What is another name for the breathing mechanism?

Semenov [28]

An intersystem crossing (ISC) is a non-radiative process that involves the transition between two electronic states with different spin multiplicity. That is, when an electron is excited in a molecule in a basal singlet state (either by absorption or radiation) into a state of greater energy, an excited singlet or triplet state can be obtained.

Therefore, ISC is understood as an a non radio active transition between states with different spin multiplicity.

Correct answer is C: a radiative transition between states with the same spin.

8 0

4 years ago

1. According to Dalton's Atomic theory which statement is correct from the following statement?

vlabodo [156]

Answer:

the mass is distributed uniformly in an atom

5 0

3 years ago

Read 2 more answers

Hello!

Mnenie [13.5K]

Answer:

$T_0=80695.17162...$

Explanation:

Given equation:

$\ln \left(\dfrac{T_0-100}{T_0}\right)=-0.00124$

To solve the given equation:

$\textsf{Apply log rules}: \quad e^{\ln (x)}=x$

$\implies \dfrac{T_0-100}{T_0}=e^{-0.00124}$

Multiply both sides by T₀:

$\implies T_0-100=T_0e^{-0.00124}$

Add 100 to both sides:

$\implies T_0=T_0e^{-0.00124}+100$

Subtract $T_0e^{-0.00124}$ from both sides:

$\implies T_0-T_0e^{-0.00124}=100$

Factor out the common term T₀:

$\implies T_0(1-e^{-0.00124})=100$

Divide both sides by $(1-e^{-0.00124})$

$\implies T_0=\dfrac{100}{1-e^{-0.00124}}$

Carry out the calculation:

$\implies T_0=\dfrac{100}{1-0.99876...}$

$\implies T_0=\dfrac{100}{0.001239231...}$

$\implies T_0=80695.17162...$

6 0

3 years ago

An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavie

IgorLugansk [536]

Answer:

The distance the lighter fragments slides is 343 m

Explanation:

Here, we have

Let the mass of the heavier fragments be m₁

Let the distance the heavy object slide be d₁

Let the mass of the lighter fragments be m₂

Let the final velocity of the heavier fragments be v₁

Let the final velocity of the lighter fragments be v₂

Let the distance the light object slide be d₂

The distance traveled by the heavy fragment = 7.00 m

Therefore since m₁ = 7 × m₂ we have

Initial total momentum = final total momentum

Since the initial total momentum = 0 we have

m₁·v₁ + m₂·v₂ = 0 or

7·m₂·v₁ = -m₂·v₂

∴ v₂ = 7·v₁

The net work done by the heavier block is

W $_{net, 7m}$ = μk × m₁ × g × d₁ = 1/2×m₂×v₂²

Also v₁² = 2μk·g·d₁ and

v₂² = 2μk·g·d₂ so that since v₂ = 7·v₁ or v₁ = v₂/7 we get

(v₂/7)² = 2μk·g·d₁

So v₂²/49 = 2μk·g·d₁

or (2μk·g·d₂)/49 = 2μk·g·d₁

∴ d₂/49 = d₁

d₂ = d₁×49 = 7 × 49 = 343 m

The distance the lighter fragments slides = 343 m.

7 0

3 years ago

What are examples of velocity and speed

inysia [295]

Explanation:

A train moves at a high velocity. Velocity is the rate of motion, speed or action. An example of velocity is a car driving at 75 miles per hour.

4 0

3 years ago