In the reaction, KHS(aq) + HCI(aq) → KCI(aq) + H2S(g), which ions are the spectator ions O HS and CI O K* and Cl O H and HS O K* and H* O K* and HS
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- Which reactant is limiting. Solve using dimensional analysis. 3.00ml of a 0.100 M Pb(NO3)2 . 12.00ml of 0.100 M K2CrO4how to calculate moles of NaHCO3 and moles of CO2 from Mass of NaHCO3= 0.081g Volume of CO2=5ml show steps 10minsplease help me The aluminum in a 1.200g sample of impure ammonium aluminum sulfate was precipitatedwith aqueous ammonia as the hydrous Al2O3 · XH2O. The precipitate was filtered and ignitedat 100°C to give anhydrous Al2O3, which weighed 0.2001 g. Express the result of this analysisin terms of % Al.
- Consider the following reaction: 4K(s)+O2(g)→2K2O(s)4K(s)+O2(g)→2K2O(s)The molar mass of KK is 39.10 gmol−1gmol−1 and that of O2O2 is 32.00 gmol−1gmol−1. The limiting reactant is 1.5 g K, 0.38 g O2 I need help with the "explain your reasoning"Please find the CO2 amount and show the calculation: Air composition (apprx) = 80% N2 and 20% O2 (mol basis only)Average molecular weight of Air = (0.8 x 28 + 0.2 x 32) = 29 g/mol (apprx) Math Sample:2kg pure charcoal (12C) is to be burnt completely with air. Find the air, CO2 andN2 amount in kg’s.Solution:C + O2 = CO2 (you must use a balanced equation)Therefore from the mole ratio of the reaction we write,C : O2 : CO2 = 1:1:1and from air composition we gotO2 : N2 : Air = 1:4:5 Given, 2 kg C = 2000 g/ (12 g/mol) = 166.67 mole CTherefore, similar mole of O2 required. So equivalent Air supposed to be 5times than the mole amount of O2 and released N2 will be 4 times than therequired O2. Therefore, Air amount = 5 x 166.67 moles= 833.34 moles= (833.33x29/1000) kg= 24.17 kg air Similarly N2 released amount will be = (166.67 x 4 x 28/1000) kg = 18.67 kgFind CO2 amount by yourself! (Isn’t it 7.34 kg?)Assume that 1.00 mL (1.02 g) of crude product mixture is obtained from the reactionbefore the washing steps, and the distribution constant of the product mixture in brine isK = (Cmixture/Cbrine) = 50.0. Assuming that after mixing the product mixture volume isstill 1.00 mL (i.e., the volume lost is small), and the volume of the first brine wash layeris 2.00 mL. What mass (grams) of product mixture is lost (dissolves) in the first brinewash? (See Technique 13.2, p 53.) Show work.
- A sample containing NaCl, NaBr, & inert material weighs 1.000 g. Excess of AgNO3 gave a whiteprecipitate consisting of only AgCl and AgBr which weighs 0.5260 g. By heating the precipitate in a currentof Cl2 gas, the AgBr (187.78 g/mol) is converted to AgCl (143.32 g/mol) and the precipitate weighs 0.4260g. Find the % NaCl (58.44 g/mol) and % NaBr (102.909 mol) in the original sample.Complete the following reaction and balance the reaction. Select all of the following that are correct coefficients for the following reaction when balanced (If no coefficient is listed, assume a value of 1) H2SO4 + NaOH ⟶ Group of answer choices H2O 2 H2O NaSO4 Na2SO4 Na(SO4)2 NaOH 2 NaOH 3 NaOH H2SO4 2 H2SO4when the following equation is balance, the coefficient of H2 is K(s)+H2O(L)-->KOH(aq)+H2(g) answers options (a)1 (b)2 (c)3 (d)4 (e)5