Molarity in Nacl
The molarity of 0.006 mole of NaCl in 100 solutions will be: (i) 0.6 (ii) 0.06 (iii) 0.006 (iv) 0.066 (v) None of theseChoose the right answer from above.Answer: The right answer is (ii) M = n/ v(l)= 0.006/ 0.1 = 0.06
The molarity of 0.006 mole of NaCl in 100 solutions will be: (i) 0.6 (ii) 0.06 (iii) 0.006 (iv) 0.066 (v) None of theseChoose the right answer from above.Answer: The right answer is (ii)
M = n/ v(l)= 0.006/ 0.1 = 0.06
The molarity of 0.006 mole of NaCl in 100 solutions will be: (i) 0.6 (ii) 0.06 (iii) 0.006 (iv) 0.066 (v) None of theseChoose the right answer from above.Answer: The right answer is (ii) M = n/ v(
Whether HCl is a base or an acid? Briefly state your comments?
Select the right answer of the question. If one mole of any substance is present in of solvent, then: (a) It shows molar concentration (b) It shows molal concentration (c) It shows normality (d) It shows strength.
Choose the right answer from following. What is the molarity of H2SO4 solution, that has a density 1.84 gm/cc at 35c and contains solute 98% by weight: (a) 4.18 M (b) 8.14 M (c)18.4 M (d)18 M
define primary secondary and tertiary alkyl halides with examples
Can someone please help me in getting through this problem. How many grams of dibasic acid (having mol. wt. 200) must be present in 100ml of its aqueous solution to provide decinormal strength: (i) 1g (ii)2g (iii) 10g (iv) 20g<
Which one is polar HCl or HF?
Superphosphate has the formula CaH4(PO4)2 H2O, what is the calculation to get the percentage of Phosphorus, I need to show the calculation. I know it is 30.9737622 u in weight and 2 atoms of the formula, but not sure how to work the calculation backwards.
Unit of measurement- These are also some systems for units: (1) C.G.S.
Provide solution of this question. If 25 ml of 0.25 M NaCl solution is diluted with water to a volume of 500ml the new concentration of the solution is : (a) 0.167 M (b) 0.0125 M (c) 0.833 M (d) 0.0167 M
18,76,764
1929842 Asked
3,689
Active Tutors
1429675
Questions Answered
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!