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10mL of 0.5M Al(NO 3 ) 3 is mixed with 3 mL of 4M NaOH. The number of moles of Al(OH) 3 obtained is: a) 3×10 -3 b) 4×10 -3 c) 5×10 -3 d) 7×10 -3

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) 3×10-3
b) 4×10-3
c) 5×10-3
d) 7×10-3

Grade:12th pass

1 Answers

Bhavya
askIITians Faculty 1281 Points
2 years ago
Hi
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