4.01 grams of Licl are needed to release this amount of heat
<h3>What does one gram weigh?</h3>
The gram is a unit of mass in the International System of Units (SI) that is equal to one thousandth of a kilogram. It was originally known as the gramme. Gram. This pen cap weighs around one gram. A weight scale like this one may provide a precise mass readout for many different things.
Mass of Licl required to release 5850J of heat to the surroundings.
Let x g Licl required.
Then x g ×∆H = 5850J
x g × 1.46 × 10³ = 5850
x = 5850/ 1.46 × 10³
x = 4.0068
so 4.01g Licl required.
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Answer: freezing and evaporation for sure
Explanation:
Answer: m = 0.42; Tb = 79°
Explanation: The relationship between boiling point of the solvent above a solution is directly proportional to the molal concentration of the solute, i.e.:
ΔT = 
where
ΔT is the change in boiling point of the solvent;
is the molal boiling point elevation constant;
m is the molal concentration of the solute in the solution;
For there two solutions:
1) Ethanol:
ΔT =
Tb - 
Tb - 78.4 = 1.22.m (1)
2) Carbon Tetrachloride:
Tb -
Tb - 76.8 = 5.03.m (2)
Solving the system of equations:
Tb - 78.4 = 1.22.m
Tb = 1.22.m + 78.4 (3)
Substituing (3) in (2)
1.22.m + 78.4 - 76.8 = 5.03m
3.81m = 1.6
m = 0.42
With m, find Tb:
T - 76.8 = 5.03.0.42
T = 2.11 + 76.8
T = 79°
<u>Molal</u> concentration is <u>0.42</u> and boiling point is 79°
Answer:
40% of the ammonia will take 4.97x10^-5 s to react.
Explanation:
The rate is equal to:
R = k*[NH3]*[HOCl] = 5.1x10^6 * [NH3] * 2x10^-3 = 10200 s^-1 * [NH3]
R = k´ * [NH3]
k´ = 10200 s^-1
Because k´ is the psuedo first-order rate constant, we have the following:
b/(b-x) = 100/(100-40) ; 40% ammonia reacts
b/(b-x) = 1.67
log(b/(b-x)) = log(1.67)
log(b/(b-x)) = 0.22
the time will equal to:
t = (2.303/k) * log(b/(b-x)) = (2.303/10200) * (0.22) = 4.97x10^-5 s
