Question

help with #30 and 32

2 solution is needed to react with 500 ml. of 0.100 M H3PO4 30) What volume of 0.30 M Cu(OH)2 solution i solution? JOHT),(aq) + 2H3PO4(aq) → Cu,(PO4 + 6H2017) A) 33.3 mL B) 60.0 mL C) 83.3 mL D) 167 mL |E) 250 ml olubility of each solute-solvent combination by writing the word soluble or insoluble Short Answer Questions: 31) Predict the solubility of each solute the blank space. A) NO2 (a polar gas) in H2O Soluble B) KCI in CCl4 (a liquid) insoluble C) BaSO4 in H2O insoluble D) LiF in H20 Lubre E) C3H18 (a nonpolar liquid) in CC Soluble - Saline I. V. solution is 0.92% (m/v) NaCl solution. A patient is to receive 0.50 L of saline solu What mass of NaCl is the patient to receive?

Answer 1

90) Kos an cles e , Cork u , 2 cles 2 = I nede CULOH will react with Inde H₂ Poa takes place equivalence between pour the may As neutralisation reaction and and base, then the be given by : - na Mava = n Me VB aud & base where - an = Ma M = Va, Vb a audity Molarity volume and of of basicity of acid & base acid & base Given =2 i H₂PO2 will release 3 Htions so na = 3 l, Cu COH) will release 2 oH ions, oo Vaa 500 me Vb c ? Ma = 0.100. 0.30 o nama va = no Mbu & 3x 0.100 x 500 = 2 x 0.30 x VG V = 34001 8500 & 150 = 250 mL 2x0.3 0.6 o volume of Cu (OH) needed is 250 ml

Saline § 0.92% (m/v) Nal solution. let the volume of solution be 100 nu, then the concentration of Nau is 0.92g. 0 0.92g Of Nall is present in 10 me solution on 100 ml solution has 0.92g Nal. - O.IL solution was 0.928 Nad (co (L=1000me 8 IL solution has 0.92 g Naul 0 0.50L solution has 0.92 x 0.50 g Nall Oil was 4.6 g Nail - 0.50 L solution Thus, 4.6g of wall is patient is to reueue. the amount the
the solutions of questions 30 and 32 are provided. Hope this helps...thank you.