Answer:
pH=10.97
Explanation:
the solution of methyl amine with methylammonium chloride will make a buffer solution.
The pH of buffer solution can be obtained using Henderson Hassalbalch's equation, which is:
![pOH=pKb+log\frac{[salt]}{[base]}](https://tex.z-dn.net/?f=pOH%3DpKb%2Blog%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bbase%5D%7D)
pH = 14- pOH
Let us calculate pOH
[Salt] = [methylammonium chloride] = 0.10 M (initial)
After adding base
![[salt] = \frac{molarityXvolume}{finalvolume}=\frac{0.1X20}{(20+50)}= 0.0286M](https://tex.z-dn.net/?f=%5Bsalt%5D%20%3D%20%5Cfrac%7BmolarityXvolume%7D%7Bfinalvolume%7D%3D%5Cfrac%7B0.1X20%7D%7B%2820%2B50%29%7D%3D%200.0286M)
[base] = [Methylamine]=0.10
After mixing with salt
![[base]= \frac{molarityXvolume}{finalvolume}=\frac{0.1X50}{(20+50)}= 0.0714M](https://tex.z-dn.net/?f=%5Bbase%5D%3D%20%5Cfrac%7BmolarityXvolume%7D%7Bfinalvolume%7D%3D%5Cfrac%7B0.1X50%7D%7B%2820%2B50%29%7D%3D%200.0714M)
pKb= -log[Kb]= 3.43
Putting values
pOH = ![3.43+log(\frac{[0.0286]}{0.0714}](https://tex.z-dn.net/?f=3.43%2Blog%28%5Cfrac%7B%5B0.0286%5D%7D%7B0.0714%7D)
Answer:
(II) only correctly rank the bonds in terms of increasing polarity.
Explanation:
Bond polarity is proportional to difference in electronegativity between bonded atoms.
Atoms Electronegativity Bond Electronegativity difference
Cl 3.0 Cl-F 1.0
Br 2.8 Br-Cl 0.2
F 4.0 Cl-Cl 0
H 2.1 H-C 0.4
C 2.5 H-N 0.9
N 3.0 H-O 1.4
O 3.5 Br-F 1.2
I 2.7 I-F 1.3
Si 1.9 Cl-F 1.0
P 2.2 Si-Cl 1.1
Si-P 0.3
Si-C 0.6
Si-F 2.1
So, clearly, order of increasing polarity : O-H > N-H > C-H
So, (II) only correctly rank the bonds in terms of increasing polarity
Answer:
-10778.95 J heat must be removed in order to form the ice at 15 °C.
Explanation:
Given data:
mass of steam = 25 g
Initial temperature = 118 °C
Final temperature = 15 °C
Heat released = ?
Solution:
Formula:
q = m . c . ΔT
we know that specific heat of water is 4.186 J/g.°C
ΔT = final temperature - initial temperature
ΔT = 15 °C - 118 °C
ΔT = -103 °C
now we will put the values in formula
q = m . c . ΔT
q = 25 g × 4.186 J/g.°C × -103 °C
q = -10778.95 J
so, -10778.95 J heat must be removed in order to form the ice at 15 °C.
Hi there
Formula
M1V1=M2V2
(455*6)/2500
Answer
1.092 M
Problem One
You will use both m * c * deltaT and H = m * heat of fusion.
Givens
m = 12.4 grams
c = 0.1291
t1 = 26oC
t2 = 1204
heat of fusion (H_f) = 63.5 J/grams.
Equation
H = m * c * deltaT + m * H_f
Solution
H = 12.4 * 0.1291 * (1063 - 26) + 12.4 * 63.5
H = 1660.1 + 787.4
H = 2447.5 or 2447.47 is the exact answer. I have to leave the rounding to you. I have no idea where to round it although I suspect 2450 would be right for 3 sig digs.
Problem Two
Formula and Givens
t1 = 14.5
t2 = 50.0
E = 5680
c = 4.186
m = ??
E = m c * deltaT
Solution
5680 = m * 4.186 * (50 - 14.5)
5680 = m * 4.186 * (35.5)
5680 = m * 148.603 * m
m = 5680 / 148.603
m = 38.22 grams That isn't very much. Be very sure you are working in joules. You'd leave that many grams in the kettle after drying it thoroughly.
m = 38.2 to 3 sig digs.
