Answer:
I would go with tododeku to leave room for kiribaku but that's my opinion
Just because I ship that doesn't mean i don't secretly ship bakudeku
Explanation:
Bakudeku is a hawt ship and I love it so friggen much, but kiribaku is kinda cute and it got me all mixed up. I'm in a mess as well trying to figure out a solid ship for each character. I'm procrastinating by drawing a bunch of random fanart.
A. The number of valence electrons increases as atomic mass increases. == Generally true for the representative elements since atomic mass generally increases with increasing Z.
B. The reactivity of alkali metals increases as atomic mass increases. == True. Atomic mass increases down the column and so does reactivity
C. The reactivity of the halogens increases as atomic mass increases. == False. Reactivity decreases down the column.
D. The number of valence electrons decreases across a period. == False. In general, the number of valence electrons increases across a period, particularly for the representative elements.
Answer:
The equilibrium will shift to the right.
Explanation:
- Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.
For the reaction: <em>CH₄ + NH₃ + 1.5O₂ ⇄ HCN + 3H₂O.</em>
<em></em>
Adding NH₃ to the reaction:
- Adding NH₃ will increase the concentration of the reactants side, so the reaction will be shifted to the right (products) side to suppress the increase in the concentration of NH₃.
Answer:
the answer is water
Explanation:
because abiotic means non living and all of the other things are living
Answer:
2.2 % and 0 %
Explanation:
The equation we will be using to solve this question is:
N/N₀ = e⁻λ t
where N₀ : Number of paricles at t= 0
N= Number of particles after time t
λ= Radioactive decay constant
e= Euler´s constant
We are not given λ , but it can be determined from the half life with the equation:
λ = 0.693 / t 1/2 where t 1/2 is the half-life
Substituting our values:
λ = 0.693 / 55 s = 0.0126/s
a) For t = 5 min = 300 s
N / N₀ = e^-(0.0126/s x 300 s) = e^-3.8 = 0.022 = 2.2 %
b) For t = 1 hr = 3600 s
N / N₀ = e^-(0.0126/s x 3600 s) = 2.9 x 10 ⁻²⁰ = 0 % (For all practical purposes)
