Answer:
One mole of a substance is equal to 6.022 × 10²³ units of that substance (such as atoms, molecules, or ions). The number 6.022 × 10²³ is known as Avogadro's number or Avogadro's constant. The concept of the mole can be used to convert between mass and number of particles
22-yes, they belong to the same group
23- they are metals
24-Atoms of group 1 elements all have one electron in their outer shell. This means Na and K have the same valence electrons.
25-Metals are solids at room temperature, therefore Na and K are solid.
26- mass number=A=235
Protons=z=92
Neutrons=mass number-protons
Neutrons=A-Z
Neutrons=235-92
Neutrons=143
Answer:
The correct option is: 4. soluble in both acidic and basic solutions
Explanation:
Oxide is a chemical compound formed by an element with oxygen. In such compounds, the oxygen is generally present in the -2 oxidation state.
Whereas, hydroxide is a chemical compound formed by an element with hydroxyl group (-OH).
<em>Oxides and hydroxides of some elements are </em><em>amphoteric</em><em> in nature.</em> <u><em>Such compounds can behave as an acid in basic medium and behave as a base in acidic medium.</em></u>
<u>Therefore, amphoteric oxides and hydroxides are soluble in both acidic and basic solutions.</u>
The complete question is;
Calculate the empirical formula for each of the following naturalflavors based on their elemental mass percent composition.
Q1)
methyl butyrate (component of apple taste andsmell): C -58.80 % H- 9.87 %
O -31.33.%Express your answer as a chemical formula.
Q2)
vanillin (responsible for the taste and smellof vanilla): C - 63.15% H- 5.30 %
O - 31.55%Express your answer as a chemical formula.
Q1)
empirical formula is the simplest ratio of whole number of elements in the compound. as the percentages have been given, lets calculate for 100 g of compound
C H O
mass 58.80 g 9.87 g 31.33
molar mass 12 g/mol 1 g/mol 16 g/mol
number of moles 58.80/12 9.87/1 31.33/16
= 4.9 =9.87 = 1.95
then divide number of moles by least number of moles - 1.95 in this case
4.9/1.95 = 2.51 9.87/1.95 = 5.06 1.95/1.95= 1
next multiply by 2 to get numbers that can be rounded off to whole numbers
2.51x2 = 5.02 5.06x2 = 10.12 1x2 = 2
when rounded off to the nearest whole number
C - 5
H - 10
O - 2
therefore empirical formula is C₅H₁₀O₂
Q2) for this too since elemental composition has been given in percentages lets calculate for 100 g of compound
C H O
mass 63.15 g 5.30 g 31.55 g
molar mass 12 g/mol 1 g/mol 16 g/mol
number of moles 63.15/12 5.30/1 31.55/16
=5.26 =5.30 =1.97
divide the number of moles by the least number of moles - 1.97
5.26/1.97 5.30/1.97 1.97/1.97
=2.67 = 2.69 = 1
multiply each by 3 to get numbers that can be rounded off to whole numbers
2.67x3 = 8.01 2.69x3 = 8.07 1x3 = 3
rounded off to the nearest whole numbers
C - 8
H - 8
O - 3
empirical formula = C₈H₈O₃
There had been new discoveries of chemicals that have been added to the periodic table in recent years
