Answer:
0.80m of KOH
Explanation:
Molality is an unit of concentration defined as the ratio between moles of solute and kg of solvent.
In the problem, the solute is KOH and solvent is water.
Moles of 36g KOH -Molar mass: 56.1g/mol- are:
36g KOH × (1mol / 56.1g) = <em>0.642 moles of KOH</em>
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Now, as density of water is 1g/mL, mass of 800mL of water is:
800mL × (1g / mL) × (1kg / 1000g) = <em>0.800kg of water</em>
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Thus, molality is:
0.642moles of KOH / 0.800kg = <em>0.80m of KOH</em>
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
Answer:
C₂ = 0.334 M
Explanation:
Given data:
Volume of HCl = 0.0780 L
Concentration of HCl = 0.12 M
Volume of LiOH = 0.0280 L
Concentration of LiOH = ?
Solution:
Formula:
C₁V₁ = C₂V₂
C₁ = Concentration of HCl
V₁ = Volume of HCl
C₂ = Concentration of LiOH
V₂ = Volume of LiOH
Now we will put the values in formula.
C₁V₁ = C₂V₂
0.12 M × 0.0780 L = C₂ × 0.0280 L
0.00936 M.L = C₂ × 0.0280 L
C₂ = 0.00936 M.L/0.0280 L
C₂ = 0.334 M
I'm not 100% sure on this, but I would go with C) NaCl.
NaCl is a salt, and that is used to melt the ice on the roads. Hope this helps!
<span>Answer:
.01 moles of D to .005 moles of L ~ so, .01+.005 = .015 total; using this total value, divide the portions of D and L.
so .01/.015 to .005/.015 ~ 67% D to 33% L.
And thus, the enantiomer excess will be 34%.</span>
