The alkali metals are so reactive that they are never found in nature in elemental form. Although some of their ores are abundant, isolating them from their ores is somewhat difficult. For these reasons, the group 1 elements were unknown until the early 19th century, when Sir Humphry Davy first prepared sodium (Na) and potassium (K) by passing an electric current through molten alkalis. (The ashes produced by the combustion of wood are largely composed of potassium and sodium carbonate.) Lithium (Li) was discovered 10 years later when the Swedish chemist Johan Arfwedson was studying the composition of a new Brazilian mineral. Cesium (Cs) and rubidium (Rb) were not discovered until the 1860s, when Robert Bunsen conducted a systematic search for new elements. Known to chemistry students as the inventor of the Bunsen burner, Bunsen’s spectroscopic studies of ores showed sky blue and deep red emission lines that he attributed to two new elements, Cs and Rb, respectively. Francium (Fr) is found in only trace amounts in nature, so our knowledge of its chemistry is limited. All the isotopes of Fr have very short half-lives, in contrast to the other elements in group 1.
Answer:
9.36
Explanation:
Sodium formate is the conjugate base of formic acid.
Also,
for sodium formate is 
Given that:
of formic acid = 
And, 
So,
Concentration = 0.35 M
HCOONa ⇒ Na⁺ + HCOO⁻
Consider the ICE take for the formate ion as:
HCOO⁻ + H₂O ⇄ HCOOH + OH⁻
At t=0 0.35 - -
At t =equilibrium (0.35-x) x x
The expression for dissociation constant of sodium formate is:
![K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}](https://tex.z-dn.net/?f=K_%7Bb%7D%3D%5Cfrac%20%7B%5BOH%5E-%5D%5BHCOOH%5D%7D%7B%5BHCOO%5E-%5D%7D)
Solving for x, we get:
x = 0.44×10⁻⁵ M
pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64
pH + pOH = 14
So,
<u>pH = 14 - 4.64 = 9.36</u>
Answer:
This element is Rubidium (Rb) and has an average atomic mass of 85.468 u
Explanation:
The average mass of an element is calculated by taking the average of the atomic masses of its stable isotopes.
The enitre atomic mass = 100 % or 1
⇒ this consists of X-85 with 72.17 % abundance with atomic massof 84.9118 g/mol
72.17 % = 0.7217
⇒ this consists of X-87 with 27.83 % abundance with atomic mass of 86.9092 g/mol
27.83 % = 0.2783
To calculate the mass of this isotope we use the following:
0.7271 * 84.9118 + 0.2783 * 86.9092 =85.468 g/mol
This element is Rubidium(Rb) and has an average atomic mass of 85.468 u
It is because in that time science was not established. This is, the experimentation to test hypotheses, which is a fundamental part of the scientific method, was not applied.
Atoms can not be seen, then they could only reflect or philosophize on this matter. This method is not able to give good answers to so complicated scientific matters.
Answer:
132g/mole
Explanation:
using the formula PV=nRT should be used to solve for the number of moles (n). R is a constant which is 62.3637 L mmHG/mole K.
Inorder for your units to match you will have to convert 125ml to .125L and the temperature of 85C to K . you do that by adding 273 to the 85C and get 358K. Once you solve for n then you use that number and divide by the number of grams from the question (.560g) since molar mass is grams/moles.
