An Arrhenius acid by definition dissociates in water to form H3O+ (or H+) ions while an arrhenius base dissociates in water to form OH- ions.
NH4+(aq) can be categorised as an arrhenius acid since it releases H3O+ ions in aqueous media
NH4+(aq) + H2O (aq) ↔ NH3 (aq) + H3O+(aq)
Answer:
trans-1,3-pentadiene is more stable than 1,4-pentadiene due to presence of a conjugated double bond.
Explanation:
Here,
H(hydrogenated pdt.) is same for both 1,4-pentadiene and 1,3-pentadiene as they both produce pentane after hydrogenation
H(diene) depends on stability of diene.
More stable a diene, lesser will be it's H(diene) value (more neagtive).
trans-1,3-pentadiene is more stable than 1,4-pentadiene due to presence of a conjugated double bond.
Hence, is higher (less negative) for trans-1,3-pentadiene
The equation for carbon-14 emission by Radium-223 nuclei is given below:
<h3>What is radioactivity?</h3>
Radioactivity is the spontaneous decay of a substance with emission of radiation.
The equation for carbon-14 emission by Radium-223 nuclei is given below:
In conclusion, the emission of carbon-14 by Radium-223 nuclei produces Lead-209 nuclei.
Learn more about radioactivity at: brainly.com/question/3603596
#SPJ1
Answer:
The molarity of the solution is 0,31 M
Explanation:
We calculate the weight of 1 mol of NaCl from the atomic weights of each element of the periodic table. Then, we calculate the molarity, which is a concentration measure that indicates the moles of solute (in this case NaCl) in 1000ml of solution (1 liter)
Weight 1 mol NaCl= Weight Na + Weight Cl= 23 g + 35, 5 g= 58, 5 g
58, 5 g-----1 mol NaCl
13,1 g ---------x= (13,1 g x 1 mol NaCl)/58, 5 g= 0, 224 mol NaCl
727 ml solution------ 0, 224 mol NaCl
1000ml solution------x= (1000ml solutionx0, 224 mol NaCl)/727 ml solution
x=0,308 mol NaCl---> <em>The solution is 0,31 molar (0,31 M)</em>
Answer : The correct option is, +91 kJ/mole
Solution :
The balanced cell reaction will be,
Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.
First we have to calculate the standard electrode potential of the cell.
Now we have to calculate the standard Gibbs free energy.
Formula used :
where,
= standard Gibbs free energy = ?
n = number of electrons = 2
F = Faraday constant = 96500 C/mole
= standard e.m.f of cell = -0.47 V
Now put all the given values in this formula, we get the Gibbs free energy.
Therefore, the standard Gibbs free energy is +91 kJ/mole