Answer:
714 nm
Explanation:
Using the equation: nλ=d<em>sin</em>θ
where
n= order of maximum
λ= wavelength
d= distance between lines on diffraction grating
θ= angle
n is 1 because the problem states the light forms 1st order bright band
λ is unknown
d=
or 0.0000014 (meters)
sin(30)= 0.5
so
(1)λ=(0.0000014)(0.5)
=0.000000714m or 714 nm
Answer:
The mass of silicon in kilograms in Earth's crust is
.
Explanation:
Mass of Earth =
(1 ton= 2000 lb)
(1 lb =453.6 g)
1 ton = 2000 × 453.6 g =907,200 g
Mass of Earth =
Percentage of earth crust = 0.50%
Mass of earth crust = M


Percentage of the silicon in Earth's crust = 27.2 %
Mass of silicon in in Earth's crust = m



1000 g = 1 kg
The mass of silicon in kilograms in Earth's crust is
.
Answer:
pH =3.8
Explanation:
Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:
HA + H₂O ⇄ H₃O⁺ + A⁻ with Ka = [ H₃O⁺] x [A⁻]/ [HA]
The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.
In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:
HA H₃O⁺ A⁻
Initial, M 0.40 0 0
Change , M -x +x +x
Equilibrium, M 0.40 - x x x
Lets express these concentrations in terms of the equilibrium constant:
Ka = x² / (0.40 - x )
Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,
7.3 x 10⁻⁶ = x² / 0.40 ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³
[H₃O⁺] = 1.71 x 10⁻³
Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.
pH = - log ( 1.71 x 10⁻³ ) = 3.8
Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.
3.11 i'm not sure about measurements maybe like 3.11kg/cm^3
<h2>Answer: C) 1s²2s²2p⁶</h2>
<h3>Explanation:</h3>
A noble gas has 8 electrons between the p and s orbitals of the outer shell. Helium is the exception because it only has two electrons.
<h3> ∴ 1s²2s²2p⁶ is the noble gas (neon)</h3>