Answer:
d= 14.007 amu
Explanation:
Abundance of N¹⁴ = 99.63%
Abundance of N¹⁵ = 0.37%
Atomic mass of N¹⁴ = 14.003 amu
Atomic mass of N¹⁵ = 15.000 amu
Average atomic mass = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (14.003 × 99.63)+(15.000× 0.37) /100
Average atomic mass = 1395.12 + 5.55 / 100
Average atomic mass = 1400.67/ 100
Average atomic mass = 14.007 amu.
Answer:
Ag+ = 47 electrons - 1 electron = 46 electrons. Finally: 47 protons , 61 neutrons and 46 electrons.
Explanation:
Answers:
______________________________________________
1) [D]: " CO₂ " .
<u>Note:</u><u> </u> This is the only answer choice given that is composed of: "at least two different elements" .
_______________________________________________
2) [D]: "carbon and hydrogen" {"C" and "H" .} .
_______________________________________________
<u>Given:</u>
Dimensions of the room= 12 ft * 15 ft * 8.60 ft
<u>To determine:</u>
The amount of HCN that gives the lethal dose in the room with the given dimensions
<u>Explanation:</u>
As per the World Health Organization, the lethal dose of HCN is around 300 ppm
300 ppm = 300 mg of HCN/ kg of inhaled air
Volume of air = volume of room = 12 * 15 *8.6 = 1548 ft³
Now, 1 ft³ = 28316.8 cm³
Therefore, the calculated volume of air corresponds to:
1548 * 28316.8 = 4.383 * 10⁷ cm3
Density of air (at room temperature 25 C) = 0.00118 g/cm3
Thus mass of air corresponding to the calculated volume is
Mass = Density * volume = 0.00118 g/cm3 * 4.383 * 10⁷ cm3
= 5.172*10⁴ g = 51.72 kg
Lethal amount of HCN corresponding to 51.72 kg of air would be.
= 51.72 kg air* 300 mg of HCN/1 kg air = 15516 mg
Ans: Lethal dose of HCN = 15.5 g
Answer: 
of PABA is 0.000022
Explanation:
cM 0 0
So dissociation constant will be:
Give c= 0.055 M and 
= ?
![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
![2.96=-log[H^+]](https://tex.z-dn.net/?f=2.96%3D-log%5BH%5E%2B%5D)
![[H^+]=1.09\times 10^{-3}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D1.09%5Ctimes%2010%5E%7B-3%7D)
![[H^+]=c\times \alpha](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Ctimes%20%5Calpha)
Putting in the values we get:
Thus of PABA is 0.000022
