Ultraviolet Rays ( UV rays) can harm humans because they can cause sunburn, and skin cance if you stay out in the sun for too long.
Answer:
The given statement - The main criterion for sigma bond formation is that the two bonded atoms have valence orbitals with lobes that point directly at each other along the line between the two nuclei , is <u>True.</u>
Explanation:
The above statement is correct , because the sigma bond is produced by the head on overlapping, the orbitals should all point in the same direction.
<u>SIGMA BONDS -</u> Sigma bonds (bonds) are the strongest type of covalent chemical bond in chemistry. They're made up of atomic orbitals that collide head-on. For diatomic molecules, sigma bonding is best characterized using the language and tools of symmetry groups.
Head-on overlapping of atomic orbitals produces sigma bonds. The concept of sigma bonding is expanded to include bonding interactions where a single lobe of one orbital overlaps with a single lobe of another. Propane, for example, is made up of ten sigma bonds, one for each of the two CC bonds and one for each of the eight CH bonds.
Hence , the answer is true .
answer: elements are on the periodic table
Answer:
CH₄
Explanation:
To determine the empirical formula of the hydrocarbon, we need to follow a series of steps.
Step 1: Determine the mass of the compound
The mass of the compound is equal to the sum of the masses of the elements that form it.
m(CxHy) = mC + mH = 7.48 g + 2.52 g = 10.00 g
Step 2: Calculate the percent by mass of each element
%C = mC / mCxHy × 100% = 7.48 g / 10.00 g × 100% = 74.8%
%H = mH / mCxHy × 100% = 2.52 g / 10.00 g × 100% = 25.2%
Step 3: Divide each percentage by the atomic mass of the element
C: 74.8/12.01 = 6.23
H: 25.2/1.01 = 24.95
Step 4: Divide both numbers by the smallest one, i.e. 6.23
C: 6.23/6.23 = 1
H: 24.95/6.23 ≈ 4
The empirical formula of the hydrocarbon is CH₄.
The balanced chemical equation for the combustion of butane is:

Δ
= Σ
Δ
-Σ
Δ
=
=[-3148kJ/mol+(-2418.2kJ/mol)]-[(-251.2kJ/mol)+0]
= -5315 kJ/mol
Calculating the enthalpy of combustion per mole of butane:

Therefore the heat of combustion per one mole butane is -2657.5 kJ/mol
Correct answer: -2657.5 kJ/mol