To get the value of ΔG we need to get first the value of ΔG°:
when ΔG° = - R*T*㏑K
when R is constant in KJ = 0.00831 KJ
T is the temperature in Kelvin = 25+273 = 298 K
and K is the equilibrium constant = 4.5 x 10^-4
so by substitution:
∴ ΔG° = - 0.00831 * 298 K * ㏑4.5 x 10^-4
= -19 KJ
then, we can now get the value of ΔG when:
ΔG = ΔG° - RT*㏑[HNO2]/[H+][NO2]
when ΔG° = -19 KJ
and R is constant in KJ = 0.00831
and T is the temperature in Kelvin = 298 K
and [HNO2] = 0.21 m & [H+] = 5.9 x 10^-2 & [NO2-] = 6.3 x 10^-4 m
so, by substitution:
ΔG = -19 KJ - 0.00831 * 298K* ㏑(0.21/5.9x10^-2*6.3 x10^-4 )
= -40
When the solubility is the amount of solute that dissolved in a unit volume of the solvent.
and when the balanced reaction equation is:
so, by using the ICE table:
Cd(OH)2(s) ↔ Cd2+ + 2OH-
initial 0 0
change +X +2X
Equ X 2X
so the Ksp expression = [Cd2+] [OH-]^2
and we have Ksp = 7.2 x 10^-15
and when we assume the solubility = X
by substitution:
7.2 x 10^-15 = (X) (2X)^2
7.2 x 10^-15 = X*4X^2
7.2 x 10^-15 = 4X^3
∴X = ∛(7.2x10^-15)/4
= 1.22 x 10^-5 M
∴ the solubility = X = 1.22 x 10^-5 M
If a carbohydrate, like xylulose, has five carbon atoms and a carbonyl group on the second carbon, it is called a(n) keto pentose.
These consist of glycogen, cellulose, as well as starch. Benedict's reagent can be used as a test to see if there are lots of simple carbohydrates present. When it interacts with lowering sugars, it changes from turquoise to yellow or orange. These contain unbound aldehyde but rather ketone groups in simple carbohydrates.
Sugars and starches are examples of carbohydrates. They contain carbon, hydrogen, and oxygen, which appear in the ratio 1:2:1. Size-based categories for carbohydrates include monosaccharides, disaccharides, or polysaccharides. Carbohydrates act as sources of power as their main purpose.
Therefore, If a carbohydrate, like xylulose, has five carbon atoms and a carbonyl group on the second carbon, it is called a(n) keto pentose.
To know more about carbohydrate
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What exactly are you looking for?
This is the balanced equation.
<span>2 C4H10g + 13 O2g ---> 8 CO2g + 10 H2Og</span>
Answer:
All of these
Explanation:
In MnO:
mass of Mn = 55 g (as 1 mole of Mn weighs 55 g)
mass of O = 16 g (as 1 mole of O weighs 16 g)
The mass percent of metal Mn= 
In MnO2:
mass of Mn = 55 g
mass of O = 2*16 = 32 g
The mass percent of metal Mn=
In Mn2O3:
mass of Mn = 2*55 = 110 g
mass of O = 3*16 = 48 g
The mass percent of metal Mn=
In all the cases the mass percent of metal is greater than 50.
