Answer:
potassium hydrogen phthalate KHP MOLAR MASS = 204.233 glmol
to get 1000 ml
Molar concentration = Mass concentration/Molar Mass
mass concentration = molar concentration x molar mass
mass concentration=0.1 M,
molar mass= 204.233 g/mol
so to get 1L
mass conc = 204.233 x 0.1
= 20.4233g for 1L or 1000 ml
to get 6.00 ml
if 20.4233g is for 1000ml
then to 6.00 ml
= 20.4233 x 6 / 1000
= 0.123g for 6.00 ml
according to the equation below
NaOH(aq) + KHC8H4O4(aq) --> KNaC8H4O4(aq) + H2O(l)
number of moles of NaOH is equal to that of KHP
so the same amount will be needed too, which is
= 0.123g
Answer:
A
Explanation:
Hmm, so we have the following in the diagram
Pt(s)
Cl2(g)
Ag(s)
NaCl(aq)
AgNO3(aq)
Pt 2+, 4+, 6+ Though it states Pt is inert
Cl 2-
Ag 1+
Na 1+
NO3-
Anode definition: the positively charged electrode by which the electrons leave an electrical device.
Electrode definition: a conductor through which electricity enters or leaves an object, substance, or region.
Cations attracted to cathode pick up electrons
Anions attracted to anode release electrodes+
Reduction at Cathode (red cat gain of e)
Oxidation at Anode (ox anode loss of e)
So from the diagram we can see that the charge is being generated through the 2 metal plates.
So the answer is A, the anode material is Pt and the half reaction is 2Cl- = Cl2 + 2e-
Answer is: 39,083kJ.
m(coal) = 2,00g.
m(water) = 500g.
ΔT = 43,7°C - 25°C = 18,7°C, <span>difference at temperatures.</span>
c(water) = 4,18 J/g·°C, <span>specific heat of water
</span>Q = m(water)·ΔT·c(water), heat of reaction.
Q = 500g·18,7°C·4,18J/g·°C.
Q = 39083J = 39,083kJ.
From the calculation, the standard free energy of the system is -359kJ.
<h3>What is the standard free-energy?</h3>
The standard free-energy is the energy present in the system. We have to first obtain the cell potential using the formula;
Ereduction - E oxidation = 0.96 V - 0.34 V = 0.62 V
Using the formula;
ΔG = -nFEcell
ΔG =-(6 * 96500 * 0.62)
ΔG =-359kJ
Learn more about free energy:brainly.com/question/15319033
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Its 4 dude i did this already
