Question

question 2
a new compound was recently discovered and found to have an atomic weight of 342.38 amu. this element has two isotopes, the lighter of which has a mass of 340.91 amu and an abundance of 68.322%. what is the mass of the heavier isotope? pay attention to significant figures
add your answer
integer, decimal, or e notation allowed
question 3
10 points
what is the correct electron configuration for na+ ion?
a 1s²2s²2p⁵
b 1s²2s²2p⁶
c 1s²2s²2p⁶3s²
d 1s²2s²2p⁶3s¹

Explanation:

Question 2

Step1: Calculate the abundance of the heavier isotope

The sum of abundances of all isotopes is 100%. Let the abundance of the heavier isotope be $x$. Then $x = 100\% - 68.322\%=31.678\% = 0.31678$ in decimal form.

Step2: Set up the atomic - weight formula

The atomic weight ($A$) of an element is the weighted - average of the masses of its isotopes. Let the mass of the heavier isotope be $m$. The atomic weight formula is $A=m_1\times a_1 + m\times a$, where $m_1 = 340.91$ amu is the mass of the lighter isotope, $a_1 = 0.68322$ is its abundance, $a = 0.31678$ is the abundance of the heavier isotope, and $A = 342.38$ amu.
So, $342.38=340.91\times0.68322+m\times0.31678$.

Step3: Solve for the mass of the heavier isotope

First, calculate $340.91\times0.68322$:
$340.91\times0.68322 = 340.91\times\frac{68322}{100000}=340.91\times0.68322 = 232.97413$.
Then, rewrite the equation as $342.38 - 232.97413=m\times0.31678$.
$342.38-232.97413 = 109.40587$.
So, $m=\frac{109.40587}{0.31678}\approx345.36$ amu.

Sodium ($Na$) has an atomic number of 11, so its neutral atom has 11 electrons with the electron - configuration $1s^{2}2s^{2}2p^{6}3s^{1}$. A $Na^{+}$ ion is formed when a sodium atom loses one electron. The electron is lost from the outermost shell, which is the 3s orbital for sodium. So the electron - configuration of $Na^{+}$ is $1s^{2}2s^{2}2p^{6}$.

Answer:

$345.36$

Question 3