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Novosadov [1.4K]
4 years ago
10

How many atoms are in lead ii bicarbonate​

Chemistry
1 answer:
sertanlavr [38]4 years ago
4 0

Answer:

11 atoms

Explanation:

The formula for lead(II) bicarbonate is: Pb(HCO3)2.

Atoms in Pb(HCO3)2:

Pb=1, H=1×2, C=1×2, O=3×2

1+1×2+1×2+3×2

=1+2+2+6

=11 atoms

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Analysis of a rock sample shows that it contains 6.25% of its original uranium-235. How old is the rock? How do you know?
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Answer:

2.82\cdot 10^9 y

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A radioactive isotope is an isotope that undergoes nuclear decay, breaking apart into a smaller nucleus and emitting radiation during the process.

The half-life of an isotope is the amount of time it takes for a certain quantity of a radioactive isotope to halve.

For a radioactive isotope, the amount of substance left after a certain time t is:

m(t)=m_0 (\frac{1}{2})^{\frac{t}{\tau}} (1)

where

m_0 is the mass of the substance at time t = 0

m(t) is the mass of the substance at time t

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In this problem, the isotope is uranium-235, which has a half-life of

\tau=7.04\cdot 10^8 y

We also know that the amount of uranium left in the rock sample is 6.25% of its original value, this means that

\frac{m(t)}{m_0}=\frac{6.25}{100}

Substituting into (1) and solving for t, we can find how much time has passed:

t=-\tau log_2 (\frac{m(t)}{m_0})=-(7.04\cdot 10^8) log_2 (\frac{6.25}{100})=2.82\cdot 10^9 y

5 0
4 years ago
A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling
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<u>Answer:</u> The mass of glucose in final solution is 0.420 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}        .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2

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M_1\text{ and }V_1 are the molarity and volume of the concentrated glucose solution

M_2\text{ and }V_2 are the molarity and volume of diluted glucose solution

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Putting values in above equation, we get:

0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M

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Final molarity of glucose solution = 0.0233 M

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Volume of solution = 100 mL

Putting values in equation 1, we get:

0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g

Hence, the mass of glucose in final solution is 0.420 grams

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