Answer: The water Inside the cell is stored inside the vacuole(s)
Explanation:
Hope this helps :)
Riley can either change the surface area of the object or can change the slipperiness of the material.
Answer:
![\frac{[F^{-}]}{[HF]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BF%5E%7B-%7D%5D%7D%7B%5BHF%5D%7D)
is larger
Explanation:
, where 
is the acid dissociation constant.
For a monoprotic acid e.g. HA, ![K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}](https://tex.z-dn.net/?f=K_%7Ba%7D%3D%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
and ![\frac{[A^{-}]}{[HA]}=\frac{K_{a}}{[H^{+}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%7D%7B%5BH%5E%7B%2B%7D%5D%7D)
So, clearly, higher the value , lower will the the 
In this mixture, at equilibrium, ![[H^{+}]](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D)
will be constant.
of HF is grater than of HCN
Hence, ![(\frac{F^{-}}{[HF]}=\frac{K_{a}(HF)}{[H^{+}]})>(\frac{CN^{-}}{[HCN]}=\frac{K_{a}(HCN)}{[H^{+}]})](https://tex.z-dn.net/?f=%28%5Cfrac%7BF%5E%7B-%7D%7D%7B%5BHF%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%28HF%29%7D%7B%5BH%5E%7B%2B%7D%5D%7D%29%3E%28%5Cfrac%7BCN%5E%7B-%7D%7D%7B%5BHCN%5D%7D%3D%5Cfrac%7BK_%7Ba%7D%28HCN%29%7D%7B%5BH%5E%7B%2B%7D%5D%7D%29)
So, is larger
It's Oxygen.
electronegativity is an arbitrary measure of how much will an electron get attracted to an atom of an element
as we know, oxygen is the best at accepting electrons from this list, so the answer is oxygen
Answer: 0.4M
Explanation:
Given that,
Amount of moles of NaOH (n) = ?
Mass of NaOH in grams = 40.0g
For molar mass of NaOH, use the atomic masses: Na = 23g; O = 16g; H = 1g
NaOH = (23g + 16g + 1g)
= 40g/mol
Since, n = mass in grams / molar mass
n = 40.0g / 40.0g/mol
n = 1 mole
Volume of NaOH solution (v) = 2.5 L
Concentration of NaOH solution (c) = ?
Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence
c = n / v
c = 1 mole / 2.5 L
c = 0.4 mol/L (Concentration in mol/L is the same as Molarity, M)
Thus, the concentration of a solution of a 40.0 g of NaOH in 2.5 L of solution is 0.4 mol/L or 0.4M
