Question

determine the number of protons in an isotope with 15 neutrons and with an atomic mass number of 49. (note: this ion might not be physically realizable.) 49 15 34 64 10 numeric 10 points a nonmetal element was discovered and named ridlium, rm. what is the molecular weight of ridlium if the compound triridlium hexafluoride has a molecular weight of 344.673 amu? answer

Explanation:

Step1: Recall atomic mass formula

Atomic mass number = number of protons + number of neutrons.

Step2: Solve for number of protons

Let the number of protons be $p$. Given atomic mass number = 49 and number of neutrons = 15. So, $p=49 - 15$.
$p = 34$

for second - part:

Step1: Determine the formula for the compound

The formula for triridium hexafluoride is $Rr_3F_6$. The atomic weight of fluorine ($F$) is approximately 18.9984 amu. The molecular weight of $F_6$ is $6\times18.9984= 113.9904$ amu.

Step2: Solve for the atomic weight of ridlium

Let the atomic weight of ridlium be $x$. The molecular weight of $Rr_3F_6$ is $3x+113.9904$. Given the molecular weight of $Rr_3F_6$ is 344.673 amu. So, $3x+113.9904 = 344.673$. First, subtract 113.9904 from both sides: $3x=344.673 - 113.9904=230.6826$. Then divide both sides by 3: $x=\frac{230.6826}{3}=76.8942$ amu.

Answer:

34