Answer:
13.5 * 10^-2 g
Explanation:
What we know:
Balanced Equation: 3Ba+Al2(SO4)3 -->2Al+3BaSO4,
Grams of Ba: 1
Grams of Al2(SO4)3: 1.8g
Calculate the # of moles of Ba and Al2(SO4)3:
1g Ba/137.3 = 7.3 *10^-3 mol Ba
1.8g Al2(SO4)3/ 342 = 5.3 *10^-3 mol Al2(SO4)3
Find the limiting reactant:
Ba has a coefficient of 3 in the balanced equation, so we divide the # of moles of Ba by 3 to get... 7.3 *10^-3 mol Ba/3 = 2.43 *10^-3
Al2(SO4)3 has a coefficient of 1, so if we divide by 1, we get the same number of 5.3 *10^-3
2.43 *10^-3 is smaller than 5.3 *10^-3, therefore Ba is the limiting reactant.
finally, we just find the number of moles of Al
The ratio of Al to Ba is 2:3 so...
7.3 * 10^-3 * (2/3) = 5 *10^-3 mol Al
CONVERT TO GRAMS
5 *10^-3 mol Al * 27 = 13.5 * 10^-2 g
<u>Hope that was helpful! </u>
Pressure is defined as the force divided by the area perpendicular to the force over which the force is applied, or. A given force can have a significantly different effect depending on the area over which the force is exerted.
Answer:
The De Broglie wavelength decreases when the momentum increases
Explanation:
The De Broglie wavelength of a particle (or any object) is given by
where
h is the Planck constant
p is the momentum of the object
As we can see, the wavelength is inversely proportional to the momentum of the object: therefore we can say that, if the momentum increases, the De Broglie wavelength will decrease.
Answer:
ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'
Explanation:
Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj
197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)
From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...
3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).
______
NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g. Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.
