i think it's I
I was confused by IV then search on gg and it said ZnSO4 should be Zn2SO4 instead but still im not sure Zn2SO4 is real
Answer:
4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.
Explanation:
The pH of the solution = 13.00
pH + pOH = 14
pOH = 14 - pH = 14 - 13.00 = 1.00
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
![1.00=-\log[OH^-]](https://tex.z-dn.net/?f=1.00%3D-%5Clog%5BOH%5E-%5D)
![[OH^-]=10^{-1.00} M=0.100 M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-1.00%7D%20M%3D0.100%20M)

![[KOH]=[OH^-]=[K^+]=0.100 M](https://tex.z-dn.net/?f=%5BKOH%5D%3D%5BOH%5E-%5D%3D%5BK%5E%2B%5D%3D0.100%20M)
Molariy of the KOH = 0.100 M
Volume of the KOH solution = 800 mL= 0.800 L
1 mL = 0.001 L
Moles of KOH = n


n = 0.0800 mol
Mass of 0.0800 moles of KOH :
0.0800 mol × 56 g/mol = 4.48 g
4.48 grams is the mass of potassium hydroxide that the chemist must weigh out in the second step.
The net ionic equation is shorter to use and already leaves out the electrons that transferred from the reducing agent to the oxidizing agent. Also, in some occasions the aqueous ions H+ and (or) OH- ions that help balance the net ionic charge are no longer shown in the net ionic equation.
<u>Answer:</u> The concentration of
required will be 0.285 M.
<u>Explanation:</u>
To calculate the molarity of
, we use the equation:

Moles of
= 0.016 moles
Volume of solution = 1 L
Putting values in above equation, we get:

For the given chemical equations:

![Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9](https://tex.z-dn.net/?f=Ni%5E%7B2%2B%7D%28aq.%29%2B6NH_3%28aq.%29%5Crightleftharpoons%20%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%2BC_2O_4%5E%7B2-%7D%28aq.%29%3BK_f%3D1.2%5Ctimes%2010%5E9)
Net equation: ![NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?](https://tex.z-dn.net/?f=NiC_2O_4%28s%29%2B6NH_3%28aq.%29%5Crightleftharpoons%20%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%2BC_2O_4%5E%7B2-%7D%28aq.%29%3BK%3D%3F)
To calculate the equilibrium constant, K for above equation, we get:

The expression for equilibrium constant of above equation is:
![K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BC_2O_4%5E%7B2-%7D%5D%5B%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%5D%7D%7B%5BNiC_2O_4%5D%5BNH_3%5D%5E6%7D)
As,
is a solid, so its activity is taken as 1 and so for 
We are given:
![[[Ni(NH_3)_6]^{2+}]=0.016M](https://tex.z-dn.net/?f=%5B%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%5D%3D0.016M)
Putting values in above equations, we get:
![0.48=\frac{0.016}{[NH_3]^6}}](https://tex.z-dn.net/?f=0.48%3D%5Cfrac%7B0.016%7D%7B%5BNH_3%5D%5E6%7D%7D)
![[NH_3]=0.285M](https://tex.z-dn.net/?f=%5BNH_3%5D%3D0.285M)
Hence, the concentration of
required will be 0.285 M.
It respresents the higher energy level than 627nm .
<h3>What is a emission line ? </h3>
Emission lines are the glowing hot gas emits lines of light whereas absorption line refers to the tendency of cool atmospheric gas to absorb the same line of light.Some lights produce dark band when the light passes through gas in the atmosphere . There are two line spectrum and absorption.
spectrum is an excitement of electrons from lower to higher energy levels and when it comes back it releases energy in the terms of colourful lights .
It represents the higher energy levels than 627nm because Energy is inversly proportional to wavelength .
to learn more about Emission lines click here
brainly.com/question/28184999
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