reactants are favored over products in the reaction
1 Equivalent = 1 Mole / (number of moles per Eq.) Eq. mass = Molar mass / (number of moles per Eq.) N = Eq. concentration (eq/L) = Molar conc.(mol/L) / (number of moles per Eq.) = M / (mol/eq)
EQ Guinea = Equatorial Guinea
To calculate the equilibrium constant ( K_{eq} ) for the reaction ( N_2 + 3H_2 \rightleftharpoons 2NH_3 ), we use the formula: [ K_{eq} = \frac{[NH_3]^2}{[N_2][H_2]^3} ] Plugging in the equilibrium concentrations: ( [NH_3] = 3 , M ), ( [N_2] = 1 , M ), and ( [H_2] = 2 , M ): [ K_{eq} = \frac{(3)^2}{(1)(2)^3} = \frac{9}{8} = 1.125 ] Thus, ( K_{eq} ) for the reaction is 1.125.
"EQ" on a telescope lens typically refers to the type of mount the telescope sits on. EQ stands for equatorial mount, which is a type of mount that can be aligned with the Earth's axis to track celestial objects as they move across the sky.
I'm taking an awesome chemistry final tomorrow. So, I'm not a massive failure at this: k=mol/liters Kc can only determine by experiment , not by evaluations of equations. so when writting the eq of Kc= [] products /[reactants], do not use units for [], as Kc has no units. Kc, only affected by temperature...
I suspect it could be " k eq 1" , or "k =1".
reactants are favored over products in the reaction
What does this mean? The rightmost digit of {eq}n^j{/eq} is the remainder when {eq}n^j{/eq} is divided by {eq}10{/eq}. yep totaly nor random :))
The CV value is the flow rate required to generate 1 psid of pressure loss through the valve. Since pressure drop follows a basic square law the relationship between Cv, flow and pressure loss is as follows: DP = k x Flow^2 eq 1 Where k is a constant that represents the flow shape in the wide open condition. Since Cv is the flow rate at 1 psi of pressure loss then it follows that DP = k x Cv^2 = 1 eq 2 solving for k from eq 2 yields k = 1/Cv^2 eq 3 substituting eq 3 into eq 1 yields DP = (Flow/Cv)^2 Now you have an equation that will tell you the pressure and flow relationship for that particular valve with a particular fully open Cv value. In short, the higher the Cv value the more flow the valve will allow for the same pressure loss or the less pressure loss for the same flow. Good luck
The difference between two numbers is 3 three times the smaller plus two times the larger is 31 what are the numbers? L = large number S = small number Eq. #1….L...- S =31 Eq. #2...2L + 3S = 31 ….Multiply eq#1 by 3 and add Eq. #1 + Eq. #2 Eq. #1…..3L - 3S =93 Eq. #2…..2L + 3S = 31 EQ.#3…..5L + 0 = 124 …………….....L = 24.8 Substitute L = 24.8 into Eq.#2 Eq. #2..............2L + 3S = 31 Eq. #2...2 * (24.8) + 3S = 31 Eq. #2……....49.6 + 3S = 31 Eq. #2……...............3S = -18.6 Eq. #2…….................S = -6.2 Check in Eq. #1 Eq. #1……….L...- S =31 Eq. #1….24.8..- -6.2 =31 Eq. #1…....24.8 + 6.2 =31 Large# = 24.8 Small# = -6.2
Equalizer
Thousand.
4x + 3y = -5 (Eq 1); 3x + 5y = -12 (Eq 2) Eliminate x Eq 1 times 3: 12x + 9y = -15 Eq 2 times 4: 12x +20y = -48 Subtract Eq 1 from Eq 2 11y = -48-(-15) ie 11y = -33 y = -3 Substitute in Eq 1: 4x - 9 = -5, ie 4x = 4 so x = 1 Check in Eq 2: (3 x 1) + (5 x -3) = 3 - 15 = -12. QED
3x + 2y = 4 (Eq 1); 4x + 3y = 7 (Eq 2) Eliminate x Eq 1 x 4 = 12x + 8y = 16 (Eq 3) Eq 2 x 3 = 12x + 9y = 21 (Eq 4) Subtract Eq 3 from Eq 4 y = 5 so x = (4 - 10)/3 = -6/3 ie -2 Check Eq 2: (4 x -2) + (3 x 5) = -8 + 15 = 7 QED
x + 7y = 39 ....... Eq.13x - 2y = 2 ....... Eq.2multiply Eq.1 with (-3) and so Eq.1 will be :-3x - 21y = -1173x - 2y = 2_________________ by Elimination method ( Eq.1 + Eq.2 ) the result will we :- 21y - 2y = - 115-23y = -115Y = 5to find the value of x, put the value of y in Eq.1 :x + 7y = 39x = 39 - 7yx = 39 - 7(5)X = 4___________________________________________________________you can solve it in another way by using substitution method:x + 7y = 39 ....... Eq.1 ,, you can write it in another form, x = 39 - 7y3x - 2y = 2 ....... Eq.2Put the value of Eq.1 on Eq.23(39 - 7y) - 2y = 2117 - 21y - 2y = 2117 - 23y = 223y = 115 ............ Y = 5to find the value of X , put the value of Y in Eq.1x = 39 - 7yx = 39 - 7(5)x = 39 - 35 .............. X = 4and so in both ways....you got the same answer choose the easiest way for you :)good luck my friend...
x + 7y = 39 ....... Eq.13x - 2y = 2 ....... Eq.2multiply Eq.1 with (-3) and so Eq.1 will be :-3x - 21y = -1173x - 2y = 2_________________ by Elimination method ( Eq.1 + Eq.2 ) the result will we :- 21y - 2y = - 115-23y = -115Y = 5to find the value of x, put the value of y in Eq.1 :x + 7y = 39x = 39 - 7yx = 39 - 7(5)X = 4___________________________________________________________you can solve it in another way by using substitution method:x + 7y = 39 ....... Eq.1 ,, you can write it in another form, x = 39 - 7y3x - 2y = 2 ....... Eq.2Put the value of Eq.1 on Eq.23(39 - 7y) - 2y = 2117 - 21y - 2y = 2117 - 23y = 223y = 115 ............ Y = 5to find the value of X , put the value of Y in Eq.1x = 39 - 7yx = 39 - 7(5)x = 39 - 35 .............. X = 4and so in both ways....you got the same answer choose the easiest way for you :)good luck my friend...
By adding the molecular weights and dividing by 1 you have the formula for Eq for salts that use only 1 chloride.