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Alenkasestr [34]
3 years ago
13

Choose the aqueous solution below with the lowest freezing point. These are all solutions of nonvolatile solutes and you should

assume ideal van't Hoff factors where applicable. Choose the aqueous solution below with the lowest freezing point. These are all solutions of nonvolatile solutes and you should assume ideal van't Hoff factors where applicable.
A. 0.075 m Li I
B. 0.075 m (NH4)3PO4
C. 0.075 m NaIO4
D. 0.075 m KCN
E. 0.075 m KNO2
Chemistry
1 answer:
Ymorist [56]3 years ago
4 0

Answer:

B. 0.075 m (NH4)3PO4

Explanation:

Our strategy here is to recall the van´t Hoff factor, i, for the colligative properties of electrolyte solutions which appears as the consequence that electrolytes disociate completely in their solutions in water.

Thus in this problem we need to determine i and then realize the one with the lowest freezing point will have the biggest  i ( all the concentrations are equal) since

ΔTf = i m Kf

Substance  van´t Hoff factor

Li I                             2

(NH4)3PO4              4

NaIO4                       2

KCN                          2

KNO2                       2

The correct answer is B. 0.075 m (NH4)3PO4

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An inventor claims to have devised a closed cyclic engine which exchanges heat with cold and hot reservoirs at 25 and 350 ◦C, re
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Explanation:

The given data is as follows.

T_{c} = 25^{o}C,  T_{h} = 350^{o}C

Work produced per kJ of heat extracted from hot reservoir = 0.45 kJ = Efficiency

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Using this relation we will calculate the efficiency as follows.

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Hence, it means that this type of device is possible and the claim is also believable.

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3 years ago
When you need to produce a variety of diluted solutions of a solute, you can dilute a series of stock solutions. A stock solutio
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Answer:

Volume of stock solution needed = 6.0299 mL

Explanation:

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This is deduced when thinking that both the dissolution at the beginning and at the end will have the same amount of moles.

<u>Data:</u>

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M2 = 0.3624 M diluted solution concentration

V2 =100 mL diluted solution volume

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3 years ago
How do you write zinc chloride as a chemical formula?
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3 years ago
A 35.0-ml sample of 0.150 m acetic acid (ch3cooh) is titrated with 0.150 m naoh solution. calculate the ph after 17.5 ml of base
Gekata [30.6K]
When the moles of CH3COOH = volume of CH3COOH * no.of moles of CH3COOH
moles of CH3COOH = 35ml * 0.15 m/1000 =0.00525 mol
moles of NaOH = volume of NaOH*no.of moles of NaOH
                           = 17.5 ml * 0.15/1000 = 0.002625
SO the reaction after add the NaOH:
                           CH3COOH(aq) +OH- (aq) ↔ CH3COO-(aq) +H2O(l)
initial                  0.00525                0                         0
change             - 0.002625         +0.002625     +0.002625
equilibrium      0.002625             0.002625        0.002625
When the total volume = 35ml _ 17.5ml = 52.5ml = 0.0525L
∴[CH3COOH] = 0.002625/0.0525 = 0.05m
and [CH3COO-]= 0.002625/0.0525= 0.05 m
when PKa = -㏒Ka
                 = -㏒1.8x10^-5 = 4.74
by substitution in the following formula:
PH = Pka + ㏒[CH3COO-]/[CH3COOH]
      = 4.74 + ㏒(0.05/0.05) = 4.74
∴PH = 4.74
6 0
3 years ago
Read 2 more answers
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