The Empirical Formula is
<span>a formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms.</span>
Answer:
Triphorphorus Pentanitride
Explanation:
since there are three phosphorus use the prefix tri-, since there are five nitrogen use the prefix penta. tri=3, penta=5. then add a suffix -ide to the end of nitrogen. that will give you triphosphorus pentanitride.
To control what molecules go in and out of the cell.
<u>Answer:</u> The average atomic mass of element X is 59.3 amu.
<u>Explanation:</u>
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
.....(1)
- <u>For isotope 1 (X-59) :</u>
Mass of isotope 1 = 59.015 amu
Percentage abundance of isotope 1 = 91.7 %
Fractional abundance of isotope 1 = 0.917
- <u>For isotope 2 (X-62) :</u>
Mass of isotope 2 = 62.011 amu
Percentage abundance of isotope 2 = (100 - 91.7) % = 8.3 %
Fractional abundance of isotope 2 = 0.083
Putting values in equation 1, we get:
![\text{Average atomic mass of X}=[(59.015\times 0.917)+(62.011\times 0.083)]](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20atomic%20mass%20of%20X%7D%3D%5B%2859.015%5Ctimes%200.917%29%2B%2862.011%5Ctimes%200.083%29%5D)

Hence, the average atomic mass of element X is 59.3 amu.
Answer:
pH = 14 - pH = 14 - 1 = 13
Explanation:
0.05M Ca(OH)₂ => 0.05M Ca⁺(aq) + 2(0.05M) OH⁻(aq)
pOH = -log[OH⁻] = -log(0.100) = 1
pH = 14 - pH = 14 - 1 = 13