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deff fn [24]
3 years ago
13

2

Chemistry
1 answer:
Zepler [3.9K]3 years ago
6 0

Answer:

Three product with are SO2, H2O and CuSO4

Explanation:

You might be interested in
Draw an electrolytic cell in which Mn2+ is reduced to Mn and Sn is oxidized to Sn2+. (Assume standard conditions).
babunello [35]

Answer:

-1.05 V

Explanation:

A detailed diagram of the setup as required in the question is shown in the image attached to this answer. The electrolytes chosen are SnCl2 for the anode half cell and MnCl2 for the cathode half cell. Tin rod and manganese rod are used as the anode and cathode materials respectively. Electrons flow from anode to cathode as indicated. The battery connected to the set up drives this non spontaneous electrolytic process.

Oxidation half equation;

Sn(s) ------> Sn^2+(aq) + 2e

Reduction half equation:

Mn^2+(aq) + 2e ----> Mn(s)

Cell voltage= E°cathode - E°anode

E°cathode= -1.19V

E°anode= -0.14 V

Cell voltage= -1.19 V - (-0.14V)

Cell voltage= -1.05 V

8 0
3 years ago
Find the molality of the solution if 42 grams of lithium chloride (LiCl) are dissolved in 3.6 kg of water.
omeli [17]
The molality is calculated using the following rule:
molality = number of moles of solute / kg of solvent

From the periodic table:
molar mass of lithium = 6.941 gm
molar mass of chlorine = 35.453 gm

molar mass of LiCl = 6.941 + 35.453 = 42.394 gm
number of moles found in 42 gm = mass / molar mass = 42 / 42.394 = 0.99

molality = 0.99 / 3.6 = 0.275 m
3 0
3 years ago
Determine the oxidation number of each element in these compounds or ions. (a) au2(seo4)3 (gold(iii) selenate) au = se = o = (b)
Stels [109]
<span>                                                    Au</span>₂(SeO₄)₃

                                         O = -2 × 4 = -8
                                             Se  =  + 6
So,
                                            (+6 - 8) = -2

Means (SeO₄) contains -2 charge, Now multiply -2 by 3
                                             
                                             -2 ₓ 3 = -6
Means,
                             Au₂ + (-6) = 0
               
                            Au₂  = +6
Or,
                            Au  =  6 / 2

                            Au  = +3
Result:
                            Au  =  +3
                            Se  =  +6
                            O   =  -2

                                                      Ni(CN)₂


Cyanide (CN⁻) contains -1 charge,
So,
                              N  =  -3
                              C  =  +2
Then,
                                         Ni + (-1)₂  =  0

                                               Ni - 2  =  0
Or,
                                                     Ni =  +2
Result:
                            N  =  -3
                            C  =  +2
                           Ni  =  +2




6 0
3 years ago
Given that the nucleophilic substitution reaction used 5.0 mL of t-pentyl alcohol and 12.0 mL of conc. hydrochloric acid to prod
Vladimir [108]

Answer:

4.90 g

Explanation:

Given that:

volume of t-pentyl alcohol = 5 mL

the standard density of t-pentyl alcohol = 0.805 g/mL

Recall that:

density = mass(in wt) /volume

mass = density × volume

mass = 0.805 g/mL × 5 mL

mass = 4.03 g

Volume of HCl used = 12 mL

The reaction for this equation is shown in the image attached below.

From the reaction,

88.15 g of t-pentyl alcohol reacts with concentrated HCl to yield 106.59 g pf t-pentyl chloride.

4.03 g of t-pentyl alcohol forms,

= \dfrac{106.59 \ g \times 4.03 \ g}{88.15 \ g} of t-pentyl chloride.

Therefore,

Theoretical yield of t-pentyl chloride = 4.90 g

8 0
3 years ago
sodium ions are moving with their concentration gradient with the use of a protein channel across a plasma membrane. Determine t
jeka57 [31]

Answer:

Explanation:

When there is difference between the concentration of the content of a cell and the solution surrounding the cell, there will be an Electrochemical gradient or Concentration gradient. Therefore, some solute will tend to move from the region of high concentration to the region of lower concentration through the cell membrane.

Such a movement is called Primary Active Transport

6 0
3 years ago
Read 2 more answers
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