2 N2O5g→4 NO2g+O2g
2PbSs+3O2g→2PbOs+2SO2g
2 H2O→2 H2+O2
4. Determine the theoretical yield of P2O5, when 3.07 g of P reacts with 6.09 g of oxygen in the following chemical equation
4 P+5O2→2P2O5
4 P+5O2→2P2O5
Instructions: Answer the following questions using the following information: ∆Hfus=6.02 kJ/mol; ∆Hvap= 40.7 kJ/mol; specific heat of water is 4.184 J/g∙˚C; specific heat of ice is 2.06 J/g∙˚C; specific heat of water vapor is 2.03 J/g∙˚C.
Acid/Base: The Water Constant and pH
Instructions: Determine if each solution is acidic, basic, or neutral.
1.[H3O+] = 1 x 10-10 M; [OH-] = 1 x 10-4 M
Instructions: Calculate [OH-] given [H3O+] in each aqueous solution and classify the solution as acidic or basic.
5. [H3O+] = 2.6 x 10-3 M
6. [H3O+] = 2.6 x 10-8 M
Instructions: Calculate the [H3O+] of solutions a and b; calculate the [OH-] solutions c and d.
9. pH = 2.76
10. pH = 3.65
Equilibrium and Le Chateleir's Principal
Instructions: Write the equilibrium expression for each chemical equation.
Oxidation-Reduction Reactions
Instructions: Assign an oxidation state to each element, ion, or molecule.
Instructions: Assign an oxidation state to each atom in each compound.
Instructions: Assign an oxidation state to each atom in each polyatomic ion.
Instructions: Identify the oxidation state of P in each ion.
For the following reaction, calculate how many moles of NO2 forms when 0.356 moles of the...
For the following reaction, calculate how many moles of NO2, forms when 0.356 moles of the reactant completely reacts. 2N2O5(g) yields 4NO2(g)+ O2(g)
For the following reaction, calculate how many moles of each product are formed when 0.356 moles of PbS completely react. Assume there is an excess of oxygen.
For the reaction shown, calculate how many moles of each product form when the given amount of each reactant com- pletely reacts. Assume there is more than enough of the other reactant. C3H2(g) + 5O2(g) — 3CO2(g) + 4H2O(8) (a) 4.6 mol C3H (b) 4.6 mol O2 (c) 0.0558 mol C3H (d) 0.0558 mol O2
For the reaction shown, calculate how many moles of NO2 form when each amount of reactant completely reacts. 2N2O5(g)→4NO2(g)+O2(g) A. 1.6 mol N2O5 B. 6.0 mol N2O5 C. 5.27×103 mol N2O5 D. 1.010×10−3 mol N2O5
For the reaction shown, calculate how many moles of NO2 form when each amount of reactant completely reacts. 2N2O5(g)→4NO2(g)+O2(g) Part A 1.4 mol N2O5 Part B 5.6 mol N2O5 Part C 12.5 g N2O5 Part D 1.75 kg N2O5
Instructions: Determine if each solution is acidic, basic, or neutral. [H3O+] = 1 x 10-10 M; [OH-] = 1 x 10-4 M [H3O+] = 1 x 10-7 M; [OH-] = 1 x 10-7 M [H3O+] = 1 x 10-1 M; [OH-] = 1 x 10-13 M [H3O+] = 1 x 10-13 M; [OH-] = 1 x 10-1 M Instructions: Calculate [OH-] given [H3O+] in each aqueous solution and classify the solution as acidic or basic. [H3O+] = 2.6 x 10-3...
Please help For the reaction shown, calculate how many moles of each product form when the given amount of each reactant completely reacts. Assume that there is more than enough of the other reactant. C3H8(g)+5O2(g)→3CO2(g)+4H2O(g) c)0.0588 molC3H8 Express your answer using three significant figures. v=_____molCO2 d)0.0588 molC3H8 Express your answer using three significant figures. v=_____molH2O e)4.2 molO2 Express your answer using two significant figures. v=______molCO2 f)4.2 molO2 Express your answer using two significant figures. v=______molH2O g)0.0588 molO2 Express your answer...
2.31 kg k g N2O5 Calculate how many moles of NO2 N O 2 form when each quantity of reactant completely reacts via the following reaction: 2N2O5(g)→4NO2(g)+O2(g)
For the reaction shown, calculate how many moles of NO, form when each amount of reactant completely reacts. 2 N,Os(9) + 4NO2(g) + O2(9) Part C 4.75 x 10' mol N205 Express your answer using three significant figures. AXO O ? % 1.02 mol Submit Previous Answers Request Answer * Incorrect; Try Again; 5 attempts remaining Part D 1.008 x 10-mol N, O, Express your answer using four significant figures. V ASD - O ? mol Submit Request Answer
#11/12 8. [H3O+] = 4.3 x 10-3M Instructions: Calculate the [H3O+] of solutions a and b; calculate the [OH-] solutions c and d. 9. pH = 2.76 - PH=-lag (H3ot) 10-PP- 69 H 10-2.76 10. pH = 3.65 11. POH = 3.65 12. pOH = 6.87 Equilibrium and Le Chateleir's Principal Instructions: Write the equilibrium expression for each chemical equation. 1. 2 H250 = 2H26 +526) = [ta?? (s. 1ST 2. NH,HS(s) = NH3(g) + H2560)