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Bookmark or Share- Sodium Bicarbonate reacts with Acetic Acid and causes a ... - SocraticSodium bicarbonate, #"NaHCO"_3#, will react with acetic acid, #"CH"_3"COOH"#, to produce aqueous sodium acetate, #"CH"_3"COONa"#, and carbonic acid, #"H"_2"CO"_3#.
- Periodic Trends in Atomic Size - Chemistry - SocraticPeriodic trends predict differences between elemental characteristics as you move across the periodic table. Trends are based on Coulomb's law which mathematically relates several characteristics of an elements. Atomic size measured the distance between the nucleus of an atom and the outermost non-valence electrons of the atom. Atomic size decreases from left to right, because the addition of ...
- How do you multiply #(a-bi)(a+bi)#? - Socratic(a-bi)(a+bi)-a^2+b^2 (a-bi)(a+bi) is the product of two complex conjugate numbers and their product is always real. Such numbers always have equal real part and their imaginary part are equal in magnitude, but have opposite in sign. While multiplying two complex numbers one should always remember that i^2=-1.
- What is the complex conjugate of 2i? - Socratic-2i To find a complex conjugate, simply change the sign of the imaginary part (the part with the i). This means that it either goes from positive to negative or from negative to positive. As a general rule, the complex conjugate of a+bi is a-bi. While 2i may not seem to be in the a+bi form, it can be written as 0+2i. Therefore, the complex conjugate of 0+2i is 0-2i, which is equal to -2i.
- How do you convert -sqrt3+i to polar form? - SocraticCurrently, #-sqrt3 + i# is written in standard/rectangular form, or #a + bi#. To convert it into polar form, we must write it in the format #r(costheta + isintheta)#, where #r# is the radius. The radius is like the hypotenuse in a right triangle. You solve it like this: #r = sqrt(a^2 + b^2)#
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