What volume (in mL) of 0.0887 M MgF_2 solution is needed to made 286.3 mL of 0.0

Question

What volume (in mL) of 0.0887 M MgF_2 solution is needed to made 286.3 mL of 0.0224 M MgF_2 solution? 72.3 mL 91.8 mL 10.9 mL 69.4 mL 14.4 mL How many milliliters of a 0.184 M NaNO_3 solution contain 0.0999 moles of NaNO_3? 543 mL 163 mL 614 mL 885 mL 326 mL Determine the concentration (in M) of a solution prepared by diluting 20.0 mL of a 0.400 M NaCl to 250.0 mL 0.160 M 0.0320 M 2.50 M 0 00800 M 0.0160 M How many grams of NaCl are required to make 250.0 ml. of a 4.25 M solution? $8.40 g 175.3 g 14.60 g 43.83 g 61.05 g Which of the following is a precipitation reaction? Zn_(s) + 2 AgNO_3(aq) rightarrow 2 Ag_(s) + Zn(NO_3)_2(aq) 2 LiI_(aq) + Hg_2(NO_3)_2(aq) rightarrow Hg_2I_2(s) + 2 LiNO_3(aq) NaCl(aq) + LiI(aq) rightarrow NaI(aq) + LiCl_(aq) HCl_(aq) + KOH rightarrow KCl_(aq) + H_2O(l) None of the above is precipitation reactions.

Explanation / Answer

(38) Volume of MgF2 required = 0.0224 M x 286.3 ml/0.0887 M

A. 72.3 ml

(39) mililiters having given NaNO3 = 0.0999 mole/0.184 M

A. 0.543

(40) concentration of solution = 0.4 M x 20 ml/250 ml

B. 0.0320 M

(41) grams of NaCl required = 4.25 M x 0.250 L x 58.44 g/mol

E. 61.05 g

(42) Precipitation reaction would be,

B.

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