A: BE has more ionization energy than LI
B: CA has more ionization energy than BA.
C: NA has more ionization energy than K
D: AR has more ionization energy than P
E: CI has a more ionization energy than SI
F: LI has more ionization energy than K
If any of these are wrong feel free to correct me in the comments.
If im correct i think the answer is 12.6 mol.
Answer:
A) yes
B) The average mass of 12 pennies should be expressed as
( 3.131 + 3.129 + -------- X12 ) g / 12
where X12 = is the mass of the 12th penny
Explanation:
A) I think the the Bureau of Mint changed the way it made pennies because from the experiment and observation carried out on 15 different pennies it can be seen that they had different weights and this difference is associated with the period/time of production of the different pennies,
B ) The average mass of any object should be expressed in the si unit of mass and not having ($) sign attached to the value instead it should be expressed in grams or kilograms (s.i unit of mass )
The average mass of 12 pennies should be expressed as
( 3.131 + 3.129 + -------- X12 ) g / 12
where X12 = is the mass of the 12th penny
Explanation:
Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:
The mixture would contain
if undergoes no hydrolysis; the solution is of volume after the mixing. The two species would thus be of concentration and , respectively.
Construct a RICE table for the hydrolysis of under a basic aqueous environment (with a negligible hydronium concentration.)
The question supplied the <em>acid</em> dissociation constant for acetic acid ; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant for its conjugate base, . The following relationship relates the two quantities:
... where the water self-ionization constant under standard conditions. Thus . By the definition of :
V(NaOH)=15 mL =0.015 L
C(NaOH)=0.1 mol/L
C(H₂SO₄)=0.05 mol/L
2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O
n(NaOH)=V(NaOH)C(NaOH)=2n(H₂SO₄)
n(H₂SO₄)=V(H₂SO₄)C(H₂SO₄)
V(NaOH)C(NaOH)=2V(H₂SO₄)C(H₂SO₄)
V(H₂SO₄)=V(NaOH)C(NaOH)/{2C(H₂SO₄)}
V(H₂SO₄)=0.015*0.1/{2*0.05}=0.015 L = 15 mL