E = hc/(lamda)
The lamda symbol is wavelength, which this site does not have. I can represent it with an "x" instead.
Plancks constant, h = 6.626×10^-32 J·s
Speed of light, c = 3.00×10^8 m/s
The energy must be greater than or equal to 1×10^-18 J
1×10^-18 J ≤ (6.626×10^-32 J·s)*(3.0×10^8 m/s) / x
x ≤ (6.626×10^-32 J·s)*(3.0×10^8 m/s) / (1×10^-18 J)
x ≤ 1.99×10^-7 m or 199 nm
The wavelength of light must be greater than or equal to 199 nm
Answer:
0.297 °C
Step-by-step explanation:
The formula for the <em>freezing point depression </em>ΔT_f is
ΔT_f = iK_f·b
i is the van’t Hoff factor: the number of moles of particles you get from a solute.
For glucose,
glucose(s) ⟶ glucose(aq)
1 mole glucose ⟶ 1 mol particles i = 1
Data:
Mass of glucose = 10.20 g
Mass of water = 355 g
ΔT_f = 1.86 °C·kg·mol⁻¹
Calculations:
(a) <em>Moles of glucose
</em>
n = 10.20 g × (1 mol/180.16 g)
= 0.056 62 mol
(b) <em>Kilograms of water
</em>
m = 355 g × (1 kg/1000 g)
= 0.355 kg
(c) <em>Molal concentration
</em>
b = moles of solute/kilograms of solvent
= 0.056 62 mol/0.355 kg
= 0.1595 mol·kg⁻¹
(d) <em>Freezing point depression
</em>
ΔT_f = 1 × 1.86 × 0.1595
= 0.297 °C
<span>The correct answer is b. Boiling point, why? because the liquid sample of napthalene is heated and remained at the temperature of 218 degrees celsius, the outcome was that the napthalene was completely vaporized, therefore we are given the scenario that at the temperature of 218 degrees celsius is considered to be the boiling pont of napthalene.</span>
Answer:
4 a minor but very important component of the atmosphere corban dioxide is released through natural processes such as reputation.;
Explanation:
a minor but very important component of the atmosphere corban dioxide is released through natural processes such as reputation
The lithosphere makes up the layer of the earth.
