n equals 1.3 x 10¹⁸ photons.
First, we calculate the energy released over the specified duration:
E = Pt,
where,
E = Energy =?,
P = Power = 4 mW = 0.004 W,
t = time = 115 s.
Thus,
E = (0.004 W)(115 s),
resulting in E = 0.46 J.
This energy can be expressed in terms of photons:
E = nhc/λ,
where,
n = Number of photons =?,
h = Planck's Constant = 6.626 x 10⁻³⁴ J.s,
c = speed of light = 3 x 10⁸ m/s,
λ = wavelength of the light = 5650 x 10⁻¹⁰ m.
Thus,
0.46 J = n(6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(5650 x 10⁻¹⁰ m),
leading to
n = (0.46 J)/(3.5 x 10⁻¹⁹ J),
culminating in n = 1.3 x 10¹⁸ photons.
Answer:
8,000 years.
Clarification:
- Radioactive isotopes are known to decay following first-order kinetics.
- The half-life is defined as the duration required for a reactant's concentration to halve.
- When a reactant starts with an initial concentration of [A₀], at the half-life it will reach a concentration of ([A₀]/2).
- Furthermore, for first-order decay, the half-life does not depend on the starting concentration.
Part 1: What is the half-life of the element? Explain how you determined this.
- The half-life of this element equals 1,600 years.
This means the reactant reduces from 56.0 g to its half (28.0 g) in 1,600 years.
Thus, the half-life for this sample is 1,600 years.
Part 2: How long would it take for 312 g of the sample to decay down to 9.75 grams? Show your work or explain your answer.
- Using the equations for first-order reactions:
k = ln(2)/(t1/2) = 0.693/(t1/2).
Where k is the reaction's rate constant.
t1/2 represents the half-life of the reaction.
∴ k =0.693/(t1/2) = 0.693/(1,600 years) = 4.33 x 10⁻⁴ year⁻¹.
- Utilizing the integral formula for first-order reaction:
kt = ln([A₀]/[A]),
with k being the reaction's rate constant (k = 4.33 x 10⁻⁴ year⁻¹).
t is the duration of the reaction (t =??? year).
[A₀] indicates the initial concentration of the sample ([A₀] = 312.0 g).
[A] shows the concentration left after decay ([A] = 9.75 g).
∴ t = (1/k) ln([A₀]/[A]) = (1/4.33 x 10⁻⁴ year⁻¹) ln(312.0 g/9.75 g) = 8,000 years.
Answer:
Mass released = 8.6 g
Explanation:
Provided information:
Starting moles of nitrogen= 0.950 mol
Starting volume = 25.5 L
Final nitrogen mass released = ?
Final volume = 17.3 L
Calculation:
Equation:
V₁/n₁ = V₂/n₂
25.5 L / 0.950 mol = 17.3 L/n₂
n₂ = 17.3 L× 0.950 mol/25.5 L
n₂ = 16.435 L.mol /25.5 L
n₂ = 0.644 mol
Starting mass of nitrogen:
Mass = moles × molar mass
Mass = 0.950 mol × 28 g/mol
Mass = 26.6 g
Ending mass of nitrogen:
Mass = moles × molar mass
Mass = 0.644 mol × 28 g/mol
Mass = 18.0 g
Mass released = initial mass - ending mass
Mass released = 26.6 g - 18.0 g
Mass released = 8.6 g
Response:The ethanol percentage is 0.1093%
Explanation:
As given:
t = time = 10 s
I = current = 320 mA
F = Faraday's constant = 96485.3365 C mol⁻¹
n = number of electrons transferred = 4
Molecular weight of ethanol is 46 g/mol
Question: What is the percent (by volume) of ethanol in the driver's breath, %E =?
First, calculate the ethanol mass:
The moles of ethanol:
Applying the ideal gas law formula:
Here:
T = 26°C = 299 K
P = 1 atm
Substituting in the values:
The percentage of ethanol:
%
