Answer:
4.81×10¹⁰ atoms.
Explanation:
We'll begin by converting 3.2 pg to Ca to grams (g). This can be obtained as follow:
1 pg = 1×10¯¹² g
Therefore,
3.2 pg = 3.2 pg × 1×10¯¹² g / 1 pg
3.2 pg = 3.2×10¯¹² g
Therefore, 3.2 pg is equivalent to 3.2×10¯¹² g
Next, we shall determine the number of mole in 3.2×10¯¹² g of Ca. This can be obtained as follow:
Mass of Ca = 3.2×10¯¹² g
Molar mass of Ca = 40.08 g/mol
Mole of ca=.?
Mole = mass /molar mass
Mole of Ca = 3.2×10¯¹² / 40.08
Mole of Ca = 7.98×10¯¹⁴ mole.
Finally, we shall determine the number of atoms present in 7.98×10¯¹⁴ mole of Ca. This can be obtained as illustrated below:
From Avogadro's hypothesis,
1 mole of Ca contains 6.02×10²³ atoms.
Therefore, 7.98×10¯¹⁴ mole of Ca will contain = 7.98×10¯¹⁴ × 6.02×10²³ = 4.81×10¹⁰ atoms.
Therefore, 3.2 pg of Ca contains 4.81×10¹⁰ atoms.
I think chemistry field of science
Answer:
A physical property is a characteristic of a substance that can be observed or measured without changing the identity of the substance. Physical properties include color, density, hardness, and melting and boiling points. A chemical property describes the ability of a substance to undergo a specific chemical change
Explanation:
I got this from chem.libretexts.org in case you wanted to know
Potassium dichromate reacts with sulfuric acid to form chromic acid, H₂CrO₄ which is a very strong oxidizing agent. The secondary alcohol, (<em>R</em>)-2-butanol will be oxidized in the presence of chromic acid, but it can only be as oxidized as far as the ketone, which is the product shown, 2-butanone.
Sodium borohydride is a reducing agent that will reduce a ketone or aldehyde to an alcohol. When sodium borohydride reacts with 2-butanone, it reduces it to 2-butanol. However, the alcohol is no longer chiral as it was in the beginning since the sodium borohydride can add a hydride to either face of the carbonyl, which results in a racemic mixture of alcohols. This explains why the product has the same refractive index and boiling point as (<em /><em>R</em>)-2-butanol, however, the product formed would no longer be optically active.