Answer:
Concentration AgBr at saturation = 7.07 x 10⁻⁷M
Explanation:
Given AgBr(s) => Ag⁺(aq) + Br⁻(aq) ; Ksp = 5 x 10⁻¹³ = [Ag⁺][Br⁻]
I --- 0 0
C --- +x +x
E --- x x
[Ag⁺][Br⁻] = (x)(x) = x² = 5 x 10⁻¹³ => x = SqrRt(5 x 10⁻¹³) = 7.07 x 10⁻⁷M
Answer:
iron (III) oxide is a gas
Three complete orders on each side of the m=0 order can be produced in addition to the m = 0 order.
The ruling separation is
d=1 / (470mm −1) = 2.1×10⁻³ mm
Diffraction lines occur at angles θ such that dsinθ=mλ, where λ is the wavelength and m is an integer.
Notice that for a given order, the line associated with a long wavelength is produced at a greater angle than the line associated with a shorter wavelength.
We take λ to be the longest wavelength in the visible spectrum (538nm) and find the greatest integer value of m such that θ is less than 90°.
That is, find the greatest integer value of m for which mλ<d.
since d / λ = 538×10⁻⁹m / 2.1×10 −6 m ≈ 3
that value is m=3.
There are three complete orders on each side of the m=0 order.
The second and third orders overlap.
Learn more about diffraction here : brainly.com/question/16749356
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First figure out how many grams must freeze and then convert the grams to moles.
<span>Hf = -334 J/g. Convert this to KJ/g by dividing by 1000. (There are 1000 Joules in a kJ). </span>
<span>Hf = -334 J/g ÷ 1000 J/kj = -0.334 kJ/g </span>
<span>Now, divide 100 kJ by -0.334 kJ/g (see how the units are lining up?) </span>
<span>100 kJ ÷ -0.334 kJ/g = 299 g </span>
<span>Now convert this to moles by dividing by the molecular weight of water (18.0g/mole). </span>
<span>299 ÷ 18.0 = 16.6 moles </span>
Answer:
See explanation below
Explanation:
In order to calculate this, we need to use the following expression to get the concentration of the base:
MaVa = MbVb (1)
We already know the volume of NaOH used which is 13.4473 mL. We do not have the concentration of KHP, but we can use the moles. We have the mass of KHP which is 0.5053 g and the molecular formula. Let's calculate the molecular mass of KHP:
Atomic weights of the elements to be used:
K = 39.0983 g/mol; H = 1.0078 g/mol; C = 12.0107 g/mol; O = 15.999 g/mol
MM KHP = (1.0078*5) + (39.0983) + (8*12.0107) + (4*15.999) = 204.2189 g/mol
Now, let's calculate the mole of KHP:
moles = 0.5053 / 204.2189 = 0.00247 moles
With the moles, we also know that:
n = M*V (2)
Replacing in (1):
n = MbVb
Now, solving for Mb:
Mb = n/Vb (3)
Finally, replacing the data:
Mb = 0.00247 / (13.4473/1000)
Mb = 0.184 M
This would be the concentration of NaOH
