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3 years ago

The half life of 226/88 Ra is 1620 years. How much of a 12 g sample of 226/88 Ra will be left after 8 half lives?

Chemistry

1 answer:

Aloiza [94]3 years ago

8 0

Answer:

0.0468 g.

Explanation:

The decay of radioactive elements obeys first-order kinetics.

For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 1620 years).

∴ k = ln2/(t1/2) = 0.693/(1620 years) = 4.28 x 10⁻⁴ year⁻¹.

For first-order reaction: kt = lna/(a-x).

where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).

t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).

a is the initial concentration (a = 12.0 g).

(a-x) is the remaining concentration.

∴ kt = lna/(a-x)

(4.28 x 10⁻⁴ year⁻¹)(12960 year) = ln(12)/(a-x).

5.54688 = ln(12)/(a-x).

Taking e for the both sides:

256.34 = (12)/(a-x).

∴ (a-x) = 12/256.34 = 0.0468 g.

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Elan Coil [88]

Answer: 40.3 L

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} Sb_2O_3=\frac{175g}{291.5g/mol}=0.600moles$

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According to stoichiometry :

1 moles of $Sb_2O_3$ produces = 3 moles of $CO$

Thus 0.600 moles of $Sb_2O_3$ will produce= $\frac{3}{1}\times 0.600=1.80moles$ of $CO$

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6 0

3 years ago

What is the % dissociation of a solution of acetic acid if at equilibrium the solution has a pH = 4.74 and a pKa = 4.74?

Ymorist [56]

Answer:

$\% diss = 50\%$

Explanation:

Hello there!

In this case, when considering weak acids which have an associated percent dissociation, we first need to set up the ionization reaction and the equilibrium expression:

$HA\rightleftharpoons H^++A^-\\\\Ka=\frac{[H^+][A^-]}{[HA]}$

Now, by introducing x as the reaction extent which also represents the concentration of both H+ and A-, we have:

$Ka=\frac{x^2}{[HA]_0-x} =10^{-4.74}=1.82x10^{-5}$

Thus, it is possible to find x given the pH as shown below:

$x=10^{-pH}=10^{-4.74}=1.82x10^{-5}M$

So that we can calculate the initial concentration of the acid:

$\frac{(1.82x10^{-5})^2}{[HA]_0-1.82x10^{-5}} =1.82x10^{-5}\\\\\frac{1.82x10^{-5}}{[HA]_0-1.82x10^{-5}} =1\\\\$

$[HA]_0=3.64x10^{-5}M$

Therefore, the percent dissociation turns out to be:

$\% diss=\frac{x}{[HA]_0}*100\% \\\\\% diss=\frac{1.82x10^{-5}M}{3.64x10^{-5}M}*100\% \\\\\% diss = 50\%$

Best regards!

6 0

2 years ago

Described the role of electrons in the formation of a covalent bond

My name is Ann [436]

Sharing of two electrons make a Covalent bond.

Explanation:

Attractions among the atoms bring them together. So the electrons from each of the atoms are attracted towards the nucleus of those two atoms, that “share” the electrons produces a covalent bond.

It is also named as molecular bond, a bond that entails the sharing of a pair of electrons among the atoms. When the atoms share the electrons among themselves, it produces a molecule, which is more stable than the atom.

If the attractions between the atoms are strong enough and if every atom has enough space for the electrons in its outermost energy level then there occurs covalent bonding. So electrons are very important in the covalent bond formation.

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3 years ago

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Alex17521 [72]

Answer:

Explanation:

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4 0

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3 years ago