Hello!
The H₃O⁺ concentration can be found using the definition of pH and clearing the equation for [H₃O⁺]. The solution has a pH lower than 7, so the Sauvignon Blanc is
acid. The calculation for [H₃O⁺] is shown below:
![pH=-log [H_3O^{+}]](https://tex.z-dn.net/?f=pH%3D-log%20%5BH_3O%5E%7B%2B%7D%5D%20)
![[H_3O^{+}]= 10^{-pH}=10^{-3,24}=0,00058M](https://tex.z-dn.net/?f=%5BH_3O%5E%7B%2B%7D%5D%3D%2010%5E%7B-pH%7D%3D10%5E%7B-3%2C24%7D%3D0%2C00058M%20)
So, the concentration of H₃O⁺ in a Sauvignon Blanc with a pH of 3,24 is
0,00058 MHave a nice day!
Answer:
The percent yield of the reaction is 35 %
Explanation:
In the reaction, 1 mol of hydrazine reacts with 1 mol O₂ to produce 1 mol of nitrogen and 2 moles of water.
Let's verify the moles that were used in the reaction.
2.05 g . 1mol/ 32 g = 0.0640 mol
In the 100% yield, 1 mol of hydrazine produce 1 mol of N₂ so If I used 0.0640 moles of reactant, I made 0.0640 moles of products.
Let's use the Ideal Gases Law equation to find out the real moles of nitrogen, I made (real yield).
1atm . 0.550L = n . 0.082 . 295K
(1atm . 0.550L) / 0.082 . 295K = n → 0.0225 moles
Percent yield of reaction = (Real yield / Theoretical yield) . 100
(0.0225 / 0.0640) . 100 = 35%
Energy is distributed not just in translational KE, but also in rotation, vibration and also distributed in electronic energy levels (if input great enough, bond breaks).
All four forms of energy are quantised and the quanta ‘gap’ differences increases from trans. KE ==> electronic.
Entropy (S) and energy distribution: The energy is distributed amongst the energy levels in the particles to maximise their entropy.
Entropy is a measure of both the way the particles are arranged AND the ways the quanta of energy can be arranged.
We can apply ΔSθsys/surr/tot ideas to chemical changes to test feasibility of a reaction:
ΔSθtot = ΔSθsys + ΔSθsurr
ΔSθtot must be >=0 for a chemical change to be feasible.
For example: CaCO3(s) ==> CaO(s) + CO2(g)
ΔSθsys = ΣSθproducts – ΣSθreactants
ΔSθsys = SθCaO(s) + SθCO2(g) – SθCaCO3(s)
ΔSθsurr is –ΔHθ/T(K) and ΔH is very endothermic (very +ve),
Now ΔSθsys is approximately constant with temperature and at room temperature the ΔSθsurr term is too negative for ΔSθtot to be plus overall.
But, as the temperature is raised, the ΔSθsurr term becomes less negative and eventually at about 800oCΔSθtot becomes plus overall (and ΔGθ becomes negative), so the decomposition is now chemically, and 'commercially' feasible in a lime kiln.
CaCO3(s) ==> CaO(s) + CO2(g) ΔHθ = +179 kJ mol–1 (very endothermic)
This important industrial reaction for converting limestone (calcium carbonate) to lime (calcium oxide) has to be performed at high temperatures in a specially designed limekiln – which these days, basically consists of a huge rotating angled ceramic lined steel tube in which a mixture of limestone plus coal/coke/oil/gas? is fed in at one end and lime collected at the lower end. The mixture is ignited and excess air blasted through to burn the coal/coke and maintain a high operating temperature.
ΔSθsys = ΣSθproducts – ΣSθreactants
ΔSθsys = SθCaO(s) + SθCO2(g) – SθCaCO3(s) = (40.0) + (214.0) – (92.9) = +161.0 J mol–1 K–1
ΔSθsurr is –ΔHθ/T = –(179000/T)
ΔSθtot = ΔSθsys + ΔSθsurr
ΔSθtot = (+161) + (–179000/T) = 161 – 179000/T
If we then substitute various values of T (in Kelvin) you can calculate when the reaction becomes feasible.
For T = 298K (room temperature)
ΔSθtot = 161 – 179000/298 = –439.7 J mol–1 K–1, no good, negative entropy change
For T = 500K (fairly high temperature for an industrial process)
ΔSθtot = 161 – 179000/500 = –197.0, still no good
For T = 1200K (limekiln temperature)
ΔSθtot = 161 – 179000/1200 = +11.8 J mol–1 K–1, definitely feasible, overall positive entropy change
Now assuming ΔSθsys is approximately constant with temperature change and at room temperature the ΔSθsurr term is too negative for ΔSθtot to be plus overall. But, as the temperature is raised, the ΔSθsurr term becomes less negative and eventually at about 800–900oC ΔSθtot becomes plus overall, so the decomposition is now chemically, and 'commercially' feasible in a lime kiln.
You can approach the problem in another more efficient way by solving the total entropy expression for T at the point when the total entropy change is zero. At this point calcium carbonate, calcium oxide and carbon dioxide are at equilibrium.
ΔSθtot–equilib = 0 = 161 – 179000/T, 179000/T = 161, T = 179000/161 = 1112 K
This means that 1112 K is the minimum temperature to get an economic yield. Well at first sight anyway. In fact because the carbon dioxide is swept away in the flue gases so an equilibrium is never truly attained so limestone continues to decompose even at lower temperatures.
Step one write the equation for dissociation of AgNO3 and NaCl
that is AgNO3-------> Ag+ + NO3-
NaCl--------> Na+ + Cl-
then find the number of moles of each compound
that is for AgNO3 = ( 1.4 x10^-3 ) x 25/1000= 3.5 x10^-5 moles
Nacl= (7.5 x10^-4)x 60/1000= 4.5 x10^-5 moles
from mole ratio the moles of Ag+= 3.5 x10^-5 moles and that of Cl-= 4.5 x10^-4 moles
then find the total volume of the mixture
that is 25ml + 60 Ml =85ml = 0.085 liters
The Ksp of Agcl = (Ag+) (cl-), let the concentration of Ag+ be represented by x and also the concentration be represented by x
ksp (1.8 x10^-10) is therefore= x^2
find the square root x=1.342 x10^-5
Ag+ in final mixture is = moles of Ag+/total volume - x
that is {(3.5 x10^-5)/0.085} - 1.342 x10^-5=3.98x10^-4
Cl- in the final mixture is =(4.5 x10^-5 /0.085) - 1.342 x10^-5= 5.16 x10^-4
<span>The solution to the problem is as follows:
125/58.69 = 2.12 mol
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Therefore, there are 2.12 moles of <span>nickel (Ni) atoms are in 125 g Ni.
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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