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3 years ago

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The solubility of AgNO3 at 10° and 20°C are 170g and 222g per 100g H2O. What is the sign of change of heat of solution for AgNO3

?

1 answer:

laiz [17]3 years ago

3 0

Answer:

The sign of change of heat of solution is positive

Explanation:

The dissolution of AgNO3 in water is represented by the equation;

AgNO3(s) --------> Ag^+(aq) + NO3^-(aq)

We can see from the question that as the temperature was increased from 10° to 20°C the solubility of the solute increased from 170g to 222g per 100g of H2O.

This implies that the solubility of the solute increases with increase in temperature.

If a reaction moves in the forward direction when the temperature is increased, then the reaction is endothermic. If the reaction is endothermic, the sign of change of heat of solution is positive.

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scandium47 has a half-life of 35s. suppose you have a 45g sample of scadium 47 how much of the sample remains unchanged after 14

Mars2501 [29]

The much of the sample that would remain unchanged after 140 seconds is 2.813 g

Explanation

Half life is time taken for the quantity to reduce to half its original value.

if the half life for Scandium is 35 sec, then the number of half life in 140 seconds

=140 sec/ 35 s = 4 half life

Therefore 45 g after first half life = 45 x1/2 =22.5 g

22.5 g after second half life = 22.5 x 1/2 =11.25 g

11.25 g after third half life = 11.25 x 1/2 = 5.625 g

5.625 after fourth half life = 5.625 x 1/2 = 2.813

therefore 2.813 g of Scandium 47 remains unchanged.

4 0

2 years ago

Can someone help me answer these ?

g100num [7]

3.3 ثنائي ميثيل الهكسان

5 0

3 years ago

Mass does not change during a chemical change. is the same true for physical change?

Mashutka [201]

Answer:

Mass states that mass is neither created nor destroyed in a chemical reaction or a physical transformation

Explanation:

4 0

3 years ago

Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+

Iteru [2.4K]

<u>Answer:</u> The spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

<u>Explanation:</u>

We are given:

$E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, iodine will always undergo reduction reaction, then copper and then iron.

The equation used to calculate electrode potential of the cell is:

$E^o_{cell}=E^o_{oxidation}+E^o_{reduction}$

The combination of the cell reactions follows:

<u>Case 1:</u>

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

$I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=0.45+0.54=0.99V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

<u>Case 2:</u>

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

$I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-$ $E^o_{oxidation}=-0.34V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=-0.34+0.54=0.20V$

Thus, this cell will give the spontaneous cell reaction with smallest $E^o_{cell}$

<u>Case 3:</u>

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

$Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)$ $E^o_{reduction}=0.34V$

$E^o_{cell}=0.45+0.34=0.79V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Hence, the spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

7 0

3 years ago

What is the molecular formula for a compound that is 44.87% potassium, 36.7%

Phoenix [80]

Answer:

Molecular formula: S4K8O16 empirical formula: SK2O4

Explanation:

First we find the moles of each by first finding grams (using the percent) and then using stoichiometry to convert into moles:

Sulfur: 696 *.18 = 125.28grams S* $\frac{1 mole S}{32.065 g S} = 3.907 moles S$

Potassium: 696 *.4487 = 312.2952 * $\frac{1 mole K}{39.08 g K}$ = 7.99117 mole K

Oxygen: 696 * .367 = 255.432 * $\frac{1 mol O}{16g O}$ = 15.9654 mole O

Then we divide each value by the atom with the smallest number of moles to find the mole ratio:

3.907/3.907= 1

7.99117 mole K/ 3.907= 2.043

15.9654 mole O/ 3.907= 4.08

The empirical formula is SK2O4

To find the molecular formula, we divide the mass given (696) by the mass of the empirical formula (174.22) to get 4. We then divide each atom by 4.

Molecular formula: S4K8O16

5 0

2 years ago