atomic mass=percentage of isotope a * mass of isotope a + percentage of isotope b * mass of isotope b+...+percentage of isotope n * mass of isotope n.
Data:
mass of isotope₁=267.8 u
percentage of isotope₁=90.3%
mass of isotope₂=270.9 u
percentage of isotope₂=9.7%
Therefore:
atomic mass=(0.903)(267.8 u)+(0.097)(270.9 u)=
=241.8234 u + 26.2773 u≈268.1 u
Answer: the mass atomic of this element would be 268.1 u
Answer:
a) Addition reaction, is your answer
Answer: Pesticide and less than 6.65
Explanation:
E2020
Answer:
[H⁺] = 1.0 x 10⁻¹² M.
Explanation:
∵ [H⁺][OH⁻] = 10⁻¹⁴.
[OH⁻] = 1 x 10⁻² mol/L.
∴ [H⁺] = 10⁻¹⁴/[OH⁻] = (10⁻¹⁴)/(1 x 10⁻² mol/L) = 1.0 x 10⁻¹² M.
∵ pH = - log[H⁺] = - log(1.0 x 10⁻¹² M) = 12.0.
∴ The solution is basic, since pH id higher than 7 and also the [OH⁻] > [H⁺].
I think- IDK
