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labwork [276]
2 years ago
8

Write balanced net ionic equations for the three reactions carried out in Part A given that the sulfur-containing product is the

S2O3
Chemistry
1 answer:
Alik [6]2 years ago
3 0

Answer:

Please find the complete question in the attached file.

Explanation:

Balanced equation: 8KMnO_4+14NaOH +Na_2S_2O_3 \longrightarrow 8Na_2MnO_4 +2K_2SO_4 +5H_2O +4KOH

Ionic equation:

8K^{+} + 8MnO_4^{-}+14Na^{+}+ 14OH^- +2Na^+ S_2O_3^{-}\longrightarrow16Na^+8MnO_4^- +4K^+ + 2SO_4 +5H_2O+4K^+ +4OH^-

Net Ionic equation:

10 OH^- +S_2O_3^- \longrightarrow 2SO_4^- +5H_2O

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beks73 [17]

Answer:

D

Explanation:

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6 0
2 years ago
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Whats the name of S4O6
IceJOKER [234]

Answer:

Sulfonic acid

Explanation:

I think this is correct

8 0
3 years ago
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ROY G BIV arranges the colors of the rainbow *
Elanso [62]

Answer:

Wavelengths..

Explanation:

The colors we see always go from red, which is least refracted, through orange, yellow, green, blue, indigo and violet -- Roy G Biv. The blue, indigo and violet wavelengths are refracted the most as sunlight passes through raindrops.

8 0
2 years ago
A 35.66g sample of copper is heated using 600j of energy. if the original temperature of the copper is 85C what is its final tem
ira [324]

This problem is providing the mass, energy, initial temperature and specific heat of a sample of copper that is required to calculate the final temperature.

Thus, we recall the general heat equation:

Q=mC(T_f-T_i)\\

Which has to be solved for the final temperature, T_f as follows:

T_f=T_i+\frac{Q}{mC}

Finally, we plug in the numbers to obtain:

T_f=85\°C+\frac{600J}{35.66g*0.38\frac{J}{g\°C} } \\\\T_f=129.3\°C

However, this result is not given in the choices.

Learn more:

  • brainly.com/question/14383794
8 0
2 years ago
from the balanced equation we see that h20 is 1 to 1 mole ratio the number of moles of acid and base. What is the number of mole
forsale [732]

Answer:

0.0498 mol

Explanation:

Number of moles = concentration in mol/L × volume in L

Concentration = 1 M = 1 mol/L

Volume = 49.8 mL = 49.8/1000 = 0.0498 L

Number of moles = 1×0.0498 = 0.0498 mol

5 0
3 years ago
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