Answer:
6.12 L
Explanation:
Given that,
Initial volume, V₁ = 5 L
Initial temperature, T₁ = 7.0°C = 343 K
Final temperature, T₂ = 147°C = 420 K
We need to find its new volume. The relation between volume and temperature is given by :

So, the new volume is 6.12 L.
Molality is the moles of solute per kg of solvent.
Moles of NH₄Cl = 2.4 / (14 + 4 x 1 + 35.5)
= 0.0448 mole
Molality = 0.0448 / (19.4 / 1000)
= 2.31 m
We have that the the liquid is
- C_2H_5OH (ethanol
- And at a condition of H_2SO4 as catalyst and temp 170
From the question we are told
- A student wished to prepare <em>ethylene </em>gas by <em>dehydration </em>of ethanol at 140oC using sulfuric acid as the <em>dehydrating </em>agent.
- A low-boiling liquid was obtained instead of ethylene.
- What was the liquid, and how might the reaction conditions be changed to give ethylene
<h3>
Ethylene formation</h3>
Generally the equation is
2C_2H_5OH------CH3CH_2O-CH_2CH_3+H_20
Therefore
with ethanol at 140oC
The product is diethyl ethen
The reaction at 170 ethylene will give
C_2H_5OH-------CH_2=CH_2+H_2O( at a condition of H_2SO4 as catalyst and temp 170)
Therefore
The the liquid is
For more information on Ethylene visit
brainly.com/question/20117360
Answer:
The solution is given below
Explanation:
Heat, q= mc∆T
q= 125g x 4.18 J/g∙°C x (21.18x- 24.28) °C
q= -1619.75J
NEGATIVE SIGN INDICATES THAT HEAT IS ABSORBED.
Enthalpy Change, ∆H = 1619.75 7/ 10.5 g
= 154.26 J/g
No. of moles of KBr = Mass of KBr/ Molecular Weight of KBr
=10.5g/119gmol-1
=0.088 mol
∆H= 1619.75 J/ 0.088 mol
= 18.41 kJ/mol
First shell from the nucleus can have a maximum of 2 electrons ! hope this helped