(4 ± i2) where i2 = -1
x + y = -2 ∴ y = -2 - x xy = 2 ∴ x(-2 - x) = 2 ∴ -2x - x2 = 2 ∴ x2 + 2x + 2 = 0 ∴ x2 + 2x + 1 = -1 ∴ (x + 1)2 = -1 ∴ x + 1 = i ∴ x = i - 1 recall: y = - 2 - x ∴ y = - 2 - (i - 1) ∴ y = -2 - i + 1 ∴ y = - i - 1 ∴ x3 + y3 = (i - 1)3 + (- i - 1)3 = (i - 1)(i2 - 2i + 1) + (- i - 1)(i2 + 2i + 2) = i3 - 2i2 + i - i2 + 2i - 1 - i3 - 2i2 - 2i - i2 - 2i - 2 = i3 - i3 - 2i2 - i2 - 2i2 - i2 + i + 2i - 2i - 2i - 1 - 2 = -6i2 - i - 3 = 3 - i
HI will be consumed. The reaction will proceed to the left. More I2 will form.
Cl2(g) + 2KI --> 2KCl(aq) + I2(s)
In math i=√(-1), in physics and engineering j=√(-1), and i2 or j2 =-1.
No.(-i)2= (-1 x i)2= (-1)2 x i2= 1 x i2= i2= -1but if it is written-i2Then= -i2= (-1) * i2= -1 * -1= 1
(4 ± i2) where i2 = -1
The reciprocal of a + bi is a - bi:1/(a + bi) since the conjugate is a - bi:= 1(a - bi)/[(a + bi)(a - bi)]= (a - bi)/[a2 - (b2)(i2)] since i2 equals to -1:= (a - bi)/(a2 + b2) since a2 + b2 = 1:= a - bi/1= a - bi
x + y = -2 ∴ y = -2 - x xy = 2 ∴ x(-2 - x) = 2 ∴ -2x - x2 = 2 ∴ x2 + 2x + 2 = 0 ∴ x2 + 2x + 1 = -1 ∴ (x + 1)2 = -1 ∴ x + 1 = i ∴ x = i - 1 recall: y = - 2 - x ∴ y = - 2 - (i - 1) ∴ y = -2 - i + 1 ∴ y = - i - 1 ∴ x3 + y3 = (i - 1)3 + (- i - 1)3 = (i - 1)(i2 - 2i + 1) + (- i - 1)(i2 + 2i + 2) = i3 - 2i2 + i - i2 + 2i - 1 - i3 - 2i2 - 2i - i2 - 2i - 2 = i3 - i3 - 2i2 - i2 - 2i2 - i2 + i + 2i - 2i - 2i - 1 - 2 = -6i2 - i - 3 = 3 - i
HI will be consumed. The reaction will proceed to the left. More I2 will form.
Cl2(g) + 2KI --> 2KCl(aq) + I2(s)
x=square root (-2) =i(square root of 2)WHERE i2 =-1
Cl2 + 2KI → 2KCl + I2
The stoichiometric relationship between IO3^- and I2 is 5:1. This means that for every one mole of IO3^- ions, five moles of I2 molecules are produced.
Yes, i2=-1, by definition.
This reaction is not at equilibrium yet since the reaction quotient, Q, is not equal to the equilibrium constant, K. In this case, Q = (0.03)^2 / ((0.01)*(0.02))^2 = 0.45, which is greater than K = 0.15. Therefore, the reaction will proceed in the reverse direction to reach equilibrium.
Iodine, I2, has a charge of 0 since it is in its elemental form and is a diatomic molecule. It does not have a net charge or ionic charge.