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.
+protons and -electrons because they would eventually equal out.
Answer:
The elements in Group 1 (lithium, sodium, potassium, rubidium, cesium, and francium) are called the alkali metals. All of the alkali metals have a single s electron in their outermost principal energy. ... For example, the electron configuration of lithium (Li), the alkali metal of Period 2, is 1s22s1.
I don’t understand the question
Answer:
Tin fork and knife from a camping set, a steel pole fence post, and a copper coin.
Explanation:
Hope this halps. UwUz
