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4 years ago

What would be a correct unit of measurement for velocity

Chemistry

2 answers:

dangina [55]4 years ago

8 0

Velocity is in units of distance per unit of time. like meters per second (m/s) or miles per hour (mph)

erastova [34]4 years ago

5 0

For ANY measurement of velocity, there must be a distance unit divided by time. So look at the units to see which ones meet that requirement, but in most scientific studies they revert to meters/second

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Valence level

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B, C, D are compounds, while A and E are just element stand-alones.

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3 years ago

How many planar and radial nodes are there in a 3p orbital?

KATRIN_1 [288]

A 3p orbital has one planar node perpendicular to the orbital's axis, and one radial node.

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4 years ago

The mass of an electron is close to:<br> A. 0.00055u. B. 0.05u. C. 10-10<br> u. D. 0.0001u

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Answer:

A. 0.0005u

Explanation:

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3 years ago

Read 2 more answers

Which of the following is the first step in balancing a chemical equation?

WARRIOR [948]

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3 years ago

How to set up the rate expressions for the following mechanism?

Anestetic [448]

Answer:

Explanation:

From the given information:

A → B k₁

B → A k₂

B + C → D k₃

The rate law = $\dfrac{d[D]}{dt}=k_3[B][C] --- (1)$

$\dfrac{d[B]}{dt}=k[A] -k_2[B] -k_3[B][C]$

Using steady-state approximation;

$\dfrac{d[B]}{dt}=0$

$k_1[A]-k_2[B]-k_3[B][C] = 0$

$[B] = \dfrac{k_1[A]}{k_2+k_3[C]}$

From equation (1), we have:

$\mathbf{\dfrac{d[D]}{dt}= \dfrac{k_3k_1[A][C]}{k_2+k_3[C]}}$

when the pressure is high;

k₂ << k₃[C]

$\dfrac{d[D]}{dt} = \dfrac{k_3k_1[A][C]}{k_3[C]}= k_1A \ \ \text{first order}$

k₂ >> k₃[C]

$\dfrac{d[D]}{dt} = \dfrac{k_3k_1[A][C]}{k_2}= \dfrac{k_1k_3}{k_2}[A][C] \ \ \text{second order}$

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3 years ago