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pav-90 [236]
3 years ago
14

Which of the following affect the discharge of ions in electrolysis​

Chemistry
2 answers:
Lina20 [59]3 years ago
7 0
Which of the following affects the discharge
dolphi86 [110]3 years ago
6 0

Answer:

Discharge of ions during electrolysis is affected by, current apllied, the time of electrolysis, electrode type, posiiton of metal in electrochemical.

Explanation:

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Stochiometry .How many grams of CO2 is produced when excess CS2 reacts with 4 mols of O2? CS2+O2-SO2 Please balance equation.
Travka [436]

Mass of CO₂ produced : 58.67 g

<h3>Further explanation</h3>

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

Reaction

CS₂ + 3O₂  -------> CO₂ + 2SO₂

mol of CO₂ based on mol of O₂ as a limiting reactant(CS₂ as an excess reactant)

From the equation, mol ratio of mol CO₂ : mol O₂ = 1 : 3, so mol CO₂  :

\tt \dfrac{1}{3}\times 4=\dfrac{4}{3}

mass  CO₂ (MW= 44 g/mol) :

\tt \dfrac{4}{3}\times 44=58.67~g

4 0
3 years ago
245 g water sample initially at at 32 oC absorbs 17 kcal of heat. What is the final temperature of water?
steposvetlana [31]

Answer:62.66°C or 235.66K

Explanation:Q=McpT, the energy was given in calories so you first convert to Joules by multiplying the value in calories by 4.184J.

17*4.184=71.128kJ.

71.128kJ=mcpT

71.128kJ=245*4.187*(T-Tm)

Tm is the final temperature of the mixture. The T is the temperature given which should be converted to Kelvin by adding 273...T=32+273=305K.

71128J=245*4.187*(305-Tm)

71128=312873.575-1025.815Tm

1025.815Tm=312873.575-71128

1025.815Tm=241745.58

Tm=241745.58/1025.815

Tm=235.66K

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4 0
3 years ago
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Michelle is trying to find the average atomic mass of a sample of an unknown
GREYUIT [131]

The average atomic mass of her sample is 114.54 amu

Let the 1st isotope be A

Let the 2nd isotope be B

From the question given above, the following data were obtained:

  • Abundance of isotope A (A%) = 59.34%
  • Mass of isotope A = 113.6459 amu
  • Mass of isotope B = 115.8488 amu
  • Abundance of isotope B (B%) = 100 – 59.34 = 40.66%
  • Average atomic mass =?

The average atomic mass of the sample can be obtained as follow:

Average \: atomic \: mass \:  =  \frac{mass \: of \: A \times A\%}{100}  + \frac{mass \: of \: B \times B\%}{100}  \\  \\ Average \: atomic \: mass \:  =  \frac{113.6459\times 59.34}{100} + \frac{115.8488\times 40.66}{100} \\  \\ Average \: atomic \: mass \:  = 114.54 \: amu  \\  \\

Thus, the average atomic mass of the sample is 114.54 amu

Learn more about isotope: brainly.com/question/25868336

3 0
2 years ago
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