Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

question 5 1 pts percent composition is also useful for evaluating the_…

Question

question 5
1 pts
percent composition is also useful for evaluating the__________of a
given element in different compounds of known formulas.
number of protons
number of electrons
relative abundance
atomic number
question 6
1 pts
what is the formula mass for nh3 in amu?
note: the molar mass is the mass in grams per mole of that substance while the
formula mass is the mass in amu.
for percent composition one can use either the formula mass or the molar mass
as the units of mass cancel out.
12.01 amu
14.01 amu
18.05 amu
17.03 amu

Explanation:

Question 5
Brief Explanations

Percent composition gives the proportion of an element in a compound. Relative abundance refers to how much of an element is present relative to others in different compounds. The number of protons (atomic number) is fixed for an element and not related to percent composition in different compounds. The number of electrons can vary (e.g., in ions) but percent composition is about mass - based proportions, not electron count.

Step1: Find the atomic masses

The atomic mass of \(N\) (nitrogen) is approximately \(14.01\ amu\) and the atomic mass of \(H\) (hydrogen) is approximately \(1.01\ amu\).

Step2: Calculate the formula mass of \(NH_3\)

For \(NH_3\), there is \(1\) \(N\) atom and \(3\) \(H\) atoms. Using the formula \(M = m_N+3\times m_H\), where \(m_N\) is the atomic mass of \(N\) and \(m_H\) is the atomic mass of \(H\). Substitute \(m_N = 14.01\ amu\) and \(m_H=1.01\ amu\) into the formula: \(M=14.01 + 3\times1.01=14.01 + 3.03\).

Step3: Perform the addition

\(14.01+3.03 = 17.04\approx17.03\ amu\) (due to rounding differences in atomic mass values used in different references).

Answer:

relative abundance

Question 6