Answer:inform the lab instructor and get instructions
Explanation:
If you come across a chemical in the laboratory which has been wrongly labelled, do not be quick to dilute it or take any further action. The laboratory instructor who may have prepared the reagent himself or has better knowledge about the reagent should be contacted immediately so that he/she can give you instructions about what to do with the wrongly labelled reagent.
<span>First divide the ionization energy by Avogadro's number to get the energy per atom of potassium;
</span>419 kj/mol / 6.023 x 10²³
= 4.19 x 10⁵ / 6.023 x 10²³ = 6.96 x 10⁻¹⁹
E = hc/λ
where lambda (λ<span>) is the wavelength, h is Planck's constant, c is the speed of light
</span>E = 6.96 x 10⁻¹⁹ j/atom<span>
h =</span>6.63x10⁻³⁴<span> Js
c = 3 x 10</span>⁸ m/s
λ = ?
λ = hc/E = (6.63x10⁻³⁴ x 3 x 10⁸ ) / 6.96 x 10⁻¹⁹ = 285.8nm = 286nm
<span>The longest wavelength of light capable of this ionization is 286nm.</span>
Answer:
6.82 kg
Explanation:
Given that the amount of water is 15L and we know that the density of water is ≈ 1kg/L. The mass of water is given by mass = volume x density, i.e,
mass = 15 x 1 = 15 kg. Also the specific heat capacity of water is 4.186 KJ/kg.
The sublimation enthalpy of dry ice is 571 KJ/kg.
Now, the amount of heat lost by water is entirely used up for the sublimation (conversion from soild to gas) of dry ice. And the heat (Q) lost by water is given as : Q = mCΔT, where m is the mass of water, C the specific heat capacity of water and ΔT the change in temperature.
Here, Q = 15 x 4.186 x (90 - 28) = 3892.98 KJ.
This amount of heat is taken up by the dry ice for its sublimation. Also the energy taken by dry ice (Q') for its sublimation is given by: Q' = m'L', where m' is the mass of dry ice, L' is the latent heat of sublimation (i.e, the amount of heat required per kg of a substance to sublime) of dry ice amd L' = 571 KJ/kg.
Now, Q' =m'L' = heat lost by water = 3892.98KJ.
And, m'L' = m' x 571 KJ/kg = 3892.98 KJ. (Dividing with 571)
Therefore, m' = 6.82 kg.
Answer:
The specific heat of water is 4.18 J/g C.
Explanation:
q
=
m
C
s
Δ
T
Never forget that!
2200
=
m
⋅
4.18
J
g
⋅
°
C
⋅
66
°
C
∴
m
≈
8.0
g
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