The mass number of an atom with 92 protons and 108 neutrons is 200.
To calculate the mass number of an atom that has 92 protons and 108 neutrons, you simply add these two numbers together. The mass number of an atom is comprised of the total number of protons and neutrons in its nucleus. So, 92 protons + 108 neutrons = 200. Therefore, the mass number of the atom is 200.
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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.
a) in this we are diluting a stock solution, so we can use the dilution formula
c1v1 = c2v2
where c1 is concentration and v1 is volume of the stock solution
c2 is concentration and v2 is volume of the diluted solution to be prepared
substituting the values
6.0 M x V = 0.500 M x 110 mL
V = 9.17 mL
9.17 mL of the stock solution should be taken and diluted upto 110 mL to prepare the 0.500 M solution
b)
In this question we are given the volume taken from the stock solution , we have to find the concentration of the diluted solution
again we use the dilution formula, c1v1 = c2v2
substituting the values
6.0 M x 10.0 mL = C x 250 mL
C = 0.24 M
the concentration of diluted solution is 0.24 M
To prepare 110 ml of 0.500 M HNO3 from a 6.0 M HNO3 solution, 9.17 ml of the stock solution would have to be used. If 10.0 ml of the stock solution is diluted to a final volume of 0.250 L, the concentration of the diluted solution will be 0.24 M.
(a) In order to prepare 110 ml of 0.500 M HNO3 from a 6.0 M HNO3 solution, we have to use the formula M1V1 = M2V2 where M and V are the molarity and volume respectively. Here, the M1 and V1 are the molarity and volume of the stock solution and M2 and V2 are the molarity and volume of the diluted solution. Filling in known values, 6.0M * V1 = 0.500M * 110ml. Solving for V1, we get V1 = (0.500 M * 110 ml) / 6.0 M = 9.17 ml. So, you would have to use 9.17 ml of the stock solution.
(b) The diluted solution's molarity is calculated using the same formula as before. Substituting the known values 6.0M * 10.0 ml = M2 * 0.250 L, rearrange the formula to get M2= (6.0M * 10.0 ml) / 0.250 L = 0.24 M or 240 mM. Therefore, the concentration of the diluted solution is 0.24 M.
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B. the epidermis consists of mostly adipose cells.
C. the epidermis is thicker.
D. the dermis is made of mostly dead cells