Answer:
i. Atomic radius of an atom is defined as the total distance from the nucleus to the outermost shell of the atom.
On moving from left to right ( Mg, K) in a period, more and more electrons get added up in the same shell and the attraction between the last electron and nucleus increases, which results in the shrinkage of size of an atom. Thus, decreasing the atomic radii of the atom on moving towards right of the periodic table.
As moving from top to bottom, there is an addition of shell around the nucleus and the outermost shell gets far away from the nucleus and hence, the distance between the nucleus and outermost shell increases.
Thus the order of atomic radii is : Ca > Mg > K
ii. The energy required to remove the last valence electron from isolated gaseous atom (first ionization energy) increases as we move from left to right in a period. It decreases on moving from top to bottom.
Thus the order of first ionization energy is : K > Mg > Ca
iii. The chemical properties depend on the valence elctrons and as the elements Mg and Ca both have two valence electrons , they have same chemical properties.
Answer:
the mass of one spoonful of sugar to nearest gram is 10g
Answer:
12.0 g/mL
Explanation:
From the question given above, the following data were obtained:
Mass of Rock = 360 g
Volume of water = 150 mL
Volume of water + Rock = 180 mL
Density of Rock =?
Next, we shall determine the volume of the rock. This can be obtained as follow:
Volume of water = 150 mL
Volume of water + Rock = 180 mL
Volume of Rock =?
Volume of Rock = (Volume of water + Rock) – (Volume of water)
Volume of Rock = 180 – 150
Volume of Rock = 30 mL
Finally, we shall determine the density of the rock as illustrated below:
Mass of Rock = 360 g
Volume of Rock = 30 mL
Density of Rock =?
Density = mass /volume
Density of Rock = 360 / 30
Density of Rock = 12.0 g/mL
Thus, the density of the rock is 12.0 g/mL
Answer:
How about......................IRON MAN
Explanation:
Answer:
Age of rock = 6.12 × 10³ years
Note: The question is incomplete.A similar but complete question is given below.
The half-life for the radioactive decay of carbon-14 to nitrogen-14 is 5.73 x 10^3 years. Suppose nuclear chemical analysis shows that there is 0.523mmol of nitrogen-14 for every 1.000 mmol of carbon-14 in a certain sample of rock.
Calculate the age of the rock. Round your answer to 2 significant digits.
Explanation:
The half-life of a radioactive material is the time taken for half the atoms in the atomic nucleus of a material to disintegrate.
The half-life for the radioactive decay of carbon-14 to nitrogen-14 is given as 5.73 x 10³ years. This means that given 1 mole of carbon-14 is present initially, after one half-life, 0.5 moles of carbon-14 would remain.
Number of millimoles of carbon-14 remaining = 1 - 0.523 = 0.477 mmol
Number of half-lives that the carbon-14 has undergone is determined as follows:
Amount remaining = (1/2)ⁿ
where nnis number of half-lives
0.5 mmol = one half-life
0.5 = (1/2)¹
O.477 = (1/2)ⁿ = (0.5)ⁿ
㏒₀.₅(0.477) = n
n = ㏒(0.477)/㏒(0.5)
n = 1.067938829
Age of the rock = number of half-lives × half-life
Age of rock = 1.067938829 × 5.73 × 10³ years
Age of rock = 6.12 × 10³ years
