Total moles of ions are released when each of the following samples dissolves completely in water?
Assuming you and Mr. V did the calculation properly then my question is what variety of homework - is this on line/computerized homework? each so often in computerized homework you may desire to have the terrific suited variety of important figures in the respond to be marked terrific suited. you have 3 sig figs in the preliminary mass, and four on your molar mass, yet purely supply 2 on your answer. the respond might desire to have 3 sig figs - so might desire to be 0.565 mol of ions
How many total moles of ions are released when the following samples dissolves completely in water?
We must also assume that they ionise 100% in water. Rb2 SO4 <=> 2Rb+ + SO4 2- You will get 2*.805 mol Rb ions = 1.61 mols Rb ions And 0.805 mol SO4 ions B Molar mass Ca(NO3)2 = 164g 3.85*10^-3 g = 2.35 *10^-5 mol Ca(NO3)2 You will have 2.35*10^-5 mol of Ca 2+ ions and 4.7*10^-5 mol NO3 + ions C. 1 mol contains 6.02*10^23 formula units. (molecules) You have (4.03*10^19 ) / (6.02^10^23) Mol = 6.69*10^-5 mol You will form: 6.69*10^-5 mol of Sr ions and 1.34*10^-4 mol of HCO3 ions
How many total moles of ions are released when the following samples dissolves completely in water?
A. The ionisation of K2HPO4 in water can be presented as: K2HPO4 = 2 K^+ + HPO4^- Thus 1 mol K2HPO4 gives 2 moles of K^+ and 1 mole of HPO4^- i.e. in total 3 moles of ions. Therefore 0.382 mol K2HPO4 will release 3x0.382 = 1.146 moles of ions. B. The ionisation of MgSO4* 7H2O in water can be presented as: MgSO4* 7H2O = Mg^2+ + SO4^2- + 7H2O I mol MgSO4* 7H2O after ionisation gives 1 mol Mg^2+ ion and 1 mol SO4^2- ion i.e. in total 2 moles of ions. Now 6.80 g MgSO4 * 7H2O = 6.80/246 = 0.028 mol So, 0.028 mol MgSO4* 7H2O will give 2x0.028 = 0.056 mol of ions.
How many total moles of ions are released when each of the following samples dissolves completely in water?
(a) NiBr2·3 H2O weighs272.5g/mole 51.9g / 272.5g/mole = .190mole of NiBr2·3 H2O all we need to care about is the .190mole of NiBr2 there 3ions per mol so .190 x 3 = .57moles of ions this ignores the 3H2O in the compound in this case its 1.71 moles of ions (b) FeCl3 has 4 ions permolecule 4.37 e22 molecules x 4 = 1.748 e23 ions 1.748 e23 ions / 6.022 e23 ions/ mole = .290 moles of ions
How many total moles of ions are released?
a) Na2HPO4 => 2 Na+ + HPO42- moles ions = ( 2 x 0.734) + 0.734=2.20 b) moles Ca(NO3)2 = 6.88 x 10^-3 g /164.085 g/mol=4.19 x 10^-5 Ca(NO3)2 = > Ca2+ + 2 NO3- moles ions = ( 2 x 4.19 x 10^-5) + 4.19 x 10^-5=1.26 x 10^-4 c) moles FeCl3 = 2.23 x 10^22/ 6.02 x 10^23=0.0370 FeCl3 => Fe3+ + 3 Cl- moles ions = 0.0370+ ( 3 x 0.0370)= 0.148
How many total moles of ions are released when each of the following samples dissolves completely in water?
1) write a balanced equation 2) calculate moles CuSO4.5H2O 3) convert moles CuSO4.5H2O to moles of ions *** 1 *** CuSO4.5H2O + H2O ------> Cu+2 + SO4-2 + 6 H2O *** 2 *** moles = mass /mw = 3.70 g x (1 mole / 249.7 g) = 0.0148 moles *** 3 *** from balanced equation, 1 mole CuSO4.5H2O ----> 2 moles ions 0.0148 mole CuSO4.5H2O x (2 moles ions / 1 mole CuSO4.5H2O) = 0.0296 moles ions
How many moles of ions are released when .32 moles of NH4Cl dissolves completely in water?
When it dissolves, .32 mol NH4Cl will split into two ions, NH4+ and Cl-. Because the amount of each of the two ions stays the same, there are .32 mol of each one. .32 mol NH4+ added to .32 mol Cl- gives you .64 mol ions total. .64 moles!
I hope this helps clear things up. Step 1: ammonium chloride dissociationNH4Cl -> NH4+ + Cl- Step 2: ammonium acts as a weak acidNH4+ + H2O <-> H3O+ + NH3Step 2 is an equilibrium with a rather small Ka, so there is only a small amount of NH3 and H3O+ made by this route. One could look up the Ka value, an exercise I leave to the reader.