15. How aralisi disebut....................................................... E) 5.87 g in an acidic solution? of NO_(3)^-(aq) and N((s) when the following redox equation is balanced A) 2,6 B) 3,6 NO_(3)^(1)/(3)(aq)+Ni(s)arrow NO(g)+Ni^2+(aq) C)3,4 D)2,3 17. Give the net ionic cquation for the reaction (if any) that Na_(2)CO_(3) and HCl 1.2 2Na^2(aq)+CO_(3)^2(aq)+2H^+(aq)+2Cl(aq)arrow H_(2)CO_(3))_(2)+CO_(2)(aq)+2Na^+(aq)+2Cr(aq) 2Na^{2 B) C) D) E)No reaction occurs. 18. The titration of 1000 mL of HCl solution of unknown concentration requires 12.54 mL of a0.100M NaOH solution to reach the equivalenc A) 0.278 M B) 0.125 M HCl(aq)+NaOH(aq)arrow H2O(l)+NaCl(aq) 5M ()0.354M D)0.143M 19. How many milliliters of 0.260MNa_(2)S are needed to react with 35.00 mL of 0.315MAgNO_(3) according to the following unbalanced precipitation reaction? E) 0.0125 M A) 21.2 mL Na_(2)S(aq)+AgNO_(3)(aq)arrow NaNO_(3)(aq)+Ag_(2)S(s) B)57.8mL D) 84.8 mL E) 35.6 mL 20. Determine the theoretical yield of H_(2)S (in moles) if 32 mol Al_(2)S_(3) and 32molH_(2)O are reacted according to the following balanced reaction. A possibly useful molar mass is Al_(2)S_(3)=150.17g/mol Al_(2)S_(3)(s)+6H_(2)O(l)arrow 2Al(OH)_(3)(s)+3H_(2)S(g) A) 16molH_(2)S B) 96molH_(2)S C) 48molH_(2)S D) 32molH_(2)S E) 64molH_(2)S 21. How many grams of NaCl are required to make 250.00 mL of 3.000 M solution? (Na:22.99g/mol,Cl: 35.45g/mol) A)58.40 g B) 43.83 g C) 175.3 g D) 14.60 g E) 1288 22. What is the concentration of a solution prepared by diluting 20.0 mL of a 0.200MLiCl to 250.0 mL A) 0.0160 M B) 0.160 M C) 2.50 M D) 0.00800 M E) 0.0320 m 23. What is the oxidation. "tate of H in H_(2)(g) in th; following belanced reaction? Mg(s)+2HBr(aq)arrow MgBr_(2)(aq)+H_(2)(g) D) A) +1 B) +2 C) 0 -1 E) -2 4. What is the oxidizing reaction? 2Al^3(aq)+2Fe(s)arrow 2Al(s)+3Fe^2+(aq) C) Al D) Fe^2+ A) Al^3+ B) Fe E) none 5. Determine the limiting reactant (LR) and the mass (ing) of nitrogen that can be formed from 50.0g below. Some possibly useful atomic a. masyarakat dangkat dan A E) No LR,45.0gN_(2) formed

15. How many moles of $NO_{3}^{-}$ and $N$ are involved in the redox reaction?<br />A) 2,6<br />B) 3,6<br />C) 3,4<br />D) 2,3<br />E) 5,87 g<br /><br />To balance the redox equation, we need to determine the number of moles of $NO_{3}^{-}$ and $N$ involved in the reaction. <br /><br />From the given equation, we can see that the oxidation state of nitrogen changes from 0 in $N$ to +5 in $NO_{3}^{-}$. Therefore, the number of moles of $N$ involved in the reaction is 5.<br /><br />Similarly, the oxidation state of nitrogen changes from +5 in $NO_{3}^{-}$ to 0 in $NO$. Therefore, the number of moles of $NO_{3}^{-}$ involved in the reaction is 3.<br /><br />Therefore, the correct answer is B) 3,6.<br /><br />17. Give the net ionic equation for the reaction between $Na_{2}CO_{3}$ and $HCl$.<br />A) $2Na^{+}(aq)+CO_{3}^{2-}(aq)+2H^{+}(aq)+2Cl^{-}(aq)\rightarrow H_{2}CO_{3}(aq)+CO_{2}(g)+2Na^{+}(aq)+2Cl^{-}(aq)$<br />B)<br />C)<br />D)<br />E) No reaction occurs.<br /><br />The net ionic equation for the reaction between $Na_{2}CO_{3}$ and $HCl$ is:<br /><br />$CO_{3}^{2-}(aq)+2H^{+}(aq)\rightarrow H_{2}CO_{3}(aq)+CO_{2}(g)$<br /><br />This equation represents the formation of carbonic acid and carbon dioxide gas from the reaction of carbonate ions and hydrogen ions.<br /><br />18. The titration of 10.00 mL of $HCl$ solution of unknown concentration requires 12.54 mL of $0.100M$ $NaOH$ solution to reach the equivalence point. What is the concentration of the $HCl$ solution?<br />A) 0.278 M<br />B) 0.125 M<br />C) 0.354 M<br />D) 0.143 M<br /><br />To find the concentration of the $HCl$ solution, we can use the formula:<br /><br />$M_{1}V_{1}=M_{2}V_{2}$<br /><br />where $M_{1}$ and $V_{1}$ are the molarity and volume of the $HCl$ solution, and $M_{2}$ and $V_{2}$ are the molarity and volume of the $NaOH$ solution.<br /><br />Substituting the given values, we have:<br /><br />$M_{1}(10.00\,mL)=0.100M(12.54\,mL)$<br /><br />Solving for $M_{1}$, we get:<br /><br />$M_{1}=0.125M$<br /><br />Therefore, the concentration of the $HCl$ solution is B) 0.125 M.<br /><br />19. How many milliliters of $0.260MNa_{2}S$ are needed to react with 35.00 mL of $0.315MAgNO_{3}$ according to the following unbalanced precipitation reaction?<br />$Na_{2}S(aq)+AgNO_{3}(aq)\rightarrow NaNO_{3}(aq)+Ag_{2}S(s)$<br />A) 21.2 mL<br />B) 57.8 mL<br />C) 84.8 mL<br />D) 35.6 mL<br /><br />To find the volume of $Na_{2}S$ needed, we can use the formula:<br /><br />$M_{1}V_{1}=M_{2}V_{2}$<br /><br />where $M_{1}$ and $V_{1}$ are the molarity and volume of the $Na_{2}S$ solution, and $M_{2}$ and $V_{2}$ are the molarity and volume of the $AgNO_{3}$ solution.<br /><br />Substituting the given values, we have:<br /><br />$0.260M(V_{1})=0.315M(35.00\,mL)$<br /><br />Solving for $V_{1}$, we get:<br /><br />$V_{1}=35.6\,mL$<br /><br />Therefore, the correct answer is D) 35.6 mL.<br /><br />20. Determine the theoretical yield of $H_{2}S$ in moles if 32 mol $Al_{2}S_{3}$ and 32 mol $H_{2}O$ are reacted according to the following balanced reaction:<br />$Al_{2}S_{3}(s)+6H_{2}O(l)\rightarrow 2Al(OH)_{3}(s)+3H_{2}S(g)$<br />