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please answer both if you can Using the values of k calculated in previous question, calculate the rate when [A]=0.1000mol/,[B]=0.200mo/L, and [C]=0.0250 mollL a. 0.0200mol

please answer both if you can
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Using the values of k calculated in previous question, calculate the rate when [A]=0.1000mol/,[B]=0.200mo/L, and [C]=0.0250 mollL a. 0.0200mol b. 0.200mol c. 5.00mol/L d. 0.0250mol e. 0.00500mol. The proposed mechanism for a reaction is (1) A(g)+B(g)=X(g) [tast] (2) X(g)+C(g)Y(g) [slow] (3) Y(g)D(g) [fasi] The experimental rate law was found to be Rate =K(AB)[C]. is the rate law from the mechanism consistent with the experimental rate law? wenet one: True Fatse For the reection. A(g)+B(g)+C(g)D(g) the following data were obtained at constant temperature. What is the order with respect to each reactant? a. A]=1st,[t]=1st,[C]=1 st b. (A)=1 st, [B=2nd,(C=1 st c. A=2 nd, {E=2nd,[C]=0 d. A=1st,B=2nd,C=0 e. A=1 st. [B]=1st,C=0 using the values of k calculated in previous question, calculate the rate when [A]=0.1000molL,[B]=0.200mol/L, and [C]=0.0250 moll a. 0.0200mol. b. 0.200moul c. 5.00mol/L d. 0.0250mol. e. 0.00500molL Using the values of k calculated in previous question, calculate the rate when [A]=0.1000mol/,[B]=0.200mo/L, and [C]=0.0250 mollL a. 0.0200mol b. 0.200mol c. 5.00mol/L d. 0.0250mol e. 0.00500mol. The proposed mechanism for a reaction is (1) A(g)+B(g)=X(g) [tast] (2) X(g)+C(g)Y(g) [slow] (3) Y(g)D(g) [fasi] The experimental rate law was found to be Rate =K(AB)[C]. is the rate law from the mechanism consistent with the experimental rate law? wenet one: True Fatse For the reection. A(g)+B(g)+C(g)D(g) the following data were obtained at constant temperature. What is the order with respect to each reactant? a. A]=1st,[t]=1st,[C]=1 st b. (A)=1 st, [B=2nd,(C=1 st c. A=2 nd, {E=2nd,[C]=0 d. A=1st,B=2nd,C=0 e. A=1 st. [B]=1st,C=0 using the values of k calculated in previous question, calculate the rate when [A]=0.1000molL,[B]=0.200mol/L, and [C]=0.0250 moll a. 0.0200mol. b. 0.200moul c. 5.00mol/L d. 0.0250mol. e. 0.00500molL

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