Answer:
The 1st and 4th options are correct
I.the oxidized form has a higher affinity for electrons
IV. the greater the tendency for the oxidized form to accept electrons
Explanation:
Half reaction can be described as the oxidation or reduction reaction in a redox reaction.it is In the redox rection there is a change in the oxidation states of Chemical species involved. the oxidized form in the redox has a higher affinity for electrons and the greater the tendency for the oxidized form to accept electrons.
Standard reduction potential which is also referred to as standard cell potential can be described as the potential difference that exist between cathode and anode of the cell. In the standard reduction potential most times the species will be reduced which is usually analysed in a reduction half reaction.
(Standard Hydrogen Electrode) is utilized when determining the Standard reduction or potentials of a chemical specie. this is because of Hydrogen having zero reduction and oxidation potentials, as a result of this a measured potential of any species is compared with that of Hydrogen, the difference helps to know the potential reduction of that particular specie.
Answer:
=> 572.83 K (299.83°C).
=> 95.86 m^2.
Explanation:
Parameters given are; Water flowing= 13.85 kg/s, temperature of water entering = 54.5°C and the temperature of water going out = 87.8°C, gas flow rate 54,430 kg/h(15.11 kg/s). Temperature of gas coming in = 427°C = 700K, specific heat capacity of hot gas and water = 1.005 kJ/ kg.K and 4.187 KJ/kg. K, overall heat transfer coefficient = Uo = 69.1 W/m^2.K.
Hence;
Mass of hot gas × specific heat capacity of hot gas × change in temperature = mass of water × specific heat capacity of water × change in temperature.
15.11 × 1.005(700K - x ) = 13.85 × 4.187(33.3).
If we solve for x, we will get the value of x to be;
x = 572.83 K (2.99.83°C).
x is the temperature of the exit gas that is 572.83 K(299.83°C).
(b). ∆T = 339.2 - 245.33/ln (339.2/245.33).
∆T = 93.87/ln 1.38.
∆T = 291.521K.
Heat transfer rate= 15.11 × 1.005 × 10^3 (700 - 572.83) = 1931146.394.
heat-transfer area = 1931146.394/69.1 × 291.521.
heat-transfer area= 95.86 m^2.
Diagram is on the picture below.
Answer is: 1).
Sodium chloride is ionic compound and in the water dissociate in sodium cation (positive charge) and chloride anion (negative charge). Water is polar compound, oxagan has negative charge and hydrogen charge. Positive interact witn negative charge and negative with positive charge.
It's a cube so the volume = edge^3
Volume = 2.5^3 cm^3 = 15.625 cm^3
density = mass / volume = 42.20 / 15.625 = 2.70 You have 3 places of accuracy.
density of object = 2.70 grams / cm^3 <<<<=== answer.
Answer:- 6984 kJ of heat is produced.
Solution:- From given information, 1367 kJ of heat is produced by the combustion of 1 mole of ethanol. We are asked to calculate the heat produced by the combustion of 235.0 g of ethanol.
Let's convert given grams to moles and multiply by the heat produced by one mole of ethanol to get the total heat produced. Molar mass of ethanol is 46 grams per mole. The set will be:
= 6984 kJ
So, 6984 kJ of heat is produced by the combustion of 235.0 g of liquid ethanol.
