Answer:
Option D is correct = 58 g
Explanation:
Data Given:
mass of LiOH = 120 g
Mass of Li3N= ?
Solution:
To solve this problem we have to look at the reaction
Reaction:
Li₃N (s) + 3H₂0 (l) -----------► NH₃ (g) + 3LiOH (l)
1 mol 3 mol
Convert moles to mass
Molar mass of LiOH = 24 g/mol
Molar mass of Li₃N = 35 g/mol
So,
Li₃N (s) + 3H₂0 (l) -----------► NH₃ (g) + 3LiOH (l)
1 mol (35 g/mol) 3 mol (24 g/mol)
35 g 72 g
So if we look at the reaction 35 g of Li₃N react with water and produces 72 g of LiOH , then how many g of Li₃N will be react to Produce by 120 g of LiOH
For this apply unity formula
35 g of Li₃N ≅ 72 g of LiOH
X of Li₃N ≅ 120 g of LiOH
By Doing cross multiplication
Mass of Li₃N = 35 g x 120 g / 72 g
mass of Li₃N = 58 g
120 g of LiOH will produce from 58 g of Li₃N
So,
Option D is correct = 58 g
Answer:
the correct answer is c
Explanation:
becuase i had the same question
Answer is: <span>name of the product nuclide is thorium-230.
</span>
<span>Alpha particle is
nucleus of a helium-4 atom, which is made of
two protons and two neutrons.
²³</span>⁴U → ²³⁰Th + α (alpha
particle).
Alpha
decay is radioactive decay in which an atomic
nucleus emits an alpha particle (helium nucleus) and transforms
into an atom with an atomic number that is reduced by
two and mass number that is reduced by four.
Answer: 44.37 degrees C
Explanation:
Use combined gas law: (P1)(V1)/T1=(P2)(V2)/T2
For most gas laws, you must convert to Kelvin:
K=deg C+273
K=25+273=298 K
Plug and chug:
(1.0 atm)(1.2 L)/(298 K)=(0.71 atm)(1.8 L)/(x)
Solve for x and get 317.37 K
Subtract 273 from this to convert to degrees Celsius. You will get 44.37 degrees Celsius.
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Answer:
24.03 J/mol.ºC
Explanation:
For a calorimeter, the heat lost must be equal to the heat gained from water plus the heat gained from calorimeter, which has the same initial temperature as the water.
-Qal = Qw + Qc (minus signal represents that the heat is lost)
-mal*Cal*ΔTal = mw*Cw*ΔTw + Cc*ΔTc
Where m is the mass, C is the specific heat, ΔT is the temperature variation, al is from aluminum. w from water and c from the calorimeter. Cw = 4.186 J/gºC
-25.5*Cal*(22.7 - 100) = 99.0*4.186*(22.7 - 18.6) + 14.2*(22.7 - 18.6)
1971.15Cal = 1699.10 + 58.22
1971.15Cal = 1757.32
Cal = 0.89 J/g.ºC
The molar mass of Al is 27 g/mol
Cal = 0.89 J/g.ºC * 27 g/mol
Cal = 24.03 J/mol.ºC
