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Papessa [141]
3 years ago
7

Which of the following liquids will havethe lowest freezing point?

Chemistry
1 answer:
VladimirAG [237]3 years ago
5 0

Answer:

The liquid which ever releases the greatest number of particles  will have the lowest freezing point.

A) pure H2O  

B) Aqueous FeI3 (0.030m) releases 4X's = 0.120 mol/litre ions

C) aqueous glucose (0.050m) releases 0.050 moles/litre glucoes

D) aqueous NaI (0.030m) releases 2X's = 0.060 mol/ litre ions

E) aqueous CoI2 (0.030m) releases 3 X's = 0.090 mol/litre ions

The answer is  

B) Aqueous FeI3 (0.030m)

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The specific heat of water is 4.184Jg ∘C. Determine the final temperature when 600.0 g water at 75.5∘C absorbs 5.90×104 J of ene
sesenic [268]

Answer:

T_2=98.5^{\circ}

Explanation:

Given that,

The specific heat of water is 4.184Jg°C

Mass, m = 600 g

Initial temperature, T₁ = 75.5°C

We need to find the final temperature. We know that heat absorbed is given by :

Q=mc\Delta T\\\\Q=mc(T_2-T_1)\\\\\dfrac{Q}{mc}=(T_2-T_1)\\\\\\T_2=\dfrac{Q}{mc}+T_1\\\\T_2=\dfrac{5.9\times 10^4}{600\times 4.184}+75\\\\T_2=98.5^{\circ}

So, the final temperature is equal to 98.5^{\circ}.

3 0
3 years ago
Be sure to answer all parts. For the titration of 10.0 mL of 0.250 M acetic acid with 0.200 M sodium hydroxide, determine the pH
DaniilM [7]

Explanation:

Molarity=\frac{moles}{Volume(L)}

Molarity of the acetic acid = 0.250 M

Volume of the acetic acid solution = 10.0 mL = 0.010 L( 1 mL =0.001L)

Moles of acetic acid ;

n=0.250 M\times 0.010 L=0.0025 mol

Molarity of the NaOH = 0.200 M

a) Volume of the NaOH solution = 10.0 mL = 0.010 L( 1 mL =0.001L)

Moles of NaOH : 0.200M\times 0.010 L=0.002 mol

CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O

1 mole NaOH neutralizes 1 mole of acetic acid , then 0.002 moles of NaOH will neutralize 0.002 mol of acetic acid.

Moles of acetic acid left un-neutralized = 0.0025 mol - 0.002 = 0.0005 mol

1 mole of acetic acid gives 1 mole of hydrogen ion, then 0.0005 mole of acetic acid will give 0.0005 mole of hydrogen ions.

Moles of hydrogen ion= 0.0005 mol

Volume of the solution = 0.010 L+ 0.010 L = 0.020 L

[H^+]=\frac{0.0005 mol}{0.020 L}=0.025 M

The pH of the 10.0 mL of base added to acetic acid solution :

pH=-\log[H^+]=-\log[0.025 M]=1.60

b) Volume of the NaOH solution = 12.0 mL = 0.012 L( 1 mL =0.001L)

Moles of NaOH : 0.200M\times 0.012 L=0.0024 mol

CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O

1 mole NaOH neutralizes 1 mole of acetic acid , then 0.0024 moles of NaOH will neutralize 0.0024 mol of acetic acid.

Moles of acetic acid left un-neutralized = 0.0025 mol - 0.0024 = 0.0001 mol

1 mole of acetic acid gives 1 mole of hydrogen ion, then 0.0001 mole of acetic acid will give 0.0001 mole of hydrogen ions.

Moles of hydrogen ion= 0.0001 mol

Volume of the solution = 0.010 L+ 0.012 L = 0.022 L

[H^+]=\frac{0.0001 mol}{0.022 L}=0.0045 M

The pH of the 12.0 mL of base added to acetic acid solution :

pH=-\log[H^+]=-\log[0.0045 M]=2.34

c) Volume of the NaOH solution = 15.0 mL = 0.015 L( 1 mL =0.001L)

Moles of NaOH : 0.200M\times 0.015 L=0.003 mol

CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O

1 mole NaOH neutralizes 1 mole of acetic acid , then 0.003 moles of NaOH will neutralize 0.003 mol of acetic acid.

All the moles of acetic acid will get neutralized by NaOH and un-neutralized sodium hydroxide will left over.

Moles of NaOH left un-neutralized = 0.003 mol - 0.0025 = 0.0005 mol

1 mole of NaOH gives 1 mole of hydroxide ion, then 0.0005 mole of NaOH acid will give 0.0005 mole of hydroxide ions.

Moles of hydroxide ion= 0.0005 mol

Volume of the solution = 0.010 L+ 0.015 L = 0.025 L

[OH^-]=\frac{0.0005 mol}{0.025 L}=0.02 M

The pOH of the 15.0 mL of base added to acetic acid solution :

pOH=-\log[OH^-]=-\log[0.02 M]=1.70

The pH of the 15.0 mL of base added to acetic acid solution :

pH=14-pOH=14-1.70=12.3

7 0
3 years ago
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