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標題:
CE CHEM Volumetric analyze
發問:
b part 唔識做,請幫忙,謝謝1. 3.60g of a solid mixture of sodium carbonate-10-water and sodium carbonate Required 23.0cm cube of 1.50M hydrochloric acid for complete reactionA(i) WRITE chemical equations of the reaction between (1) sodium carbonate and hydrochloric acid, and (2) sodium hydrogencarbonate and... 顯示更多 b part 唔識做,請幫忙,謝謝 1. 3.60g of a solid mixture of sodium carbonate-10-water and sodium carbonate Required 23.0cm cube of 1.50M hydrochloric acid for complete reaction A(i) WRITE chemical equations of the reaction between (1) sodium carbonate and hydrochloric acid, and (2) sodium hydrogencarbonate and hydrochloric acid B Calculate the percentage by mass of each substance in the mixture (R.A.M : H =1 , C=12 ,O=16,Na=23) Ans : (B) 47.3% sodium carbonate-10-water 52.3% sodium hydrogencarbonate
最佳解答:
b. let x be the mass of hydrated sodium carbonate (3.6-x ) is the mass of sodium bicarbonate. the no. of mole of sodium carbonate used=x/286 the no. of mole of sodium bicarbonate used=(3.6-x)/84 the no. of mole of acid used=23/1000 x 1.5 therefore, 2(x/286) + (3.6-x)/84 = 23/1000 x 1.5 after solving, x=1.7 the %of sodium carbonate by mass =1.7/3.6x100=47.2% the % of sodium bicarbonate by mass=1.9/3.6x100%=52.7%
其他解答:
CE CHEM Volumetric analyze
發問:
b part 唔識做,請幫忙,謝謝1. 3.60g of a solid mixture of sodium carbonate-10-water and sodium carbonate Required 23.0cm cube of 1.50M hydrochloric acid for complete reactionA(i) WRITE chemical equations of the reaction between (1) sodium carbonate and hydrochloric acid, and (2) sodium hydrogencarbonate and... 顯示更多 b part 唔識做,請幫忙,謝謝 1. 3.60g of a solid mixture of sodium carbonate-10-water and sodium carbonate Required 23.0cm cube of 1.50M hydrochloric acid for complete reaction A(i) WRITE chemical equations of the reaction between (1) sodium carbonate and hydrochloric acid, and (2) sodium hydrogencarbonate and hydrochloric acid B Calculate the percentage by mass of each substance in the mixture (R.A.M : H =1 , C=12 ,O=16,Na=23) Ans : (B) 47.3% sodium carbonate-10-water 52.3% sodium hydrogencarbonate
最佳解答:
b. let x be the mass of hydrated sodium carbonate (3.6-x ) is the mass of sodium bicarbonate. the no. of mole of sodium carbonate used=x/286 the no. of mole of sodium bicarbonate used=(3.6-x)/84 the no. of mole of acid used=23/1000 x 1.5 therefore, 2(x/286) + (3.6-x)/84 = 23/1000 x 1.5 after solving, x=1.7 the %of sodium carbonate by mass =1.7/3.6x100=47.2% the % of sodium bicarbonate by mass=1.9/3.6x100%=52.7%
其他解答:
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