The molar mass of carbon is 12, hydrogen is 1, and
nitrogen is 14, hence the ratio are:
C = 38.65 / 12 = 3.22
H = 16.25 / 1 = 16.25
N = 45.09 / 14 = 3.22
Divide the three by the lowest ratio which is 3.22:
C = 3.22 / 3.22 = 1
H = 16.25 / 3.22 = 5
N = 3.22 / 3.22 = 1
So the empirical formula is:
CHN
This is a hard question because BEING BOLD cant be light ya know... but the light bulb.. thomas edison and joseph swan
Answer: Low melting points
Explanation:
Answer:
O, N, C, H
Explanation:
Electronegativity of an element is the property that combines the ability of its atom to lose or gain electrons. It measures the relative tendency with which the atoms of the element attracts valence electrons in a chemical bond.
On the periodic table, Electronegativity increases across the period but decreases down a group.
To solve the given problem, let us use thE Pauling's table of electronegativities to compare the electronegativities of the elements.
On the table:
C = 2.5
H = 2.1
O = 3.5
N = 3.0
In terms of decreasing electronegativities, the atoms are arranged as:
O N C H
Answer:
the stronger light 5.5 m apart from the total illumination
Explanation:
From the problem's statement , the following equation can be deducted:
I= k/r²
where I = intensity of illumination , r= distance between the point and the light source , k = constant of proportionality
denoting 1 as the stronger light and 2 as the weaker light
I₁= k/r₁²
I₂= k/r₂²
dividing both equations
I₂/I₁ = r₁²/r₂²=(r₁/r₂)²
solving for r₁
r₁ = r₂ * √(I₂/I₁)
since we are on the line between the two light sources , the distance from the light source to the weaker light is he distance from the light source to the stronger light + distance between the lights . Thus
r₂ = r₁ + d
then
r₁ = (r₁ + d)* √(I₂/I₁)
r₁ = r₁*√(I₂/I₁) + d*√(I₂/I₁)
r₁*(1-√(I₂/I₁)) = d*√(I₂/I₁)
r₁ = d*√(I₂/I₁)/(1-√(I₂/I₁)) =
r₁ = d/[√(I₁/I₂)-1)]
since the stronger light is 9 times more intense than the weaker
I₁= 9*I₂ → I₁/I₂ = 9 →√(I₁/I₂)= 3
then since d=11 m
r₁ = d/[√(I₁/I₂)-1)] = 11 m / (3-1) = 5.5 m
r₁ = 5.5 m
therefore the stronger light 5.5 m apart from the total illumination
