Answer: 24.1 L
Explanation:
To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
where,
are the initial volume and temperature of the gas.
are the final volume and temperature of the gas.
We are given:
Putting values in above equation, we get:
Thus the volume of the sample when heated to 220.0oC and the pressure is constant is 24.1 L
Answer:
Explanation:
NH₄NO₃ = NH₄⁺ +NO₃⁻
heat released by water = msΔ T
m is mass , s is specific heat and ΔT is fall in temperature
= 50 x 4.18 x ( 22 - 16.5 ) ( mass of 50 mL is 50 g )
= 1149.5 J .
This heat will be absorbed by the reaction above .
q for the reaction = + 1149.5 J
2 )
molecular weight of NH₄NO₃ = 80
No of moles reacted = 5/80 = 1 / 16 moles.
3 )
5 g absorbs 1149.5 J
80 g absorbs 1149.5 x 16 J
= 18392 J
= 18.392 kJ.
= + 18.392 kJ
ΔH = 18.392 kJ / mol
Answer:
It is an ether so the nomenclature rule for naming the above compound will be:Alkoxy alkane , generally Alkoxy represent the smaller alkyl group and as in the above question both the alkyl group are same so it's IUPAC Name will be: Ethoxy ethane or Diethylether both are correct
Explanation:
happy to help ya!
Answer:
Explanation:
1. Calculate the initial moles of acid and base
2. Calculate the moles remaining after the reaction
OH⁻ + H₃O⁺ ⟶ 2H₂O
I/mol: 0.0053 0.005 00
C/mol: -0.00500 -0.005 00
E/mol: 0.0003 0
We have an excess of 0.0003 mol of base.
3. Calculate the concentration of OH⁻
Total volume = 53 mL + 25.0 mL = 78 mL = 0.078 L
![\text{[OH}^{-}] = \dfrac{\text{0.0003 mol}}{\text{0.078 L}} = \textbf{0.0038 mol/L}\\\\\text{The final concentration of OH$^{-}$ is $\large \boxed{\textbf{0.0038 mol/L}}$}](https://tex.z-dn.net/?f=%5Ctext%7B%5BOH%7D%5E%7B-%7D%5D%20%3D%20%5Cdfrac%7B%5Ctext%7B0.0003%20mol%7D%7D%7B%5Ctext%7B0.078%20L%7D%7D%20%3D%20%5Ctextbf%7B0.0038%20mol%2FL%7D%5C%5C%5C%5C%5Ctext%7BThe%20final%20concentration%20of%20OH%24%5E%7B-%7D%24%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.0038%20mol%2FL%7D%7D%24%7D)
