Answer:
ΔH⁰(11.4g NH₄NO₃) = -30.59Kj (4 sig. figs. ~mass of NH₄NO₃(s) given) (exothermic)
Explanation:
3NH₄NO₃(s) + C₁₀H₂₂(l) + 14O₂(g) => 3N₂(g) + 17H₂O(g) + 10CO₂(g)
ΔH⁰(f): 3(-365.6)Kj 1(-301)Kj 14(0)Kj 3(0)Kj 17(-241.8)Kj 10(-393.5)Kj
= -1096.8Kj = -301Kj = 0Kj = 0Kj = -4110.6Kj = -3930.5Kj
ΔHₙ°(rxn) = ∑
(ΔH˚(f)products) - ∑(ΔH˚(f)reactants)
= [3(0)Kj + 17(-241.8)Kj + (-393.5)Kj] - [(-(1096.8)Kj + (-301)Kj + (0)Kj]
= [-(8041.1) - (-1397.8)]Kj
= -6643.3Kj (for 3 moles NH₄NO₃ used in above equation)
∴ Standard Heat of Rxn = -6643.3Kj/3moles = -214.8Kj/mole NH₄NO₃(s)
ΔH°(rxn for 14.11g NH₄NO₃(s)) = (11.4g/80.04g·mol⁻¹)(-214.8Kj/mol) = 30.5937Kj ≅ 30.59Kj (4 sig. figs. ~mass of NH₄NO₃(s) given)
Explanation:
The given data is as follows.
Mass of apple sauce mixture = 454 kg
Heat added (Q) = 121300 kJ
Heat capacity (
) of apple sauce at 
= 4.0177 
So, Heat given by heat exchanger = heat taken by apple sauce
Q = 
or, Q = 
Putting the given values into the above formula as follows.
Q =
121300 kJ = 
= 
Thus, we can conclude that outlet temperature of the apple sauce is .
You will want to find how many grams are in a whole mole so you know which element it is. To do this, find out how much of a mole you have.
4.95 x 10^23 atoms / 6.022 x 10^23 atoms (one whole mole of any element) = .8219860511 or ~82% of 1 mole
Now we know that, find what to multiply 20 g by to get the rest of the mole.
1 mole / .8219860511 mole = 1.216565657
20 g x 1.216565657 = ~24.33 g / mol
Now that you have grams per mole, you can look at the periodic table and the molar masses to see which this number is closely aligned.
Your answer is Magnesium (Mg), which has a molar mass of 24.305 g
Answer:
The first option: Strontium Fluorate.
Explanation:
because Fluorine and oxygen combines to make fluorate, Strontium stays the same.
p.s: i need help in geo and there's an exam tomorrow.
Answer:
Ka = 6.02x10⁻⁶
Explanation:
The equilibrium that takes place is:
We <u>calculate [H⁺] from the pH</u>:
- [H⁺] =
Keep in mind that [H⁺]=[A⁻].
As for [HA], we know the acid is 0.66% dissociated, in other words:
We <u>calculate [HA]</u>:
Finally we <u>calculate the Ka</u>:
- Ka =
![\frac{[9.12x10^{-4}]*[9.12x10^{-4}]}{[0.138]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B9.12x10%5E%7B-4%7D%5D%2A%5B9.12x10%5E%7B-4%7D%5D%7D%7B%5B0.138%5D%7D)
= 6.02x10⁻⁶
