1 molecule of NaCl contains 1 sodium ion (Na+), that's why if we have 3.0 moles of.
NaCl, we have 3.0 moles of Na+.
N(ions) = n(mol) · NA.
N(ions) = 3.0 moles · 6.02·1023 = 18.06 ·1023 ions.
The motion of particles can be changed by the temperature you put it at.
Answer:
The 1st and 4th options are correct
I.the oxidized form has a higher affinity for electrons
IV. the greater the tendency for the oxidized form to accept electrons
Explanation:
Half reaction can be described as the oxidation or reduction reaction in a redox reaction.it is In the redox rection there is a change in the oxidation states of Chemical species involved. the oxidized form in the redox has a higher affinity for electrons and the greater the tendency for the oxidized form to accept electrons.
Standard reduction potential which is also referred to as standard cell potential can be described as the potential difference that exist between cathode and anode of the cell. In the standard reduction potential most times the species will be reduced which is usually analysed in a reduction half reaction.
(Standard Hydrogen Electrode) is utilized when determining the Standard reduction or potentials of a chemical specie. this is because of Hydrogen having zero reduction and oxidation potentials, as a result of this a measured potential of any species is compared with that of Hydrogen, the difference helps to know the potential reduction of that particular specie.
Answer: There are molecules in 63.00 g of
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number of particles.
To calculate the moles, we use the equation:
1 mole of contains = molecules
Thus 3.5 moles of contains = molecules.
There are molecules in 63.00 g of
Answer:
It will decrease by 2 units.
Explanation:
The Henderson-Hasselbalch equation for a buffer is
pH = pKa + log(base/acid)
Let's assume your acid has pKa = 5.
(a) If the base: acid ratio is 1:1,
pH(1) = 5 + log(1/1) = 5 + log(1) = 5 + 0 = 5
(b) If the base: acid ratio is 1:100,
pH(2) = 5 + log(1/100) = 5 + log(0.01) = 5 - 2 = 3
(c) Difference
ΔpH = pH(2) - pH(1) = 5 - 3 = -2
If you increase the acid:base ratio to 100:1, the pH will decrease by two units.