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3 years ago

What is the sentence that we use to remember how to convert metric prefixes?

Physics

1 answer:

Marrrta [24]3 years ago

8 0

Answer:

The mnemonic I can use to memorize the metric prefixes in this order is: Gigantic Monsters Killed One Million Men Napping Peacefully. All right, so again, gigantic monsters killed one million men napping peacefully.

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Please help me with this question :

aalyn [17]

Answer:

514.27 ( wavelength )

the color is green

602.93 nm ( orange color )

the observation is that there is a change of visible color

Explanation:

A) wavelength of visible light that is most strongly reflected from a point on a soap

refraction n = 1.33

wall thickness (t) = 290 nm

2nt = (2m +1 ) ∝/2 -----equation 1

note when m = 0

therefore ∝ = 4nt/ 1 = 4 * 1.33 * 290 = 1542.8nm we will discard this

when m = 1

equation 1 becomes

∝ = 4nt/3 =( 4 * 1.33 * 290) / 3 = 1542.8 / 3 = 514.27 ( wavelength )

the color is green

B) the wavelength when the wall thickness is 340 nm

∝ = 4nt / 2m +1

where m = 1

∝ = (4 * 1.33 * 340 ) / 3 = 1808.8 / 3 = 602.93 nm ( orange color )

the observation is that there is a change of visible color

7 0

3 years ago

ome metal oxides can be decomposed to the metal and oxygen under reasonable conditions. 2 Ag2O(s) → 4 Ag(s) + O2(g) Thermodynami

yuradex [85]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

Explanation :

The given balanced chemical reaction is,

$2Ag_2O(s)\rightarrow 4Ag(s)+O_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{product}$

$\Delta H^o=[n_{Ag}\times \Delta H_f^0_{(Ag)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta H_f^0_{(Ag_2O)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

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

$\Delta H^o=[4mole\times (0kJ/mol)+1mole\times (0kJ/mol)}]-[2mole\times (-31.1kJ/mol)]$

$\Delta H^o=62.2kJ=62200J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{product}$

$\Delta S^o=[n_{Ag}\times \Delta S_f^0_{(Ag)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta S_f^0_{(Ag_2O)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

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

$\Delta S^o=[4mole\times (42.55J/K.mole)+1mole\times (205.07J/K.mole)}]-[2mole\times (121.3J/K.mole)]$

$\Delta S^o=132.67J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 298 K.

$\Delta G^o=(62200J)-(298K\times 132.67J/K)$

$\Delta G^o=22664.34J=22.66kJ$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

6 0

3 years ago

A plot of land has been surveyed for a new housing development with borders AB, BC, DC, and DA. The plot of land is a right trap

balandron [24]

Angle ABC is 119

Length of DC is 105

6 0

3 years ago

Read 2 more answers

Your image appears 3.0 m from a plane mirror. How far IS YOUR IMAGE IN<br> RELATION TO YOU?*

Vsevolod [243]

Answer:

Being a plane mirror the Image is formed 3 metres beyond the mirror . So total distance is 3+3 = 6metres

8 0

3 years ago

Read 2 more answers

A 650-kg elevator starts from rest. It moves upward for 3.00 s with constant acceleration until it reaches its cruis- ing speed

Nataly_w [17]

Answer: $P=5573.43\ W$

Explanation:

Given

Mass of the elevator is $M=650\ kg\\\$

Time period of ascension $t=3\ s$

cruising speed $v=1.75\ m/s$

Distance moved by elevator during this time

Suppose Elevator starts from rest

$\Rightarrow v=u+at\\\Rightarrow 1.75=0+a(3)\\\Rightarrow a=0.583\ s$

Distance moved

$\Rightarrow h=ut+0.5at^2\\\Rightarrow h=0+0.5\times 0.5833\times (3)^2\\\Rightarrow h=2.62\ m$

Gain in Potential Energy is

$\Rightarrow E=mgh\\\Rightarrow E=650\times 9.8\times 2.62\\\Rightarrow E=16,720.3\ N$

Average power during this period is

$\Rightarrow P=\dfrac{E}{t}\\\\\Rightarrow P=\dfrac{16,720.3}{3}\\\\\Rightarrow P=5573.43\ W$

7 0

3 years ago

Read 2 more answers