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V125BC [204]
2 years ago
9

Most of the ultraviolet radiation reaching the surface of the earth is UV-A radiation, which has a wavelength range of 315 nm to

400 nm. UV-A radiation is a cause of skin aging and also contributes to development of skin cancer. Determine the energy of a mole of photons of UV-A radiation that have a wavelength of 325 nm.
Chemistry
1 answer:
posledela2 years ago
4 0

Answer:

6.1 ×10^-19 J

Explanation:

From E= hc/λ

h= planks constant = 6.6×10^-34 Js

c= speed of light = 3×10^8 ms^1

λ= wavelength = 325 nm

E= 6.6 × 10^-34 × 3×10^8/325 × 10^-9

E= 0.061 × 10^ -17 J

E= 6.1 ×10^-19 J

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The average adult human burns 2.00x20^3 kcal per day in energy. What is this rate in kJ per hour?
IRISSAK [1]

Answer:

1.61 × 10⁶ kJ

Explanation:

The human burns energy so as to be healthy.

The amount of energy burnt per day by an adult human is 2 × 20³ kcal per day. Since there is 24 hours in a day, the amount of energy burnt per hour is 2 × 20³ × 24 = 48 × 20³ kcal per hour.

The conversion rate of kcal to kJ is 1 kcal = 4.184 kJ. Therefore converting the kcal per hour to kJ per hour gives:

48 × 20³ × 4.184 = 200.882 × 20³ kJ = 1.61 × 10⁶ kJ

7 0
2 years ago
Acetylene burns in air according to the following equation: C2H2(g) + 5 2 O2(g) → 2 CO2(g) + H2O(g) ΔH o rxn = −1255.8 kJ Given
professor190 [17]

Answer:  -227 kJ

Explanation:

The balanced chemical reaction is,

C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(g)

The expression for enthalpy change is,

\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+ n_{H_2O}\times \Delta H_{H_2O})]-[(n_{C_2H_2}\times \Delta H_{C_2H_2})+(n_{O_2}\times \Delta H_{O_2})]

where,

n = number of moles

\Delta H_{O_2}=0 (as heat of formation of substances in their standard state is zero

Now put all the given values in this expression, we get

-1255.8=[(2\times -393.5)+(1\times -241.8)]-[(1\times \Delta H_{C_2H_2})+(\frac{5}{2}\times 0)]

-1255.8=[(-787)+(-241.8)]-[(1\times \Delta H_{C_2H_2})+(0)]

\Delta H_{C_2H_2}=-227kJ

Therefore, the enthalpy change for C_2H_2 is -227 kJ.

7 0
2 years ago
A solution is made by dissolving 9.74 g of sodium sulfate in water to a final volume of 165 mL of solution. What is the weight/w
MatroZZZ [7]

Answer: The weight/weight % or percent by mass of the solute is 5.41 %.

Explanation:

Mass of the sodium sulfate,w = 9.74 g

Volume of the water = 165 mL

Density of the water = 1 g/mL

Density=1 g/mL=\frac{\text{Mass of water}}{\text{Volume of water}}

Mass of the water =1 g/mL\times 165 mL=165 g

Mass of the solution, W:

Mass of solute + Mass of solvent =9.47 g + 165 g=174.47 g

w/w\%=\frac{w\times 100}{W}=\frac{9.45 g\times 100}{174.47 g}=5.41 \%

The weight/weight % or percent by mass of the solute is 5.41 %.


8 0
2 years ago
Which of the following actions cannot induce voltage in a wire?
MariettaO [177]
Your answer is D. Since there is little to no magnetic field to  wire, if it is copper which most wires are, there will be no voltage in a wire.
7 0
2 years ago
Read 2 more answers
Calculate the amount, in moles, of PO43- present at equilibrium when excess Sr3(PO4)2 is added to 750. mL 1.2 M Sr(NO3)2(aq). As
Crank

Answer:

1.8 × 10⁻¹⁶ mol  

Explanation:

(a) Calculate the solubility of the Sr₃(PO₄)₂

Let s = the solubility of Sr₃(PO₄)₂.

The equation for the equilibrium is

Sr₃(PO₄)₂(s) ⇌ 3Sr²⁺(aq) + 2PO₄³⁻(aq); Ksp = 1.0 × 10⁻³¹

                         1.2 + 3s          2s

K_{sp} =\text{[Sr$^{2+}$]$^{3}$[PO$_{4}^{3-}$]$^{2}$} = (1.2 + 3s)^{3}\times (2s)^{2} =  1.0 \times 10^{-31}\\\text{Assume } 3s \ll 1.2\\1.2^{3} \times 4s^{2} = 1.0 \times 10^{-31}\\6.91s^{2} = 1.0 \times 10^{-31}\\s^{2} = \dfrac{1.0 \times 10^{-31}}{6.91} = 1.45 \times 10^{-32}\\\\s = \sqrt{ 1.45 \times 10^{-32}} = 1.20 \times 10^{-16} \text{ mol/L}\\

(b) Concentration of PO₄³⁻

[PO₄³⁻] = 2s = 2 × 1.20× 10⁻¹⁶ mol·L⁻¹ = 2.41× 10⁻¹⁶ mol·L⁻¹

(c) Moles of PO₄³⁻

Moles = 0.750 L × 2.41 × 10⁻¹⁶ mol·L⁻¹ = 1.8 × 10⁻¹⁶ mol

7 0
2 years ago
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