They are experimentally determined exponents.
It is simply a straight line equation in the form of: y = mx+c whereas m is the slope and c is the y intercept
It is the graphical representation of a linear relationship between two variables. In its most general form, the relationship is of the form a1x1 + a2x2 + ... + anxn + c = 0 where x1, x2 etc are variables, and the as are constants. This equation would be a straight line in n-dimensional space. In 2-dimensional space, and renaming the variables, you get ax + by + c = 0 where x and y are the variables and a, b and c are constants. This equation can be expressed in the form y = mx + d where m and d are constants. m is called the slope or gradient. For every increase of 1 unit in x, you will get an increase of m units in y. Also, d is called the intercept and it represents the value of y when x is 0.
The percentage of red M&M's in a bag of M&M's can vary depending on the specific mix and packaging. However, historically, red M&M's usually make up about 13% of the total M&M's in a standard bag. It's important to note that this ratio can differ based on the product line or seasonal varieties offered by the brand.
there are 1,000 m and m's
No, M&M's are better because they contain chocolate.
5.4 (apex)
The rate of the reaction can be calculated using the rate law equation rate = k[A]^m[B]^n. Plugging in the given values k = 0.2, m = 1, n = 2, [A] = 3 M, and [B] = 3 M into the equation gives rate = 0.2 * (3)^1 * (3)^2 = 16.2 M/s.
In the context of chemistry, "k Rate kAmBn" refers to the rate constant (k) of a reaction involving reactants A and B, where "m" and "n" represent the stoichiometric coefficients of these reactants in the rate law. The rate of the reaction can be expressed as proportional to the concentrations of A and B raised to their respective powers, leading to the equation: rate = k [A]^m [B]^n. This relationship helps in understanding how changes in concentration affect the speed of the reaction.
4.5 (mol/L)/s
The rate of the reaction can be calculated using the rate law rate = k[A]^m[B]^n. Plugging in the given values: rate = 0.02*(3)^3*(3)^3 = 0.022727 = 14.58 M/s.
Rate = k[A]m[B]n
They are experimentally determined exponents
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.
r=[A]m[B]n APPLEX
To write a rate law for a chemical reaction, one must determine the order of the reaction with respect to each reactant by conducting experiments and analyzing the rate of reaction at different concentrations. The rate law is then expressed as rate kAmBn, where k is the rate constant, A and B are the concentrations of the reactants, and m and n are the orders of the reaction with respect to each reactant.
The equation is called the rate law equation. For the reaction aA+bB =>cC+dD the rate law would be rate = k[A]^m[B]^n where k is the rate constant and m and n are the "order" with respect to each reactant. m and n must be determined experimentally and may or may not be the same as the coefficients a and b.
To find the rate of the reaction, we can use the given rate law: ( \text{rate} = k[A]^m[B]^n ). Substituting the values, we have ( k = 0.02 , \text{M}^{-1}\text{s}^{-1} ), ( [A] = 3 , \text{M} ), and ( [B] = 3 , \text{M} ) with ( m = 1 ) and ( n = 2 ). Thus, the rate is calculated as: [ \text{rate} = 0.02 \times (3)^1 \times (3)^2 = 0.02 \times 3 \times 9 = 0.54 , \text{M/s} ] Therefore, the rate of the reaction is 0.54 M/s.