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AlexFokin [52]
4 years ago
15

A voltaic cell is constructed in which the following cell reaction occurs. The half-cell compartments are connected by a salt br

idge. F2(g) + 2I-(aq) 2F-(aq) + I2(s)
The anode reaction is: + +
The cathode reaction is: + +
In the external circuit, electrons migrate the F-|F2 electrode the I-|I2 electrode. In the salt bridge, anions migrate the F-|F2 compartment the I-|I2 compartment.
Chemistry
1 answer:
ddd [48]4 years ago
7 0

Answer:

See explanation below

Explanation:

The anode reaction is :

2I^-(aq) -------> I2(g) +2e

Cathode reaction

F2(g) + 2e------> 2F^-(aq)

In the external circuit, electrons migrate from the I-|I2 electrode (anode) to the F-|F2 electrode (cathode)

In the salt bridge, anions migrate from the F-|F2

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One of the hydrates of MnSO4 is manganese(II) sulfate tetrahydrate . A 71.6 gram sample of MnSO4 4 H2O was heated thoroughly in
Dennis_Churaev [7]

Answer:

48.32 g of anhydrous MnSO4.

Explanation:

Equation of dehydration reaction:

MnSO4 •4H2O --> MnSO4 + 4H2O

Molar mass = 55 + 32 + (4*16) + 4((1*2) + 16)

= 223 g/mol

Mass of MnSO4 • 4H2O = 71.6 g

Number of moles = mass/molar mass

= 71.6/223

= 0.32 mol.

By stoichiometry, since 1 mole of MnSO4 •4H2O is dehydrated to give 1 mole of anhydrous MnSO4

Number of moles of MnSO4 = 0.32 mol.

Molar mass = 55 + 32 + (4*16)

= 151 g/mol.

Mass = 151 * 0.32

= 48.32 g of anhydrous MnSO4.

3 0
3 years ago
A sample of a compound contains 41.33 g of carbon and 8.67 g of hydrogen. The molar mass of the
mihalych1998 [28]

Answer:

C6H15

Explanation:

First, we calculate the empirical formula as follows:

C = 41.33 g

H = 8.67 g

We convert each mass value to mole by dividing each element by its molar mass (C = 12g/mol, H = 1g/mol)

C = 41.33g ÷ 12g/mol = 3.44mol

H = 8.67g ÷ 1g/mol = 8.67mol

Next, we divide each mole value by the smallest (3.44mol)

C = 3.44mol ÷ 3.44mol = 1

H = 8.67mol ÷ 3.44mol = 2.52

We multiply this ratio by 2 to get a simple whole number ratio

C = 1 × 2 = 2

H = 2.52 × 2 = 5.04

Based on this, the whole number ratio of C and H is 2:5, hence, the empirical formula is C2H5.

The molecular mass of the compound is given as 87.18 g/mol, hence, the molecular formula is calculated as follows:

(C2H5)n = 87.18

[12(2) + 1(5)]n = 87.18

[24 + 5]n = 87.18

(29)n = 87.18

n = 87.18 ÷ 29

n = 3.006

Approximately to whole number, n = 3

Hence, the molecular formula of the compound is [C2H5]3

= C6H15

4 0
3 years ago
This is for a study guide, I can't figure it out!
miskamm [114]
<span>The answer is "D" where the number of collisions per unit area is reduced by one-half. Drawing back on the piston means the volume is increased. The pressure is reduced. There are fewer collisions when the pressure is reduced.</span>
3 0
3 years ago
What volume is needed to store 0.80 moles of helium gas at 204.6 kpa and 300 k?
shepuryov [24]
Answer is: 9,7 L is needed to store helium gas.
n(He) = 0,80 mol.
p(He) = 204,6 kPa.
T = 300 K.
R = 8,314 J/K·mol; universal gas constant.
Use ideal law eqaution: p·V = n·R·T.
V = n·R·T / p.
V(He) = 0,80 mol · 8,314 J/K·mol · 300 K ÷ 204,6 kPa.
V(He) = 9,75 L.


3 0
3 years ago
What mass of CO was used up in the reaction with an excess of oxygen gas if 24.7g of carbon dioxide is formed? 2 CO + O2 &gt; 2
Zolol [24]
Balance Chemical Equation,
                                      2 CO  +  O₂   →   2CO₂
Acc. to this reaction,
88 g (2 mole) of CO₂ was produced when  =  56 g (2 mole)of CO was reacted
So, 
24.7 g of CO₂ will be produced by reacting =  X g of CO

Solving for X,
                                    X  =  (56 g × 24.7 g) ÷ 88 g

                                    X  =  2.26 g ÷ 88 g

                                    X  =  0.0257 g of CO

Result:           
            0.0257 g of CO is required to be reacted with excess of O₂ to produce 24.7 g of CO₂.
7 0
3 years ago
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