools from 90.5°C to 25.7°C, it releases 7020 J of energy. Usi
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Answer:

Explanation:

sheesh i wouldn’t know

6 0
2 years ago
If 0.100 moles of AgNO₃ react with 0.155 moles of H₂SO₄ according to this UNBALANCED equation below, what is the mass in grams o
AlekseyPX

Answer:

15.6g Ag2SO4

Explanation:

2AgNO3 + H2SO4 --> Ag2SO4 + 2HNO3

-2x              -x

0.1-2x.       0.155-x

x=0.05      x=0.155

0.05mol Ag2SO4 x 311.78g = 15.6g Ag2SO4

4 0
3 years ago
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