Answer:
never true is the answer
explanation:
compounds are formed by two or more elements that are chemically combined with one another. They can be broken down only be chemical or electro chemical means not by physical means.
hence, the answer i have given above is correct
please mark as brainliest
Answer:
True
Explanation:
Ethers react with HI to form the corresponding alcohols and alkyl iodides.
Similarly, ethyl ether react with excess of HI to form ethanol and ethyl iodide. But in the excess of HI as mentioned in the question, ethanol too undergoes 
reaction with HI to form ethyl iodide.
<u>Hence, ethyl iodide is the only product when ethyl ether reacts with excess of HI for several hours.</u>
The mole fraction of solute in a 3.87 m aqueous solution is 0.0697
<h3>
calculation</h3>
molality = moles of the solute/Kg of the solvent
3.87 m dissolve in 1 Kg of water= 1000g
find the moles of water= mass/molar mass
that is 1000 g/ 18 g/mol= 55.56 moles
mole of solute = 3.87 moles
mole fraction is = moles of solute/moles of solvent
that is 3.87/ 55.56 = 0.0697
Answer:
Q14: 17,140 g = 17.14 kg.
Q16: 504 J.
Explanation:
<u><em>Q14:</em></u>
- To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat absorbed by ice (Q = 3600 x 10³ J).
m is the mass of the ice (m = ??? g).
c is the specific heat of the ice (c of ice = 2.1 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 100.0°C - 0.0°C = 100.0°C).
∵ Q = m.c.ΔT
∴ (3600 x 10³ J) = m.(2.1 J/g.°C).(100.0°C)
∴ m = (3600 x 10³ J)/(2.1 J/g.°C).(100.0°C) = 17,140 g = 17.14 kg.
<u><em>Q16:</em></u>
- To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat absorbed by ice (Q = ??? J).
m is the mass of the ice (m = 12.0 g).
c is the specific heat of the ice (c of ice = 2.1 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 0.0°C - (-20.0°C) = 20.0°C).
∴ Q = m.c.ΔT = (12.0 g)(2.1 J/g.°C)(20.0°C) = 504 J.
