6.023 x 10²³ particles in 1 mole of Boron
1.61 x 10³⁴ particles in = 1.61 x 10³⁴ / 6.02 x 10²³
= 2.61 x 10³⁰ moles of Boron.
Answer:
3.97 M
Explanation:
Given data:
Initial volume V₁ = 318 mL
Initial molarity M₁ = 5.75 M
New volume V₂= 461 mL
New concentration M₂= ?
Solution:
New volume V₂= 143 mL+ 318 mL
New volume V₂= 461 mL
Formula:
M₁V₁ = M₂V₂
M₂ = M₁V₁ / V₂
M₂ = 5.75 M × 318 mL / 461 mL
M₂ = 1828.5 M. mL/ 461 mL
M₂ = 3.97 M
The answer would have to be letter b. Energy
Answer:
Covalent bonds
Explanation:
The type of bond present in methane is covalent bonds.
In covalent bonds, two species share electrons to give an interatomic interaction.
The two species are always non-metals with little to no electronegativity differences between the two of them.
Carbon is made of 4 valence electrons. It will require an additional 4 electrons to attain stability
Hydrogen has one electron its outer shell. When each of the four hydrogen shares their single electrons with carbon, the central carbon will attain an octet configuration. The binding hydrogen will have a complete orbital too.