a) Magnetite
b) CaO<sub>(s)</sub> + SiO<sub>2(s) $\rightarrow$ CaSiO<sub>3(s)</sub>
johnmulu answered the question on February 21, 2017 at 06:46
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Date posted: February 21, 2017. Answers (1)
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Date posted: February 18, 2017. Answers (1)
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Date posted: February 18, 2017. Answers (1)
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Date posted: February 18, 2017. Answers (1)
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Date posted: February 16, 2017. Answers (1)
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Date posted: February 16, 2017. Answers (1)
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Date posted: February 16, 2017. Answers (1)
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Date posted: February 16, 2017. Answers (1)
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Date posted: February 16, 2017. Answers (1)
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When a current of 0.82 A was passed for 5 hours through an aqueous solution of metal Z, 2.65 g of the metal were deposited. (1 Faraday = 96500 Coulombs: Relative atomic mass of Z = 52)
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(c) Write the equation for the formation of ion of P.
Date posted: February 16, 2017. Answers (1)
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Date posted: February 16, 2017. Answers (1)
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During purification of copper by electrolysis, 1.48 g of copper were deposited when current was passed through aqueous copper (II) sulphate for 2$\frac{1}{2}$ hours. Calculate the amount of current used. (Cu = 63.5; 1 F = 96500 C).
Date posted: February 16, 2017. Answers (1)
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An element P has a relative atomic mass of 88. When a current of 0.5 amperes was passed through the fused chloride of P for 32 minutes and 10 seconds, 0.44 g of P were deposited at the cathode. Determine the charge on an ion of P.
(1 Faraday = 96500 Coulombs)
Date posted: February 16, 2017. Answers (1)
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When a current of 1.5 amperes was passed through a cell containing M3+ ions of metal M for 15 minutes, the mass of the cathode increases by 0.26 g. (1 Faraday = 96500 coulombs).
a) Calculate the quantity of electricity used
b) Determine the relative atomic mass of metal M.
Date posted: February 16, 2017. Answers (1)