Asked by Ogunsanwo eyitayo
A is a solution of Hcl containing 7.3gdm-3 B is a solution of X2CO3 containing 10.6g1dm3,if 2343cm3 of a neutrized 25.0 cm3 of 13 calculate the
(a)conc n of A in mold3
(b)conc of B in mol.1dm3
(C)molar mass of X2Co3
(H=1, c=12,o=16,cl=35.5)
C2Co3+2Hck------->2×cl+H20+C02
(a)conc n of A in mold3
(b)conc of B in mol.1dm3
(C)molar mass of X2Co3
(H=1, c=12,o=16,cl=35.5)
C2Co3+2Hck------->2×cl+H20+C02
Answers
Answered by
DrBob222
I can't decipher this problem, particularly the "if 2343cm3 of a neutrized 25.0 cm3 of 13". Part A is 7.3 g HCl/dm^3
mols = g/molar mass = 7.3 g/36.5 = 0.2 mols/dm^3
When you repost please make sure that 2343 cc is correct. It seems large to me. Perhaps a decimal was omitted.
mols = g/molar mass = 7.3 g/36.5 = 0.2 mols/dm^3
When you repost please make sure that 2343 cc is correct. It seems large to me. Perhaps a decimal was omitted.
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