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V125BC [204]
3 years ago
5

If the volume is increased the number of collisions per second will decrease or increase

Chemistry
1 answer:
Alinara [238K]3 years ago
7 0

Answer

Decrease

Explanation:

As surface area will also increase the particles are less likely to collide on a regular basis.

You might be interested in
The wavelength of a particular color of violet light is 433 nm. The energy of this wavelength of light is kJ/photon. (109 nm = 1
mash [69]

Answer:

4.59 × 10⁻³⁶ kJ/photon

Explanation:

Step 1: Given and required data

  • Wavelength of the violet light (λ): 433 nm
  • Planck's constant (h): 6.63 × 10⁻³⁴ J.s
  • Speed of light (c): 3.00 × 10⁸ m/s

Step 2: Convert "λ" to meters

We will use the conversion factor 1 m = 10⁹ nm.

433 nm × 1 m/10⁹ nm = 4.33 × 10⁷ m

Step 3: Calculate the energy (E) of the photon

We will use the Planck-Einstein's relation.

E = h × c/λ

E = 6.63 × 10⁻³⁴ J.s × (3.00 × 10⁸ m/s)/4.33 × 10⁷ m

E = 4.59 × 10⁻³³ J = 4.59 × 10⁻³⁶ kJ

5 0
3 years ago
What is moment and its calculations ​
PIT_PIT [208]

Answer:

Moment=Force x Pivot

Explanation:

A moment is the turning effect of a force. Moments act about a point in a clockwise or anticlockwise direction.

Law of moments:

When an object is balanced (in equilibrium) the sum of the clockwise moments is equal to the sum of the anticlockwise moments.

How to calculate moments:

Moment=Force x Pivot

3 0
2 years ago
Force necessary to change an object's motion
igomit [66]
Yes force is necessary
8 0
3 years ago
What is the unit of measurement for mass?
Licemer1 [7]
Grams. It is a smaller unit. 
4 0
3 years ago
A) Find the gas speed of sulfur dioxide at 100.0 degrees Celsius? ______________
gtnhenbr [62]

a. 381.27 m/s

b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide

<h3>Further explanation</h3>

Given

T = 100 + 273 = 373 K

Required

a. the gas speedi

b. The rate of effusion comparison

Solution

a.

Average velocities of gases can be expressed as root-mean-square averages. (V rms)  

\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}

R = gas constant, T = temperature, Mm = molar mass of the gas particles  

From the question  

R = 8,314 J / mol K  

T = temperature  

Mm = molar mass, kg / mol  

Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol

\tt v=\sqrt{\dfrac{3\times 8.314\times 373}{0.064} }\\\\v=381.27~m/s

b. the effusion rates of two gases = the square root of the inverse of their molar masses:  

\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }

M₁ = molar mass sulfur dioxide = 64

M₂ =  molar mass nitrogen triodide = 395

\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{395}{64} }=\dfrac{20}{8}=2.5

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide

4 0
3 years ago
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