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NNADVOKAT [17]
3 years ago
7

The symbol for xenon xe would be part of the noble gas notation for the element

Chemistry
2 answers:
Allushta [10]3 years ago
8 0
The symbol for xenon xe would be part of the noble gas notation for the element of radium, it is a chemical element with the symbol of Ra and the atomic number of radium is 88. It is the sixth element in group 2 of the periodic table, also known as the alkaline earth metals.<span> </span>
Mademuasel [1]3 years ago
4 0

Answer : The noble gas notation is often given to the short or abbreviated form of the electron configuration.

The use of symbol of the previous noble gas as part of the electron configuration of an element is the shortcut way to write the electron configuration.

Xenon has electron configuration as 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2} 5p^{6}.

Cesium has the electron configuration as 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{10} 4s^{2} 4p^{6} 4d^{10} 5s^{2} 5p^{6} 6s^{1}

So, Cesium can be abbreviated using the symbol of xenon [Xe] 6s^{1}.

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2) Dillute acids feel slippery is not correct. Bases, for example solution of sodium hydroxide feels slipery.

3) Acids have a distinctly sour taste is correct. For example, vinegar is mixture of acetic acid (CH₃COOH) and water (H₂O). Vinegar is colourless liquid with sour taste and pungent smell, freezing point of the vinegar is lower than glacial acetic acid.

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2 years ago
Find the mole fraction of benzene and toluene in solution containing​
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3 years ago
Elimination of the pharmaceutical IV antibiotic gentamicin follows first-order kinetics. If the half-life of gentamicin is 1.5 h
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Explanation:

The given data is:

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The concentration of reactant after 8 hrs can be calculated as shown below:

The formula of the half-life of the first-order reaction is:

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Where k = rate constant

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So, the rate constant k value is:

k=\frac{0.693}{1.5 hrs}

The expression for the rate constant is :

k=\frac{2.303}{t} log \frac{initial concentration}{concentration after time "t"}

Substitute the given values and the k value in this formula to get the concentration of the reactant after time 8 hrs is shown below:

\frac{0.693}{1.5 hrs} =\frac{2.303}{8 hrs} x log \frac{8.4x10^-^5}{y} \\ log \frac{8.4x10^-^5}{y} =1.604\\\frac{8.4x10^-^5}{y}=10^1^.^6^0^4\\\frac{8.4x10^-^5}{y}=40.18\\y=\frac{8.4x10^-^5}{40.18} \\=>y=2.09x10^-^6

Answer: The concentration of reactant remains after 8 hours is 2.09x10^-6M.

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