SIMULATION lonic Compound Formulas Cations Anions O NH4 O Lit O cF O Br O OH O s?- O co, O so,2- PO,- O NO3 O C104 Na Name: ammonium fluoride O Ca²+ 2+ Formula: NHF O Ag O Fe2+ O Fe+ O AlP+ O Pb2- Solubility: soluble in water Choose Na as the cation of the salt and chloride ion as the anion. This is sodium chloride. How many CF ions are there for each Na ion in the formula? (1 of 9) Recheck 2nd attempt Incorrect Incorrect FLV rch Cop

OWLV2 | Online teachin X According To The Following+ My Home gagenow.com/ilm/takeAssignment/takeCovalentActivity.do?locator=assignment-take&takeAsignmentSessionLocator=assignment-take [Review Topics] [References] E 2req Use the References to access important values if needed for this question. 3 2reg For the following reaction, 0.397 grams of hydrogen gas are allowed to react with 54.1 grams of iodine. S 2req hydrogen(g) + iodine(s) – → hydrogen iodide(g) What is the maximum mass of hydrogen iodide that can be formed? grams What is the FORMULA for the limiting reagent? Es 2req is 2req What mass of the excess reagent remains after the reaction is complete? grams ts 2req ts 2req Submit Answer Retry Entire Group 9 more group attempts remaining ts 2req ots 2req ots 2reg ots 2req ots 2req pts 2req Previor Show Hint a $ hp Ins prt sc f9 12 delete 16

4) lonic vs Covalent: If ionic, give ionic formula (showing charges) & if covalent, give structural formula (bonds), then answer if it were a good electrolyte. g) C3F4. vs MgCl2

Br Li Et₂0 Cul I CHCI3 (CH3)3CO-K

ed Complete the following for the compound ammonium phosphate. formula = (NH4)3PO4 atom nitrogen hydrogen phosphorus oxygen Submit Answer number in formula important values if needed for this question Retry Entire Group 9 more group attempts remaining

please answer quickly

Only typed solution

Match the left 'cation' column with the right 'anion' columnto create a formula of a soluble ionic compound. Use each anion and cation only once. Na+ S2- Sr2+ CO32- Co2+ NO-3 Pb2+ SO42-

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(c) 4.27 M NaH2PO4 lons: O PO43- H2PO4 PO4 3+ Na O Na* NaH* [cation*] = i M [anion] = M