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Liquation is a method for concentrating ores, which have a lower melting point than the impurities. The ores of antimony are concentrated by this method. The powdered ore is heated upon a sloping floor of the furnace. The temperature is raised above the melting point of the ore: this causes the or..
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(i) Create laws/legislation to force the reduction of air pollution e.g. strict automobile emission standards. In the National Capital Region of Delhi, the pollution laws state that all vehicles being sold have to confirm to Euro-II emission standards, vehicles have to use unleaded fuel ..
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The mixture of lead sulphide and Zinc sulphide ores are concentrated by the method of electrostatic concentration. The powdered ore is fed upon a roller in a thin layer and subjected to the influence of a electrostatic field. Lead sulphide being a good conductor gets charged immediately and is thr..
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The method used to purify impurities that are easily oxidizable is cupellation. Silver is refined by this method. The impure metal is fused in small boat shaped dishes made of bone ash called as cupels. The cupels are heated in a suitable furnace by a blast of air blown over them. The im..
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. The form of energy, which cannot be recreated, is the amount that goes into increasing disorder. In reality a lot of effort and time is needed by nature to recreate energy once it goes through a spontaneous process. In the above example it is done through photosynthesis, carbonization of wood etc..
. The form of energy, which cannot be recreated, is the amount that goes into increasing disorder. In reality a lot of effort and time is needed by nature to recreate energy once it goes through a spontaneous process. In the above example it is done through photosynthesis, carbonization of wood etc..Solution:
Initial velocity (u) = 0 Time taken (t) = 6 seconds Acceleration due to gravity (g) = 9.8 m/s 2 We make use of second equation of motion h = 0 + 4.9 x 36 h = 176.4 m 03. A stone projected vertically upward, takes 5 seconds to reach the highest point. What is the initial velocity of ..
Initial velocity (u) = 0 Time taken (t) = 6 seconds Acceleration due to gravity (g) = 9.8 m/s 2 We make use of second equation of motion h = 0 + 4.9 x 36 h = 176.4 m 03. A stone projected vertically upward, takes 5 seconds to reach the highest point. What is the initial velocity of ..Solution
Calculation of the percentages of different elements. Percentage of nitrogen can be calculated thus, Volume of H 2 SO 4 taken = 50 cm 3 N/4 of solution Volume of NaOH required for excess acid = 77 cm 3 N/10 of solution To calculate volume of N/4 H 2 SO 4 used..
Calculation of the percentages of different elements. Percentage of nitrogen can be calculated thus, Volume of H 2 SO 4 taken = 50 cm 3 N/4 of solution Volume of NaOH required for excess acid = 77 cm 3 N/10 of solution To calculate volume of N/4 H 2 SO 4 used..Solution
Mass of copper oxide taken in experiment 1. = 1.375 g Mass of copper obtained = 1.098 g Mass of copper oxide produced in experiment 2 = 1.476 g Mass of copper used = 1.179 g Since the percentage of copper in the two samples of copper oxide is the same, the law of definite proportion ..
Mass of copper oxide taken in experiment 1. = 1.375 g Mass of copper obtained = 1.098 g Mass of copper oxide produced in experiment 2 = 1.476 g Mass of copper used = 1.179 g Since the percentage of copper in the two samples of copper oxide is the same, the law of definite proportion ..Solution
According to the equation, if 2 moles of SO 3 are formed, 2 moles of SO 2 and 1 mole of O 2 are used up. Hence 1.6 moles of SO3 are formed from 1.6 moles of SO 2 and 0.8 mole of O 2 . Therefore, SO 2 left over at equilibrium = 2 - 1.6 = 0.4 mole O 2 left over at equilibrium = 1 -..
According to the equation, if 2 moles of SO 3 are formed, 2 moles of SO 2 and 1 mole of O 2 are used up. Hence 1.6 moles of SO3 are formed from 1.6 moles of SO 2 and 0.8 mole of O 2 . Therefore, SO 2 left over at equilibrium = 2 - 1.6 = 0.4 mole O 2 left over at equilibrium = 1 -..Solution
(a) Molecular formula of butane = C 4 H 1 0 Molecular mass of butane = 4 x 12 + 10 x 1 = 58 Heat of combustion of butane = 2658kJmol - 1 1 mole of 58 g of butane on complete combustion give heat = 2658 kJ 14 x 10 3 g of butane on complete combustion gives heat = The family needs 20,000 kJ o..
(a) Molecular formula of butane = C 4 H 1 0 Molecular mass of butane = 4 x 12 + 10 x 1 = 58 Heat of combustion of butane = 2658kJmol - 1 1 mole of 58 g of butane on complete combustion give heat = 2658 kJ 14 x 10 3 g of butane on complete combustion gives heat = The family needs 20,000 kJ o.. Result
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