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K2B with 1s typically refers to a specific notation or abbreviation in a particular context, but it's not a widely recognized term. It could represent a variety of things depending on the field, such as a mathematical expression, a programming code, or a shorthand in a specific industry. More context is needed to provide a precise interpretation.
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What is the rate constant of a reaction if rate 1 x 10-2 (MolL)s A is 2 M B is 3 M m 2 and n 1?
To find the rate constant (k) of the reaction, we can use the rate equation: Rate = k[A]^m[B]^n. Given that the rate is (1 \times 10^{-2} , \text{(mol L)}^{-1} , \text{s}^{-1}), [A] = 2 M, [B] = 3 M, m = 2, and n = 1, we substitute these values into the equation: [ 1 \times 10^{-2} = k \cdot (2^2) \cdot (3^1) ] This simplifies to: [ 1 \times 10^{-2} = k \cdot 4 \cdot 3 \implies 1 \times 10^{-2} = k \cdot 12 ] Solving for k gives: [ k = \frac{1 \times 10^{-2}}{12} \approx 8.33 \times 10^{-4} , \text{(mol L)}^{-1} , \text{s}^{-1} ]
What is the rate constant of a reaction if rate 1 10 2 mol L s A is 2 M B is 3 M m 2 and n 1?
The rate law for this reaction is rate = k[A]^m[B]^n. From the given information, substituting the values for rate, [A], [B], and the exponents m and n, you can solve for the rate constant k. In this case, k = rate / ([A]^m[B]^n), so k = 2 / (10^2 * 3^1).
What is the rate of a reaction if the value of k is 0.1 A is 1 M and B is 2 M?
0.4 (mol/L)/s
What is the rate constant of a reaction if rate 0.2 mol L s A and B are each 3 M m 1 and n 2?
To find the rate constant ( k ) of the reaction, we can use the rate law equation: [ \text{rate} = k [A]^m [B]^n ] Given that the rate is 0.2 mol L⁻¹ s⁻¹, with concentrations ( [A] = 3 , \text{M} ), ( m = 1 ), ( [B] = 3 , \text{M} ), and ( n = 2 ), we can substitute these values into the equation: [ 0.2 = k (3)^1 (3)^2 ] This simplifies to: [ 0.2 = k (3)(9) = 27k ] Solving for ( k ), we find: [ k = \frac{0.2}{27} \approx 0.00741 , \text{L}^2 \text{mol}^{-2} \text{s}^{-1} ]
What is the rate of a reaction that follows the rate law rate kAmBn if k 02 A and B are each 3 M m 2 and n3?
To determine the rate of the reaction that follows the rate law rate = k[A]^m[B]^n, where k = 3 M^(-2) s^(-1), [A] = 2 M, and [B] = 3 M, we first need to substitute these values into the rate law. Given that m = 2 and n = 3, the rate can be calculated as follows: Rate = k[A]^m[B]^n = 3 M^(-2) s^(-1) * (2 M)^2 * (3 M)^3 = 3 * 4 * 27 = 324 M/s. Thus, the rate of the reaction is 324 M/s.
What is K-2 made of?
Rock, thicko
Which number is between point S and point T on the number line?
k*S + (1-k)*T where k is any number between 0 and 1.
What is the rate of a reaction if the value of k is 0.1 A is 1 M and B is 2 M Rate kA2B2?
To calculate the rate of the reaction using the rate law ( \text{Rate} = k[A]^m[B]^n ), where ( k = 0.1 ), ( [A] = 1 , \text{M} ), and ( [B] = 2 , \text{M} ), we first need to determine the values of ( m ) and ( n ). Assuming the reaction is second-order in A and second-order in B (i.e., ( m = 2 ) and ( n = 2 )), the rate would be calculated as: [ \text{Rate} = 0.1 \times (1)^2 \times (2)^2 = 0.1 \times 1 \times 4 = 0.4 , \text{M/s}. ] Thus, the rate of the reaction is 0.4 M/s.
Find equilibrium constant of the reaction?
eqb constant k For a general EQN A+B=S+T the equilibrium constant can be defined by[1] k={S}{T}/{A}{B} {S} = MOLAR CONC. OF S{T} = MOLAR CONC. OF T{A} = MOLAR CONC. OF A{B} = MOLAR CONC. OF B
What is the rate constant of a reaction if rate 1 10-2 molLs A is 2 M B is 3 M m 2 and n 1?
The rate law for the reaction is given by Rate = k[A]^m[B]^n. Plugging in the given concentrations and rate into the rate law, we have 10^-2 = k*(2)^2*(3)^1. Solving for k gives k = 10^-2 / (43) = 10^-2 / 12 = 8.33 x 10^-4 L/mols.
What is the smallest integer k such that r3 equals s2 equals k where are and s are positive integers?
If 'r' and 's' can be the same integer, then (1)3 = (1)2 , and k = 1 .If 'r' and 's' must be different integers, then (4)3 = (8)2 , and k = 64 .
What rate of a reaction that follows the rate law rate kAmBn where k 1.5 A 1 M B 3 M m 2 n 1?
To determine the rate of the reaction using the rate law ( \text{rate} = k[A]^m[B]^n ), we can substitute the values given. With ( k = 1.5 , \text{M}^{-2}\text{s}^{-1} ), ( [A] = 1 , \text{M} ), ( [B] = 3 , \text{M} ), ( m = 2 ), and ( n = 1 ), the rate can be calculated as follows: [ \text{rate} = 1.5 \times (1)^2 \times (3)^1 = 1.5 \times 1 \times 3 = 4.5 , \text{M/s} ] Thus, the rate of the reaction is ( 4.5 , \text{M/s} ).
