1) If 1 drop (0.050 mL) of .20 M NaBr is added to 100.00 mL ofwater, what is the

Question

1) If 1 drop (0.050 mL) of .20 M NaBr is added to 100.00 mL ofwater, what is the molarity of the NaBr in the resultingsolution? 2) Concentrated HNO3 is 15.4 M . How would you prepare 2.50 Lof 0.375 M HNO3? 3) What is the concentration of CaCl2 produced when 55.0 mL of0.300 M HCl is mixed with 80.0 mL of 0.550 M CaCl2? 4) If 75.0 mL of 0.200 M Na3PO4 is added to 25.0 mL of 0.800 MK3PO4, what is the concentration of Na3Po4 in the mixture? 1) If 1 drop (0.050 mL) of .20 M NaBr is added to 100.00 mL ofwater, what is the molarity of the NaBr in the resultingsolution? 2) Concentrated HNO3 is 15.4 M . How would you prepare 2.50 Lof 0.375 M HNO3? 3) What is the concentration of CaCl2 produced when 55.0 mL of0.300 M HCl is mixed with 80.0 mL of 0.550 M CaCl2? 4) If 75.0 mL of 0.200 M Na3PO4 is added to 25.0 mL of 0.800 MK3PO4, what is the concentration of Na3Po4 in the mixture?

Explanation / Answer

For the first question, what you do is times the volume ofNaBr by the concentration, then divide it by the volume ofsolution: 0.00005L NaBr x [0.20mol NaBr] x[   1L   ] = 0.0001mol/L NaBr                                       1L                 0.10L For the second question, find the moles of the givenconcentration and volume. Next divide the moles found the theconcentration (15.4mol/L): 2.5L HNO3 x [0.375mol HNO3] x[         1L         ] = 0.0609L HNO3                                       1L               15.4molHNO3 60.9mL of 15.4mol/L HNO3 is added to enough water tomake 2.5L of solution. The for third question, find the limiting reagent and usethose moles to find the needed concentration: 0.055L HCl x [0.300mol HCl] = 0.0165mol     *L.R.*                            1LHCl 0.080L CaCl2 x [0.550mol CaCl2] = 0.044molCaCl2                                  1LCaCl2 0.0165mol HCl x [1mol CaCl2] x[   1L   ] = 0.206mol/L CaCl2                               1molHCl         0.08L Number 4 is the sane thing as 3: 0.075L Na3PO4 x [0.200mol/L Na3PO4] = 0.015mol Na3PO4    *L.R.* 0.025L K3PO4 x [0.80mol/L K3PO4] = 0.02molK3PO4 0.015mol Na3PO4 x [1mol Na3PO4] x[   1    ] = 0.200mol/L Na3PO4                                    1molNa3PO4      0.75L Hope this helps!!
-PilsnerPrincess                                    1molNa3PO4      0.75L Hope this helps!!
-PilsnerPrincess

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