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27) how many neutrons are present in c - 14? a) 14 b) 8 c) 12 d) 6 e) n…

Question

  1. how many neutrons are present in c - 14? a) 14 b) 8 c) 12 d) 6 e) none of the above 28) how many protons and neutrons are in cl - 37? a) 17 protons, 20 neutrons b) 17 protons, 37 neutrons c) 37 protons, 17 neutrons d) 20 protons, 17 neutrons e) none of the above 29) a fictional element has two isotopes, each making up 50% of the population. isotope 1 has a mass of 80.0 amu, isotope 2 has a mass of 85.0 amu. calculate the atomic mass of the fictional element. a) 42.5 amu b) 165 amu c) 40 amu d) 82.5 amu e) none of the above 30) chlorine has two stable isotopes, cl - 35 and cl - 37. if their exact masses are 34.9689 amu and 36.9695 amu, respectively, what is the natural abundance of cl - 35? (the atomic mass of chlorine is 35.45 amu) a) 75.95% b) 35.00% c) 24.05% d) 37.00% e) 50.00%

Explanation:

Step1: Recall carbon - 14 properties

Carbon has an atomic number of 6. The mass number of C - 14 is 14. The number of neutrons $n$ is calculated by the formula $n = \text{Mass number}-\text{Atomic number}$. So for C - 14, $n=14 - 6=8$.

Step2: Recall chlorine - 37 properties

Chlorine has an atomic number of 17. The mass number of Cl - 37 is 37. The number of protons $p$ is equal to the atomic number, so $p = 17$. The number of neutrons $n$ is $n=37 - 17 = 20$.

Step3: Calculate atomic mass of fictional element

The formula for the average atomic mass $A$ of an element with two isotopes is $A=x_1m_1 + x_2m_2$, where $x_1$ and $x_2$ are the fractional abundances and $m_1$ and $m_2$ are the masses of the isotopes. Given $x_1=x_2 = 0.5$, $m_1 = 80.0$ amu and $m_2=85.0$ amu. Then $A=0.5\times80.0+0.5\times85.0=40 + 42.5=82.5$ amu.

Step4: Calculate natural abundance of Cl - 35

Let the abundance of Cl - 35 be $x$, then the abundance of Cl - 37 is $1 - x$. The average atomic mass of chlorine is given by $35.45=34.9689x+36.9695(1 - x)$. Expand: $35.45=34.9689x+36.9695-36.9695x$. Combine like - terms: $36.9695x-34.9689x=36.9695 - 35.45$. $2.0006x = 1.5195$. Solve for $x$: $x=\frac{1.5195}{2.0006}\approx0.7595$ or $75.95\%$.

Answer:

  1. B. 8
  2. A. 17 protons, 20 neutrons
  3. D. 82.5 amu
  4. A. 75.95%