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stepladder [879]
3 years ago
15

A certain metal forms a soluble nitrate salt M(NO3)3. Suppose the left half cell of a galvanic cell apparatus is filled with a 3

.0mM solution of M(NO3)3 and the right half cell with a 3.0M solution of the same substance. Electrodes made of M are dipped into both solutions and a voltmeter is connected between them. The temperature of the apparatus is held constant at 20.0 C.
Required:
a. Which electrode will be positive?
b. What voltage will the voltmeter show? Assume its positive lead is connected to the positive electrode.

Chemistry
1 answer:
DiKsa [7]3 years ago
7 0

Answer:

1.The electrode on the right is positive

2. 0.058V

Explanation:

The above cell is a concentration cell.

A concentration cell is an electrolytic cell that is made of two half-cells with the same electrodes, but differs in concentrations of the solutions. A concentration cell functions by diluting the more concentrated solution and concentrating the more dilute solution, creating a voltage as the cell reaches an equilibrium thereby transferring the electrons from the cell with the lower concentration to the cell with the higher concentration.

In the above cell, electrons flow from the left electrode (less concentrated) to the right electrode (more concentrated). Therefore, the right electrode is the positive electrode (cathode).

Part 2: Please, see the attachment below for the calculations.

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The degree to which a weak base dissociates is given by the base-ionization constant, Kb. For the generic weak base, B B(aq)+H2O
mamaluj [8]

Answer:

a) pH = 11.37

b) %ionization = 0.774%

Explanation:

a) the equilibrium reaction is as follows:

                                  NH3   +    H2O       =       NH4+         +        OH-

equilibrium (moles)     0.3           x                       x                          x

The equilibrium equation is equal to:

Kb = ([NH4+]*[OH-])/[NH3]

Since the value of x is very small compared to 0.3, we can simplify the expression to:

Kb = x^2/0.3

Clearing x, we have:

x = (Kb * 0.3)^1/2 = (0.3 * 1.8x10^-5)^1/2 = 0.00232 M

Due [OH-] = x = 0.00232 M, we can calculate the pOH

pOH = -log[OH-] = -log(0.00232) = 2.63

the pH is equal to:

pH = 14 - pOH = 14 - 2.63 = 11.37

b) Since NH3 is a weak base, we can use the Ostwald equation to calculate the degree dissociation

α = (Kb/C)^1/2, where C is the molar concentration

α = (1.8x10^-5/0.3)^1/2 = 0.00774

%α = 0.00774 * 100 = 0.774%

6 0
3 years ago
When sodium metal is added to water, the following reaction occurs:
zheka24 [161]

Answer:

d. n H2(g) = 0.034 mol

Explanation:

  • 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

                              2 - Na - 2

                              4 - H - 4

                               2 - O - 2

∴ n Na(s) = 0.068 mol

⇒ n H2(g) = ( 0.068 mol Na(s) )( mol H2(g) / 2 mol Na(s) )

⇒ n H2(g) = 0.034 mol

8 0
3 years ago
What is the chemical formula for ammonium carbonate?
xenn [34]

(NH4),CO

(NH₄)₂ CO₃  

Explanation:

The chemical formula for ammonium carbonate is (NH4),CO

Ammonium carbonate is made up of:

  Ammonium ion

  Carbonate ion

Ammonium ion is represented by NH₄⁺

Carbonate ion is represented by CO₃²⁻

Now to write the formula, we are going to use their combining powers;

                   

                              NH₄⁺                     CO₃²⁻  

Combining

power                       1                             2

Exchange

of power                    2                            1

  Formula of compound              (NH₄)₂ CO₃  

learn more:

Chemical compound brainly.com/question/10585691

#learnwithBrainly

3 0
4 years ago
If you reacted 88.9 g of ammonia with excess oxygen, what mass of nitric oxide would you expect to make? You will need to balanc
KonstantinChe [14]
1) Balanced chemical equation

4NH3(g) + 5O2(g) ---> 4NO(g) + 6H2O(g)

2) State the molar ratios

4 mol NH3 : 5 mol O2 : 4 mol NO : 6 mol H2O

3) Convert 88.9 g of ammonia to moles, using the molar mass

molar mass of NH3 = 14 g/mol + 3 * 1 g/mol = 17 g/mol

number of moles = mass in grams / molar mass = 88.9 g / 17 g/mol = 5.23 mol NH3

4) Make the proportion

4 mol NO / 4 mol NH3 = x / 5.23 mol NH3=> x = 5.23 mol NO

5) Convert 5.23 mol NO to grams

molar mass NO = 14 g/mol + 16g/mol = 30 g/mol

mass = number of moles * molar mass = 5.23 mol * 30 g/mol = 156.9 g ≈ 157 g

Answer: 157 grams
5 0
4 years ago
How many moles are in 2.12g Cr2O3
ohaa [14]

Answer:

0.01395mol Cr2O3

Explanation:

the molar mass of Cr2O3 is 151.9904 g/mol

2.12g Cr2O3 x 1 mol/151.9904g = 0.01395mol Cr2O3

4 0
3 years ago
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