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.
Answer:
A.The sound becomes louder.
And
C.The sound waves get further.
Explanation:
Louder the sound it will travel long.
Answer:
0.84kg of gatorade powder is needed.
Explanation:
From the question given above, we were told that the team manager mixes 0.6kg of gatorade powder with 5 gallons of water.
Now, to obtain the desired amount of the powder of gatorade needed for 7 gallons of water, we simply do the following:
If 0.6kg of gatorade powder required 5 gallons of water,
Then Xg of gatorade will require 7 gallons of water i.e
Xg of gatorade = (0.6 x 7)/5
Xg of gatorade = 0.84kg
Therefore, to maintain the same gatorade mixture, the team manager must mix 0.84kg of gatorade powder with the 7 gallons of water.
The balanced chemical equation would be as follows:
<span>
Mg + O2 → MgO2
</span>We are given the amounts of the reactants. We need to determine first which one is the limiting reactant. We do as follows:
62.0 g Mg (1 mol / <span>24.31 g ) = 2.55 mol Mg
</span>55.5 g O2 ( 1 mol / 32 g ) = 1.74 mol O2 -----> <span>consumed completely and therefore the limiting reactant
2.55 mol - 1.74 mol O2( 1 mol Mg / 1 mol O2 ) = 0.81 mol Mg excess</span>
Answer:
4 m/s
Explanation:
formula is v = (KE/.5m)^1/2
there is a calculator
https://www.calculatorsoup.com/calculators/physics/kinetic.php
