b. CO2
c. H2
d. H2O
To determine how much of a 144g sample of carbon-14 will remain after 1.719 x 10^4 years, you can use the formula for exponential decay:
\[N(t) = N_0 \cdot \left(\frac{1}{2}\right)^{\frac{t}{T}}\]
Where:
- \(N(t)\) is the remaining amount after time \(t\).
- \(N_0\) is the initial amount.
- \(t\) is the time that has passed.
- \(T\) is the half-life.
In this case, \(N_0\) is 144g, \(t\) is 1.719 x 10^4 years, and \(T\) is the half-life of carbon-14, which is 5,730 years.
Plug these values into the formula:
\[N(t) = 144g \cdot \left(\frac{1}{2}\right)^{\frac{1.719 \times 10^4\text{ years}}{5,730\text{ years}}}\]
Now, calculate:
\[N(t) = 144g \cdot \left(\frac{1}{2}\right)^{\frac{3}{2}}\]
\[N(t) = 144g \cdot \left(\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\right)\]
\[N(t) = 144g \cdot \frac{1}{8}\]
Now, multiply 144g by 1/8 to find the remaining amount:
\[N(t) = \frac{144g}{8} = 18g\]
So, after 1.719 x 10^4 years, only 18g of the 144g sample of carbon-14 will remain.
Explanation:
An atom or object that occupies space also has mass. Molar mass or molecular weight is the mass of 1 mole of a substance.
The formula to calculate molar mass is as follows.
For example, molar mass of will be as follows
= ) g/mol
= (45.98 + 12.01 + 48.00) g/mol
= 105.99 g/mol
Therefore, it can be concluded that the sum of mass of all the atoms in grams make up 1 mole of a particular molecule. This mass is known as the molar mass or molecular weight.
Answer : The limiting reagent is .
Solution : Given,
Mass of = 4.0 g
Mass of = 8.0 g
Molar mass of = 17 g/mole
Molar mass of = 32 g/mole
First we have to calculate the moles of and .
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 5 mole of react with 4 mole of
So, 0.25 moles of react with moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.
Hence, the limiting reagent is .