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rusak2 [61]
2 years ago
9

Iodine-131 decays with a half-life of 8.02

Chemistry
1 answer:
dlinn [17]2 years ago
3 0
Radioactive material undergoes 1st order decay kinetics.

For 1st order decay, half life = 0.693/k

where k = rate constant

k = 0.693/half life = 0.693/8.02 = 0.0864 day-1

Now, for 1st order reaction,
k = \frac{2.303}{t} X log \frac{initial.conc}{final.conc}

Given: t = 6.01d, initial conc. = 5mg

∴0.0864 = \frac{2.303}{6.01} X log \frac{5}{final.conc}
∴ final conc. = 2.975 mg
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A sample of a compound containing only carbon and oxygen decomposes and produces 24.50g of carbon and 32.59g of oxygen. what is
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<span>Carbon Monoxide. First, determine the relative number of moles of each element by looking up the atomic weights of carbon and oxygen Atomic weight carbon = 12.0107 Atomic weight oxygen = 15.999 Moles of Carbon = 24.50 g / 12.0107 g/mol = 2.039847802 mol Moles of Oxygen = 32.59 g / 15.999 g/mol = 2.037002313 mol Given that the number of moles of both carbon and oxygen are nearly identical, it wouldn't be unreasonable to think that the empirical formula for the compound is CO which also happens to be the formula for Carbon Monoxide.</span>
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2 years ago
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What type of bond would form between two atoms of selenium?
siniylev [52]

Answer:

the correct answer is B

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2 years ago
Albus Dumbledore provides his students with a sample of 19.3 g of sodium sulfate. How many oxygen atoms are in this sample
Dimas [21]

Answer:

<em>3.27·10²³ atoms of O</em>

Explanation:

To figure out the amount of oxygen atoms in this sample, we must first evaluate the sample.

The chemical formula for sodium sulfate is <em>Na₂SO₄, </em>and its molar mass is approximately 142.05\frac{g}{mol}.

We will use stoichiometry to convert from our mass of <em>Na₂SO₄ </em>to moles of <em>Na₂SO₄</em>, and then from moles of <em>Na₂SO₄ </em>to moles of <em>O </em>using the mole ratio; then finally, we will convert from moles of <em>O </em>to atoms of <em>O </em>using Avogadro's constant.

19.3g <em>Na₂SO₄</em> · \frac{1 mol Na^2SO^4}{142.05g Na^2SO^4} · \frac{4 mol O}{1 mol Na^2SO^4} ·\frac{6.022x10^2^3}{1 mol O}

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A student heats a sample of Copper (II) sulfate in a crucible and records the data shown in the table. What is the complete form
liberstina [14]

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Mass of water loss during heating

= 152.00g - 147.60g = 4.40g.

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Hence we have CuSO4 * 2H2O.

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