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4vir4ik [10]
3 years ago
14

Suppose you have 100 grams of a radioisotope with a half-life of 100 years. How much of the isotope will you have after 200 year

s?
Chemistry
2 answers:
Ulleksa [173]3 years ago
8 0
        Amount Remaining     Years       #half lives 
             100g                            0                 0
             50 g                           100                1                  
             25g                            200               2                      




Brilliant_brown [7]3 years ago
8 0

Answer:

m = 25 g

Explanation:

To do this, we need to use the general expression for Half life:

<em>A = Ao e^-tλ (1)</em>

<em>Where:</em>

<em>A: concentration or mass of the substance after t time has passed</em>

<em>Ao: Initial concentration or mass of the substance</em>

<em>t: time that has passed.</em>

<em>λ: lambda that is relationed to half life time.</em>

The value of λ can be calculated with the following expression:

λ = ln2 / t(1/2)   (2)

So, let's calculate first the value of lambda, and then, we replace it in expression (1) to know the mass of the radioisotope:

λ = ln2/100

λ = 6.93x10^-3

Now, let's use (1) to calculate the mass after 200 years:

m = 100e^(-200*6.93x10^-3)

m = 100e^(-1.386)

<em>m = 25 g</em>

<em>And this is the mass of the isotope after 200 years.</em>

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Answer:

The interior plains

5 0
3 years ago
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Arrange the following substances in order of decreasing magnitude of lattice energy. Rank the compounds in order of decreasing m
True [87]

The lattice energy of the compounds is distributed in the following decreasing order of magnitude: MgO > CaO > NaF > KCl.

<h3>KCl or NaF, which has a higher lattice energy?</h3>

The lattice energy increases with increasing charge and decreasing ion size.(Refer to Coulomb's Law.)MgF2 > MgO.Following that, we can examine NaF and KCl (both of which have 1+ and 1-charges), as well as atomic radii.NaF will have a larger LE than KCl since Na is smaller then K and F was smaller than Cl.

<h3>MgO or CaO, which has a larger lattice energy?</h3>

MGO is more difficult than CaO, hence.This is because "Mg" (two-plus) ions are smaller than "Ca" (two-plus) ions in size.MgO has higher lattice energy as a result.

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3 0
1 year ago
Chromium(III) oxide can be prepared by heating chromium(IV) oxide in vacuo at high temperature: 4Cr02 —2Cr2O3 +02 The reaction o
kkurt [141]

<u>Answer:</u> The theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of CrO_2 = 480.1 g

Molar mass of CrO_2 = 84 g/mol

Putting values in equation 1, we get:

\text{Moles of }CrO_2=\frac{480.1g}{84g/mol}=5.72mol

For the given chemical equation:

4CrO_2\rightarrow 2Cr_2O_3+O_2

By Stoichiometry of the reaction:

4 moles of CrO_2 produces 2 moles of chromium (III) oxide

So, 5.72 moles of CrO_2 will produce = \frac{2}{4}\times 5.72=2.86mol of chromium (III) oxide

Now, calculating the mass of chromium (III) oxide from equation 1, we get:

Molar mass of chromium (III) oxide = 152 g/mol

Moles of chromium (III) oxide = 2.86 moles

Putting values in equation 1, we get:

2.86mol=\frac{\text{Mass of chromium (III) oxide}}{152g/mol}\\\\\text{Mass of chromium (III) oxide}=(2.86mol\times 152g/mol)=434.72g

To calculate the percentage yield of chromium (III) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of chromium (III) oxide = 402.4 g

Theoretical yield of chromium (III) oxide = 434.72 g

Putting values in above equation, we get:

\%\text{ yield of chromium (III) oxide}=\frac{402.4g}{434.72g}\times 100\\\\\% \text{yield of chromium (III) oxide}=\%

Hence, the theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

7 0
3 years ago
Which of the following measurements is expressed to three significant figures? 7.30 × 10–7 km 0.007 m 7077 mg 0.070 mm
Sonbull [250]

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4 0
3 years ago
An oxybromate compound, kbrox, where x is unknown, is analyzed and found to contain 43.66 % br.
aniked [119]

Answer is: an oxybromate compound is KBrO₄ (x = 4).

ω(Br) = 43.66% ÷ 100%.

ω(Br) = 0.4366; mass percentage of bromine.

If we take 100 grams of compound:

m(Br) = ω(Br) · 100 g.

m(Br) = 0.4366 · 100 g.

m(Br) = 43.66 g; mass of bromine.

n(Br) = m(Br) ÷ M(Br).

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n(Br) = 0.55 mol; amoun of bromine.

From chemical formula (KBrOₓ), amount of potassium is equal to amount of bromine: n(Br) = n(K).

m(K) = 0.55 mol · 39.1 g/mol.

m(K) = 21.365 g; mass of potassium in the compound.

m(O) = 100 g - 21.365 g - 43.66 g.

m(O) =34.97 g; mass of oxygen.

n(O) = 34.97 g ÷ 16 g/mol.

n(O) = 2.185 mol.

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n(K) : n(Br) : n(O) = 1 : 1 : 4.

6 0
3 years ago
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