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*Reaction is in image*
Questions based on reaction in image:
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The reducing agent is:
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- The following chemical reaction takes place in aqueous solution: 2FeCl3 (aq) + 3Na2S (aq) → Fe2S3 (s) + 6NaCl (aq)Write the net ionic equation for this reaction.Given: You weigh out exactly 0.200 g of Fe(NH4)2(SO4)2·6H2O and dissolve it in the 100.00 mL volumetric flask. You then pipette 2.00 mL of this solution into the 50.00 mL volumetric flask to prepare the stock standard tris-bipyridyl-iron(II) solution. a. Calculate the molar concentration of iron(II) in this solution in the 50.00 mL volumetric flask. (The MW of Fe(NH4)2(SO4)2·6H2O is 392.14 g/mol) (answer a given the information above)What is the redox equation for K2Cr2O7(aq) (ACIDIC) + Co(NO3)2(aq)?
- What happens in this reaction? 3Zn(s) + 2Cr3*(aq) 2Cr(s) + 3Zn2*(aq) O The Cr(s) is reduced O Zn(s) is oxidized O The Cr(s) is oxidized O The Cr+(aq) is oxidizedComplete and balance the following equation. I2(s)+OCl−(aq)→IO3−(aq)+Cl−(aq) (acidic solution)The overall reaction for the synthesis of iron(III) oxalate complex is 6 KOH(aq) + 3 H2C2O4(aq) + FeCl3(aq) → K3[Fe(C2O4)3(aq) + 3 KCl(aq) + 2 H2O(aq) A student weighed 4.118 g oxalic acid dihydrate (H2C2O4 2 H2O), how many grams of iron(III) chloride hexahydrate does he need to completely react with the oxalic acid?
- The molecular weight of sperm whale myoglobin is 17.8 kDa.17.8 kDa. The myoglobin content of sperm whale muscle is about 80 g · kg−1.80 g · kg−1. In contrast, the myoglobin content of some human muscles is about 8 g · kg−1.8 g · kg−1. Compare the amounts of O2O2 bound to myoglobin in human muscle and in sperm whale muscle. Assume that the myoglobin is saturated with O2,O2, and that the molecular weights of human and sperm whale myoglobin are the same. How much O2O2 is bound to myoglobin in human muscle? How much O2O2 is bound to myoglobin in whale muscle? The amount of oxygen dissolved in tissue water at 37°C37°C is about 3.5×10−5 M.3.5×10−5 M. What is the ratio of myoglobin‑bound oxygen to dissolved oxygen in the tissue water of sperm whale muscle?Consider the series of reactions to synthesize the alum (KAl(SO4 )2 · xH2O(s)) from the introduction. (a) Assuming an excess of the other reagents, from one mole of aluminum Al (s), how many moles of alum will be produced? (b) Assuming an excess of the other reagents, from one mole of potassium hydroxide KOH, how many moles of alum will be produced? (c) Assuming an excess of the other reagents, from one mole of sulfuric acid H2SO4 , how many moles of alum will be produced? (d) If you start the synthesis with 1.00 g of Al, 40.0 mL of 1.50 M KOH, and 20.0 mL of 9.00 M H2SO4 , which of the three will be the limiting reagent? (e) Assuming that the product is anhydrous (that there are no waters of hydration), calculate the theoretical yield of alum, in grams, based on the amounts of reagents in part (d). 3. Consider the nickel salt: (NH4 )2Ni(SO4 )2 ·y H2O (Ammonium Nickel Sulfate Hydrate), where y is the number of coordinated waters. (a) Assuming that the product is anhydrous (y = 0),…Nitrogen of ammonia can be determined by treatment of the sample with chloroplatinic acid; the product is slightly soluble ammonium chloroplatinate: H₂PtCl 6 + 2NH4+ → (NH4)2PtCl6 + 2H+ The precipitate decomposes on ignition, yielding metallic platinum and gaseous products: (NH4)2PtCl6 3 → Pt(s) + 2Cl₂(g) + 2NH3(g)↑ + 2HCl(g)↑ Calculate the percentage of ammonia in a sample if 0.2115 g of the impure sample gave rise to 0.4693 g of platinum. Given: M(NH3) = 17.0306 g/mol; M(Pt) = 195.08 g/mol. O O b. 14.25% O O d. 82.38% a. 56.52% c. 38.74%