Ca (Clo3)2(s) ⇒ Ca Cl2 (s) + 3 O2 (g)
I believe the correct answer from the choices listed above is option B. The best explanation for this happening would be that air <span>particles speed up and collide with the tire walls more often. As the particles are heated, kinetic energy increases. Hope this answers the question. Have a nice day.</span>
Answer:
44.6millilitres
Explanation:
Using the general gas law equation as follows:
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (atm)
V1 = initial volume (L)
T1 = initial temperature (K)
P2 = final pressure (atm)
V2 = final volume (L)
T2 = final temperature (K)
According to this question;
V1 = 30mL
T1 = 273K (STP)
P1 = 1 atm (STP)
V2 = ?
T2 = 300K
P2 = 75.0 kPa = 75 × 0.00987 = 0.74atm
Using P1V1/T1 = P2V2/T2
1 × 30/273 = 0.74×V2/300
30/273 = 0.74V2/300
Cross multiply
300 × 30 = 273 × 0.74V2
9000 = 202.02V2
V2 = 9000/202.02
V2 = 44.55
V2 = 44.6millilitres.
To determine the pH of a solution which has 0.195 M hc2h3o2 and 0.125 M kc2h3o2, we use the ICE table and the acid dissociation constant of hc2h3o2 <span>to determine the concentration of the hydrogen ion present at equilibrium. We do as follows:
HC2H3OO = H+ + </span>C2H3OO-
KC2H3OO = K+ + C2H3OO-
Therefore, the only source of hydrogen ion would be the acid. We use the ICE table,
HC2H3OO H+ C2H3OO-
I 0.195 0 0.125
C -x +x +x
------------------------------------------------------------------
E 0.195-x x 0.125 + x
Ka = <span>1.8*10^-5 = (0.125 + x) (x) / 0.195 -x
x = 2.81x10^-5 M = [H+]
pH = - log [H+]
pH = -log 2.81x10^-5
pH = 4.55
Therefore, the pH of the resulting solution would be 4.55.</span>
