What mass of calcium chloride in grams would be enough to produce 14.35g of Agcl ?
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Answer:
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Explanation:
equivalent of calcium chloride = equivalent of 14.32 agcl
molarity*f=14.32/143.5*f
molarity*2=1/10
molarity=1/20
w/111=1/20
w=5.55gm
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