I. Balance the following equations: 1. Mg + Mgo - *o 2. Zn + AGCI ZnCl, + Ag 3. So: NAOH + H;SO. Na,SO. 4. 5. co H,0 кон . 6. K- 7. H:O2 H20 + 8. FezO3 + co - Fe + CO2
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- 7. Consider a1Msolution ofNa3AsO4. Write the charge and mass balance equations for this system. (please type answer not write by hend)Which of the following is a spontaneous reaction.? a. Rxn with ΔH =- 10Kj/mol ΔS= -5J/mol T= 300K b. NaCl +H20 -> NaOH + HCl 25C c. H20(l) -> H2O(s) Temp: 25C d. Dissolution of 100g of solid sugar in 100 mL ice tea. Consider following reaction: HgO (s) -> Hg(l) + ½ O2 (g) Delta H = +90.7 kj/mol. What quantity of heat in kj/mol is required to produce one mole HgO? Write your answer without units. Given the following data 2ClF(g) + O2(g) --> Cl2O(g) + F2O (g) Delta H= 167.4 kJ I 2ClF3(g) + 2O2(g) --> Cl2O(g) + 3F2 O (g) Delta H= 341.4 kJ II 2 F2(g) + O2(g) ---> 2F2O (g) Delta H= -43.4 kJ III Calculate the delta H in kJ for below reaction: ClF(g) + F2(g) ---> ClF3(g)Can you work our KSP for me? thanks Liquid Amount - How many grams of each of the following substances will dissolve in 4.70×102 mL of cold water? Substance - The solubility is 0.123 g/100 mL at 20 oC. ( Ce(IO3)4 )
- What is ΔHsys for a reaction at 16.9 °C with ΔSsurr = -159 J mol-1 K-1 ? Express your answer in kJ mol-1 to at least two significant figures. (Please type answer no write by hend)In the synthesis of hydrocarbons, the carbon source is carbon dioxide. Although the CO2 concentra?on in the atmosphere raises at a drama?c speed, point sources are probably the easier sources for a PtX process. Iden?fy 3 possible point sources, explain why CO2 is formed and what challenges each of the three CO2 streams presentsFor the reduction 2FeCl3 + SnCl2 =====➔ 2 FeCl2 + SnCl4 in aqueous solution the following data were obtained at 25oC t(min) 1 3 7 11 40 Y 0.01434 0.02664 0.03612 0.04102 0.05058 Where y is the amount of FeCl3 reacted in moles per liter. The initial concentrations of SnCl3 and FeCl3 were respectively, 0.03125, 0.0625 moles/L. a.)Show that the reaction is third order (derive the rate law), and b.) calculate the average specific rate constant.
- Potassium dichromate has several industrial applications. To determine the purity of the salt that will be used in different industrial processes, a sample mass equal to 2.660 g was dissolved and quantitatively transferred to a 500.00 mL flask. An aliquot of 25.00 mL of this solution was treated with excess KI and the released iodine was titrated with 0.1000 mol L-1 sodium thiosulfate, spending 27.00 mL. Calculate the purity of the analyzed salt. Data:K = 39.10 O = 16.00 Cr = 52.00 I = 126.9 S = 32.07Show all steps leading to the final answer po. Here’s a pdf file in accordance with the topic po: https://drive.google.com/file/d/1_FnDtXCrFKSol3RNWIG_9tNQ7IxgxD6t/view?usp=drivesdkCalculate Trial I, Trial II, Trial III Vol.Ag(NO3)Added.
- If all other variables were kept constant, determine theeffect that the following errors would have on the calculatedpercent yield of the product. Would the yield be expected toincrease, decrease, or would there be no effect? Explainyour reasoning.– The product was insufficiently dried before weighing.– Some of the product was lost during the transfer fromthe Buchner funnel to the evaporating dish.– 7.5 mL of FeCl3 was added instead of 3.0 mL asoutlined in the procedure.– 4.587g of K2C2O4H2O was used instead of exactly4.000g .– The recrystallization step was skipped and theexperiment went straight to vacuum filtration.Lactated Ringer’s/5% Dextrose solution contains: 6 g/L of Sodium Chloride (NaCl MW 58.5) 3.1 g/L of Sodium Lactate (C3H5O3Na MW 112) 0.3g/L of Potassium Chloride (KCl MW 74.5) 0.2g/L of Calcium Chloride (CaCl2•2H2O MW 147) 50g/L of Dextrose (C6H12O6 MW 180) You receive an order to increase the Potassium ion concentration to 0.045 mEq/mL. How many mL of 14.9% Potassium chloride injection should be added to 1L of the above solution to increase the potassium ion concentration to 0.045 mEq/mL ____________________mL 14.9% KCl injectionWhich of the following are equivalent to 2,500 ppm Cu2+? (There may be more than one answer) MW: Cu (63.55) a) 2.5 ppb Cu2+ b) 2,500,000 ppb Cu c) 2.5 ppt Cu2+ d) 39.34 mM e) 0.03934 M f) 0.07868 N (in precipitation reaction) g) 0.07868 N (in redox into Cu+)