1.how many ml of 0.401 m hi are needed to dissolve 5.97 g of baco3?
2hi(aq) + baco3(s) bai2(aq) + h2o(l) + co2(g)
2.calculate the number of milliliters of 0.757 m naoh required to precipitate all of the co2+ ions in 167 ml of 0.548 m coso4 solution as co(oh)2. the equation for the reaction is:
coso4(aq) + 2naoh(aq) co(oh)2(s) + na2so4(aq)

The answer to your question is:

1.- volume = 0.151 l or 151 ml

2.- 0.241 l  or 241 ml of NaOH

Explanation:

1.-

Data

V = ? HI = 0.401 M

BaCO3 = 5.97 g

                     2HI(aq)    +    BaCO3(s)   ⇒   BaI2(aq) + H2O(l) + CO2(g)

MW BaCO3 = 137 + 12 + 48 = 197 g

                     197 g of BaCO3 1 mol

                     5.97 g                   x

                     x = (5.97 x 1) /197

                    x = 0.03 mol of BaCO3

                    2 moles of HI  1 mol of BaCO3

                    x                      0.03 mol of BaCO3

                    x = (0.03 x 2) / 1

                   x = 0.060 mol of HI

Molarity = moles / volume

volume = moles / molarity

volume = 0.060 / 0.401

volume = 0.151 l or 151 ml

2.-

V = ?    NaoH 0.757 M

Co⁺² Volume = 167 ml   0.548 M

             CoSO4(aq) + 2NaOH(aq)   ⇒   Co(OH)2(s) + Na2SO4(aq)

Moles of Co = Molarity x  volume

Moles of Co = 0.548 x 0.167

Moles of Co = 0.092

                                 1 mol of CoSO4 2 moles of NaOH

                                0.092 moles         x

                                x = (0.092 x 2) /1

                               x = 0.183 moles of NaOH

Volume of NaOH = moles / molarity

                             = 0.183 / 0.757

                            = 0.241 l  or 241 ml of NaOH

answer

fusion: the energy source of stars. the energy released from the collapse of the gas into a protostar causes the center of the protostar to become extremely hot. when the core is hot enough, nuclear fusion commences. fusion is the process where two hydrogen atoms combine to form a helium atom, releasing energy.

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