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Yuki888 [10]
2 years ago
15

Balance the following skeleton ionic reaction, and calculate E°cell to decide whether the reaction is spontaneous: Ag(s) + Cu2+

(aq) LaTeX: \longrightarrow⟶ Ag+ (aq) + Cu(s) Enter a number so 1, 2, 3 etc E°cell = spontaneous? enter yes or no (MTS 3/3/2019)
Chemistry
2 answers:
Pachacha [2.7K]2 years ago
7 0

Answer:

Explanation:

Electrochemical cell is composed of 2 half cells, one where reduction reaction takes place which is the cathode and the other is where oxidation reaction takes place.

Reduction reaction

Cu2+(aq) + 2e- ---> Cu(s)

E° = 0.34 V

Oxidation reaction

2Ag(s) --> 2Ag+(aq) + 2e-

E° = -0.80 V

Combining both equations,

2Ag(s) + Cu2+ (aq) ⟶ 2Ag+ (aq) + Cu(s)

Standard eletrode potential of the cell, E°cell = E°cath - E°anode

= 0.34 - (-0.80)

= +1.14 V.

If the value of E°cell is positive, then the reaction is spontaneous, and it will occur as written under standard conditions; it will, however, not proceed spontaneously in the opposite direction.

tia_tia [17]2 years ago
5 0

Answer:

2Ag(s) + Cu^2+(aq) ----------> 2Ag^+(aq) + Cu(s)

Explanation:

Ag(s)/Ag^+ (aq) is the anode as shown while Cu^2+(aq)/Cu^2(s) is the cathode.

E°cell= E°cathode -E°anode= 0.34 -0.80= -0.5V

The cell is not spontaneous as written because E°cell is negative. This implies that the electrodes of the cell must be interchanged to make the cell spontaneous.

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A 0.500 g sample of tin (Sn) is reacted with oxygen to give 0.534 g of product. What is the empirical formula of the oxide?
REY [17]

Answer:

Sn_2O

Explanation:

Hello,

In this case, given that the mass of the product is 0.534 g, we can infer that the percent composition of tin is:

\%Sn=\frac{0.500g}{0.534g}*100\%\\ \\\%Sn=93.6\%

Therefore, the percent composition of oxygen is 6.4% for a 100% in total. Thus, with such percents we compute the moles of each element in the oxide:

n_{Sn}=93.6gSn*\frac{1molSn}{118.8gSn} =0.788molSn\\\\n_O=6.4gO*\frac{1molO}{16gO}=0.4molO

In such a way, for finding the smallest whole number we divide the moles of both tin and oxygen by the moles of oxygen as the smallest moles:

Sn:\frac{0.788}{0.4}=2\\ \\O:\frac{0.4}{0.4}=1

Therefore, the empirical formula is:

Sn_2O

Best regards.

8 0
2 years ago
Calculate the mass of 25,000 molecules of nitrogen gas. (1 mole = 6.02 x 1023 molecules)
Ainat [17]

Hey there!

Molar mass N2 = 28.01 g/mol

Therefore:

28.01 g N2 -------------- 6.02*10²² molecules N2

( mass N2 ?? ) ----------- 25,000 molecules N2

mass N2 =  ( 25,000 * 28.01 ) /  ( 6.02*10²³ )

mass N2 = 700250 / 6.02*10²³

mass N2 = 1.163*10⁻¹⁸ g


Hope that helps!

7 0
2 years ago
Read 2 more answers
Which of the following was most clearly, even uniquely perhaps, revealed by the discovery of the photoelectric effect?
nika2105 [10]

Answer : Option 3) Wave/Particle duality.

Explanation : The experiment on discovery of photoelectric effect revealed about the photoelectrons of light that can behave as particle or waves.

The photoelectric effect is observed when the emission of electrons or other free carriers occurs on shining a light radiation on a material. The electrons emitted from this can be called photo electrons. These photoelectrons may behave as wave or particle in duality which holds that light and matter exhibit properties of both waves and of particles.

7 0
2 years ago
Read 2 more answers
A 0.25 m solution of the sugar sucrose (c12h22o11) in water is tested for conductivity using the type of apparatus shown. Bulb w
Anvisha [2.4K]

Sucrose is a non ionic compound. It does liberates ion when dissolved in water unlike NaCl or other salts which dissolve in water and produce respective cations and anions.

Thus if any amount of sucrose is dissolved in water, it will form non ionic aqueous solution (it will dissolve completely). Thus sucrose solution being non electrolytic will not conduct electricity in aqueous solution.

the bulb will not light up as sucrose will remain in molecular form only

6 0
2 years ago
A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (
djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

Kb = \frac{[ HSO3-][0H-]}{SO3-2}

Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation Kb=\frac{[HSO3-][OH-]}{[SO3-2]}

We noe have:

1.485*10^-^7=\frac{x*x}{(0.056-x)}

x = [OH-] = 9.11*10^-^5

pOH = -log(OH) = -log(9.11*10^-^5)

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0
2 years ago
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