Use the ideal gas law:
PV = nRT
so, T = PV / nR
n=0.5
V= 120 dm^3 = 120 L (1 dm^3 = 1 L)
R = 1/12
P = 15,000 Pa = 0.147 atm (1 pa = 9.86 10^{-6} )
Put the values:
T = PV / nR
T = (0.147) (120) / (0.5) (1/12)
T= 426 K
Answer:
105 grams PbI₂
Explanation:
Pb(NO₃)₂ + 2KI => 2KNO₃ + PbI₂(s)
moles Pb(NO₃)₂ = 0.265L(1.2M) = 0.318 mole
moles KI = 0.293(1.55M) = 0.454 mole => Limiting Reactant
moles PbI₂ from mole KI in excess Pb(NO₃)₂ = 1/2(0.454 mole) = 0.227 mol PbI₂
grams PbI₂ = 0.227 mol PbI₂ x 461 g/mole = 104.68 g ≈ 105 g PbI₂(s)
Answer:
This question is incomplete
Explanation:
This question is incomplete. However, the beaker that contained some water before NaOH were added means that the resulting solution in that beaker will be more dilute. When this diluted sodium hydroxide solution is added to HCl (not hci), the reaction below occurs
HCl + NaOH ⇒ NaCl + H₂O
The reaction above is a neutralization reaction. <u>The concentration of the acid (HCl) will reduce when a base (sodium hydroxide) is added and will also reduce more because of the presence of more water (in the base) which normally reduces the concentration of ions present in an acid or a base to become more dilute.</u>
