Protons, neutrons, and electrons.
Answer:
Pb(NO₂)₂(aq) + 2 LiCl(aq) ⇒ PbCl₂(s) + 2 LiNO₂(aq)
Explanation:
Let's consider the reaction between aqueous lead (II) nitrite and aqueous lithium chloride to form solid lead (II) chloride and aqueous lithium nitrite.
Pb(NO₂)₂(aq) + LiCl(aq) ⇒ PbCl₂(s) + LiNO₂(aq)
This is a double displacement reaction. We will start balancing Cl by multiplying LiCl by 2.
Pb(NO₂)₂(aq) + 2 LiCl(aq) ⇒ PbCl₂(s) + LiNO₂(aq)
Now, we have to balance Li by multiplying LiNO₂ by 2.
Pb(NO₂)₂(aq) + 2 LiCl(aq) ⇒ PbCl₂(s) + 2 LiNO₂(aq)
The equation is now balanced.
Answer:
#Molecules XeF₆ = 2.75 x 10²³ molecules XeF₆.
Explanation:
Given … Excess Xe + 12.9L F₂ @298K & 2.6Atm => ? molecules XeF₆
1. Convert 12.9L 298K & 2.6Atm to STP conditions so 22.4L/mole can be used to determine moles of F₂ used.
=> V(F₂ @ STP) = 12.6L(273K/298K)(2.6Atm/1.0Atm) = 30.7L F₂ @ STP
2. Calculate moles of F₂ used
=> moles F₂ = 30.7L/22.4L/mole = 1.372 mole F₂ used
3. Calculate moles of XeF₆ produced from reaction ratios …
Xe + 3F₂ => XeF₆ => moles of XeF₆ = ⅓(moles F₂) = ⅓(1.372) moles XeF₆ = 0.4572 mole XeF₆
4. Calculate number molecules XeF₆ by multiplying by Avogadro’s Number (6.02 x 10²³ molecules/mole)
=> #Molecules XeF₆ = 0.4572mole(6.02 x 10²³ molecules/mole)
= 2.75 x 10²³ molecules XeF₆.
