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10mL of 0.5M Al(NO3)3 is mixed with 3 mL of 4M NaOH. The number of moles of Al(OH)3 obtained is:a) 3×10-3b) 4×10-3c) 5×10-3d) 7×10-3

Jobin C J , 6 Years ago
Grade 12th pass
anser 1 Answers
Bhavya

Last Activity: 5 Years ago

Hi
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