<span>First we need to find the energy of one photon with a wavelength of 589 nm.
E = hc / wavelength
E = (6.63 x 10^{-34} J s)(3 x 10^8 m/s) / (589 x 10^{-9} m)
E = 3.3769 x 10^{-19} Joules
To find N, the number of photons, we need to divide the total energy by the energy of each photon.
N = 623000 J / 3.3769 x 10^{-19} Joules
N = 1.84 x 10^{24} photons
There are 1.84 x 10^{24} photons in the burst of yellow light.</span>
Answer:
1. 35 mg of H₃PO₄
2. 27 mol AlF₃; 82 mol F⁻
3. 300 mL of stock solution.
Explanation:
1. Preparing a solution of known molar concentration
Data:
V = 80 mL
c = 4.5 × 10⁻³ mol·L⁻¹
Calculations:
(a) Moles of H₃PO₄
Molar concentration = moles of solute/litres of solution
c = n/V
n = Vc = 0.080L × (4.5 × 10⁻³ mol/1 L) = 3.60 × 10⁻⁴ mol
(b) Mass of H₃PO₄
moles = mass/molar mass
n = m/MM
m = n × MM = 3.60 × 10⁻⁴ mol × (98 g/1 mol) = 0.035 g = 35 mg
(c) Procedure
Dissolve 35 mg of solid H₃PO₄ in enough water to make 80 mL of solution,
2. Moles of solute.
Data:
V = 4900 mL
c = 5.6 mol·L⁻¹
Calculations:
Moles of AlF₃ = cV = 4.9 L AlF₃ × (5.6 mol AlF₃/1L AlF₃) = 27 mol AlF₃
Moles of F⁻ = 27 mol AlF₃ × (3 mol F⁻/1 mol AlF₃) = 82 mol F⁻.
3. Dilution calculation
Data:
V₁= 750 mL; c₁ = 0.80 mol·L⁻¹
V₂ = ? ; c₂ = 2.0 mol·L⁻¹
Calculation:
V₁c₁ = V₂c₂
V₂ = V₁ × c₁/c₂ = 750 mL × (0.80/2.0) = 300 mL
Procedure:
Measure out 300 mL of stock solution. Then add 500 mL of water.
Answer:
A lava lamp consists of oil, and wax in a glass, and a heat source (a light bulb) placed underneath the glass. When the lamp is turned on the bulb gets hot.
Explanation:
Answer:
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Explanation:
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Answer:
High temperatures
Explanation:
NaHCO₃ (8) + HCH,O₂ (aq) → H₂O (l) + CO₂ (g) + NaC,H₃O₂ (aq)
As the flask gets cooler to the touch as the reaction proceeds, the reaction is endothermic. This means that ΔH is positive (ΔH>0).
As a gas is formed (bubbles are formed), ΔS is positive (ΔS>0).
<em>In terms of ΔG:</em>
<em>In order for the reaction to be thermodynamically favorable, ΔH has to be negative</em>, thus:
- The reaction is favorable if TΔS > ΔH.
The greater the temperature, the easier it would be for TΔS to be greater than ΔH.
