Answer:
d) 0.1202 M
Explanation:
Let's consider the neutralization reaction between NaOH and a generic monoprotic acid.
NaOH + HA → NaA + H₂O
The used volume of NaOH is 41.63 mL - 19.63 mL = 22.00 mL. The moles of NaOH are:
22.00 × 10⁻³ L × 0.1093 mol/L = 2.405 × 10⁻³ mol
The molar ratio of NaOH to HA is 1:1. The moles of HA that reacted are 2.405 × 10⁻³ moles.
The molar concentration of HA is:
2.405 × 10⁻³ mol / 20.00 × 10⁻³ L = 0.1202 M
Answer:
0.169
Explanation:
Let's consider the following reaction.
A(g) + 2B(g) ⇄ C(g) + D(g)
We can find the pressures at equilibrium using an ICE chart.
A(g) + 2 B(g) ⇄ C(g) + D(g)
I 1.00 1.00 0 0
C -x -2x +x +x
E 1.00-x 1.00-2x x x
The pressure at equilibrium of C is 0.211 atm, so x = 0.211.
The pressures at equilibrium are:
pA = 1.00-x = 1.00-0.211 = 0.789 atm
pB = 1.00-2x = 1.00-2(0.211) = 0.578 atm
pC = x = 0.211 atm
pD = x = 0.211 atm
The pressure equilibrium constant (Kp) is:
Kp = pC × pD / pA × pB²
Kp = 0.211 × 0.211 / 0.789 × 0.578²
Kp = 0.169
carbon, hydrogen, nitrogen, and oxygen. the number infront is how many of each element there are, they are increasing and decreasing
