Answer:
sorry
Explanation:
I don't know the answer this is really confusing but I am really sorry you have to do this.
Filtration can be used to separate an insoluble solid from a liquid, or a precipitate from the reaction mixture in which it formed. The solid which collects in the filter paper<span> is called the residue. The clear liquid which passes through the </span>filter paper<span> is called the filtrate.</span>
Answer:
2AlCl3 + 3H2SO4 → Al2(SO4)3 + 6HCl
Explanation:
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.
Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
Next, we identify the limiting reactant by computing the available moles of ethane and the moles of ethane consumed by 60.0 grams of oxygen:
Thus, we notice there are less available moles, for that reason, the ethane is the limiting reactant. Finally, we can compute the produced moles of water by:
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