Answer:
The Barium flame is green because it is a difficult flame to excite, therefore for it to trigger a flame it is necessary that it be too excited for it to occur.
The reddish color of calcium is due to its high volatility and it is sometimes very difficult to differentiate it from strontium.the compression of these elements is due to being able to make them work during combustion
Explanation:
The flame test is a widely used qualitative analysis method to identify the presence of a certain chemical element in a sample. To carry it out you must have a gas burner. Usually a Bunsen burner, since the temperature of the flame is high enough to carry out the experience (a wick burner with an alcohol tank is not useful). The flame temperature of the Bunsen burner must first be adjusted until it is no longer yellowish and has a bluish hue to the body of the flame and a colorless envelope. Then the tip of a clean platinum or nichrome rod (an alloy of nickel and chromium), or failing that of glass, is impregnated with a small amount of the substance to be analyzed and, subsequently, the rod is introduced into the flame, trying to locate the tip in the least colored part of the flame.
The electrons in these will jump to higher levels from the lower levels and immediately (the time that an electron can be in higher levels is of the order of nanoseconds), they will emit energy in all directions in the form of electromagnetic radiation (light) of frequencies characteristics. This is what is called an atomic emission spectrum.
At a macroscopic level, it is observed that the sample, when heated in the flame, will provide a characteristic color to it. For example, if the tip of a rod is impregnated with a drop of Ca2 + solution (the previous notation indicates that it is the calcium ion, that is, the calcium atom that has lost two electrons), the color observed is brick red .
Answer:
correct option is (a)
The solution would be using this: C6H5COOH = H+ + C6H5COO Ka = 6.5 x 10^-5 = (H+)(C6H5COO-) over
(C6H5COOH)
Let X = moles per liter (H+) and also = moles per liter (C6H5COO-)
Ka = 6.5 x 10^-5 = (X)(X) over .350 molar = acid solution 6.5 x 10^-5 = X^2 over .350
X^2 = 6.5 x 10^-5 times .350 which = 2.275 x 10^-5
x = V2.275 x 10^-5
X = 1.5083 x 10^-5 moles per liter H+
pH = -log(H+) = -log 1.5083 x 10^-5 which
= 4.6215
Answer: 4.12
Explanation:
we know that the given mol is 8.23 mol and they are 2NaI and I2 so we will write the equation like this.
8.23mol NaI x 1mol of I2 ÷ 2molNaI = 4.115≅ 4.12 mol of I2
we placed NaI at the bottom to cancel out with the 8.23 mol of NaI
