Describe the rool each element plays in the life of a plant. a)soil b)water c)light

Rank these photons in terms of decreasing energy: (b) microwave (v=9.8 ×10¹¹s⁻¹)

maw [93]

The order of the energy of the photons of given wave will be

= Ultraviolet waves > infrared waves > microwaves

$E = hv$

where,

E = energy of photon

$v$ = frequency of the radiation

h = Planck's constant = $6.63$ × $10$ ⁻³⁴ $J s$

We have :

(a) Frequency of infrared waves = $v$ ₁ = $6.5$ × $10$ ¹³ $J s$

(b) Frequency of microwaves= $v$ ₂ = $9.8$ × $10$ ¹¹ $J s$

(c) Frequency of ultraviolet waves = $v$ ₃ = $8$ × $10$ ¹⁵ $J s$

So, the decreasing order of the frequencies of the waves will be :

$v$ ₃ $>$ $v$ ₁ $> v$ ₂

As we can see from the formula that energy is directly proportional to the frequency of the wave.

$E$ ∝ $v$

So, the order of the energy of the photons of given wave will be same as their order of frequencies:

$E$ ₃> $E$ ₁ > $E$ ₂

= Ultraviolet waves > infrared waves > microwaves

Also the complete question is :

Rank the following photons in terms of decreasing energy:

(a) IR (v = 6.5 x 10¹³ s⁻¹)

(b) microwave (v = 9.8 x 10¹¹ s⁻¹)

(c) UV (v = 8.0 x 10¹⁵ s⁻¹)

To learn more about microwaves visit the link:

brainly.com/question/19811153

#SPJ4

8 0

1 year ago

It takes 11.2 kj of energy to raise the temperature of 145 g of benzene from 22.0°c to 67.0°c. what is the specific heat of benz

Cloud [144]

We can use the heat equation,
Q = mcΔT

where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J g⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).
Q = 11.2 kJ = 11200 J
m = 145 g
c = ?
ΔT = (67 - 22) °C = 45 °C
By applying the formula,
11200 J = 145 g x c x 45 °C
c = 1.72 J g⁻¹ °C⁻¹

Hence, specific heat of benzene is 1.72 J g⁻¹ °C⁻¹.

7 0

3 years ago

Which statement is true regarding the circuit schematic below?

LenKa [72]

Unless you're able to provide a diagram representing the problem, there's not much I can do so solve this problem.

4 0

3 years ago

The protein lysozyme unfolds at a transition temperature of 75.5°C, and the standard enthalpy of transition is 509 kJ mol-1. Cal

spin [16.1K]

Answer:

0.4774 KJ/K.mol

Explanation:

We are told that the transition at 25.0°C occurs in three steps. Steps i, ii and iii.

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii

Now,

C_p,m(unfolded protein) = C_p,m(folded protein) + 6.28 kJ/K.mol

Now, for the first process, ΔS_i is given as;

ΔS_i = C_p,m × In(T2/T1)

We are given;

T1 = 25°C = 25 + 273.15K = 298.15 K

T2 = 75.5°C = 75.5 + 273.15 K=348.65 K

Thus;

ΔS_i = C_p,m × In(348.65/298.15)

Now, for the third process, ΔS_iii is given as;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(T1/T2)

Thus;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)

Now, we don't know C_pm. So, we have to find a way to eliminate it. We will do it by rewriting In(298.15/348.65) in such a way that when ΔS_iii is added to ΔS_i, C_p,m will cancel out. Thus;

In(298.15/348.65) can also be written as;

In(348.65/298.15)^(-1) or

- In(348.65/298.15)

Thus;

ΔS_iii = - [(C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)]

Now, let's add ΔS_iii to ΔS_i to get;

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] + [(-C_p,m - 6.28 kJ/K.mol) × In(348.65/298.15)]

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] - [C_p,m × In(348.65/298.15)] - [6.28In(348.65/298.15)]

First 2 terms will cancel out to give;

ΔS_i + ΔS_iii = -6.28In(348.65/298.15)

ΔS_i + ΔS_iii = -0.9826 KJ/K.mol

Now,for process ii;

ΔS_ii = standard enthalpy of transition/Transition Temperature

Thus;

ΔS_ii = (509 KJ/K.mol)/348.65

ΔS_ii = 1.46 KJ/K.mol

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii = -0.9826 + 1.46 = 0.4774 KJ/K.mol

5 0

3 years ago

Which of these is an acid?

julsineya [31]

My best guess would be c

8 0

3 years ago