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aleksklad [387]
2 years ago
14

If the half life of a radioactive isotope is 1 day, then how much of the original isotope remains at the end of two days?

Chemistry
2 answers:
ikadub [295]2 years ago
8 0

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25%

Explanation:

In a half life of a radioactive isotope is 1 day,it means it loses its half mass each day

We a formula for N half life

\sf  { \frac{1}{2} }^{n}

where n is the number of days

Here the isotope is kept for 2 days

so it's left over mass will be

\sf { (\frac{1}{2}) }^{2}  \\  \\  \\  \frac{1}{2}  \times  \frac{1}{2}  \\  \\  \sf  \frac{1}{4}

It's left over mass 1/4th of the original mass

Now, we need to find it's percentage by multiplying with 100

\sf  \frac{1}{4}  \times 100 \%\\  \\  \sf   \cancel\frac{100}{4} \% \\  \\  \sf 25\%

<u>So</u><u> </u><u>2</u><u>5</u><u>%</u><u> </u><u>mass</u><u> </u><u>will</u><u> </u><u>be</u><u> </u><u>left</u><u> </u><u>after</u><u> </u><u>2</u><u> </u><u>day</u><u> </u><u>of</u><u> </u><u>half</u><u> </u><u>life</u><u> </u><u>radioactive</u><u> </u><u>isotope</u><u>.</u><u> </u>

JulijaS [17]2 years ago
8 0
25% because the half life of 100 is 50 and the half life of 50 is 25
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0.14 M

Explanation:

To determinate the concentration of a new solution, we can use the equation below:

C1xV1 = C2xV2

Where C is the concentration, and V the volume, 1 represents the initial solution, and 2 the final one. So, first, the initial concentration is 1.50 M, the initial volume is 55.0 mL and the final volume is 278 mL

1.50x55.0 = C2x278

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The portion of 139 mL will be the same concentration because it wasn't diluted or evaporated. The final volume will be the volume of the initial solution plus the volume of water added, V2 = 139 + 155 = 294 mL

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0.30x139 = C2x294

C2 = 0.14 M

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A sample of solid sodium hydroxide, weighing 13.20 grams is dissolved in deionized water to make a solution. What volume in mL o
Andreas93 [3]
<h3>Answer:</h3>

2.809 L of H₂SO₄

<h3>Explanation:</h3>

Concept tested: Moles and Molarity

In this case we are give;

Mass of solid sodium hydroxide as 13.20 g

Molarity of H₂SO₄ as 0.235 M

We are required to determine the volume of H₂SO₄ required

<h3>First: We need to write the balanced equation for the reaction.</h3>
  • The reaction between NaOH and H₂SO₄ is a neutralization reaction.
  • The balanced equation for the reaction is;

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

<h3>Second: We calculate the umber of moles of NaOH used </h3>
  • Number of moles = Mass ÷ Molar mass
  • Molar mass of NaOH is 40.0 g/mol
  • Therefore;

Moles of NaOH = 13.20 g ÷ 40.0 g/mol

                          = 0.33 moles

<h3>Third: Determine the number of moles of the acid, H₂SO₄</h3>
  • From the equation, 2 moles of NaOH reacts with 1 mole of H₂SO₄
  • Therefore, the mole ratio of NaOH: H₂SO₄ is 2 : 1.
  • Thus, Moles of H₂SO₄ = moles of NaOH × 2

                                    = 0.33 moles × 2

                                   = 0.66 moles of H₂SO₄

<h3>Fourth: Determine the Volume of the acid, H₂SO₄ used</h3>
  • When given the molarity of an acid and the number of moles we can calculate the volume of the acid.
  • That is; Volume = Number of moles ÷ Molarity

In this case;

Volume of the acid = 0.66 moles ÷ 0.235 M

                                = 2.809 L

Therefore, the volume of the acid required to neutralize the base,NaOH is 2.809 L.

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