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iren2701 [21]
2 years ago
8

The question is below

Chemistry
1 answer:
Nezavi [6.7K]2 years ago
3 0
The correct conditions for measuring reduction potentials (the tendency to acquire electrons and become reduced), is 25C and 1M (or 1 mole/litre) for reactants - ANSWER B.
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Hello summer roses~<br>help me with these questions ​
Olegator [25]

Answer:

  • The process of "slaking" of lime refers to the use of excess water for hydration. balanced equation: CaO + H₂0 = Ca(OH)₂ + heat released

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3 0
2 years ago
Why is coke sprinkled on the top of electrolyte in the bayer's process?​
Vesnalui [34]

Powdered coke prevent the heat loss. As a result, it prevents burning of carbon electrodes.

4 0
1 year ago
Problem 12.002 the molar analysis of a gas mixture at 30°c, 2 bar is 40% n2, 50% co2, 10% ch4. determine
andreev551 [17]
The problem is incomplete. However, there can only be two probable questions for this problem. First, you can be asked the individual partial pressures of each gas. Second, you can be asked the volume occupied by each gas. I can answer both cases for you.

1.

Let's assume ideal gas.
Pressure for N₂: 2 bar*0.4 = 0.8 bar
Pressure for CO₂: 2 bar*0.5 = 1 bar
Pressure for CH₄: 2 bar*0.1 = 0.2 bar

2. For the volume, let's find the total volume first.

V = nRT/P = (1 mol)(8.314 J/mol-K)(30 +273 K)/(2 bar*10⁵ Pa/1 bar)
V = 0.0126 m³
Hence,
Volume for N₂: 0.0126 bar*0.4 = 0.00504 m³
Volume for CO₂: 0.0126*0.5 = 0.0063 m³
Volume for CH₄: 0.0126*0.1 = 0.00126 m³
3 0
2 years ago
(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g
Anarel [89]

Answer:

Part A

 The volume of the gaseous product  is  V = 787L

Part B

The volume of the the engine’s gaseous exhaust is  V_e = 2178 \ L

Explanation:

Part A

From the question we are told that

    The temperature is  T = 350^oC = 350 +273 =623K

     The pressure is  P = 735 \ torr = \frac{735}{760} =  0.967\ atm

     The of  C_8 H_{18} = 100.0g

The chemical equation for this combustion is

               2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}

 The number of moles of  C_8 H_{18} that reacted is mathematically represented as

               n = \frac{mass \ of \  C_8H_{18}  }{Molar \  mass \ of  C_8H_{18} }

The molar mass of  C_8 H_{18} is constant value which is

                  M = 114.23 \ g/mole  

So          n = \frac{100  }{114.23} }

             n = 0.8754 \ moles

The gaseous product in the reaction is CO_2_{(g)} and water vapour

Now from the reaction

    2 moles of C_8 H_{18}  will react with 25 moles of O_2 to give (16 + 18) moles of CO_2_{(g)} and  H_2 O_{(g)}

So

    1 mole of C_8 H_{18} will  react with 12.5 moles of  O_2 to give 17 moles of CO_2_{(g)} and  H_2 O_{(g)}

This implies that

    0.8754 moles of C_8 H_{18} will react with (12.5 * 0.8754 ) moles of O_2 to give  (17 * 0.8754) of CO_2_{(g)} and  H_2 O_{(g)}

So the no of moles of gaseous product is

         N_g = 17 * 0.8754

         N_g = 14.88 \ moles

From the ideal gas law

       PV = N_gRT

making V the subject

        V = \frac{N_gRT}{P}

Where R is the gas constant with a value R = 0.08206 \  L\cdot atm /K \cdot mole

Substituting values

          V = \frac{14.88* 0.08206 *623}{0.967}

          V = 787L

Part B

From the reaction the number of moles of oxygen that reacted is

         N_o = 0.8754 * 12.5

         N_o = 10.94 \ moles

The volume is

      V_o  = \frac{10.94 * 0.08206 *623}{0.967}

      V_o  = 579 \ L

No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

         V_e = V_o * \frac{0.79}{0.21}

Substituting values

       V_e = 579 * \frac{0.79}{0.21}

       V_e = 2178 \ L

3 0
3 years ago
Before the development of electrophoresis to separate macromolecules, ____________ was used.
max2010maxim [7]

Before the development of electrophoresis to separate macromolecules, high-speed centrifugation was used to isolate DNA.

A laboratory procedure called electrophoresis is used to divide DNA, RNA, or protein molecules according to their size and electrical charge. The molecules are moved by an electric current through a gel or other matrix. The technology of electrophoresis is crucial for the separation and examination of nucleic acids. At the lab bench, cloned DNA fragments are frequently isolated and worked with using nucleic acid electrophoresis.

High-speed centrifugation employs centrifugal force to separate particles with various densities or masses suspended in a liquid. High-speed rotation of the solution inside the tube causes each particle's angular momentum to experience centrifugal forces inversely proportionate to its mass.

To know more about electrophoresis refer to: brainly.com/question/28709201

#SPJ4

4 0
1 year ago
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