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3 years ago

Why is the sky blue and doesn't change color

1 answer:

7 0

The statement is <u>false</u> because the sky <u>can change </u>colors during sunsets, sun rises, etc. The sky is not always blue.

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Jada is rowing a boat across a river that has a current of 5 m/s in the ˆ j direction. Leanne, standing on the shore, observes J

olchik [2.2K]

Answer: d. 8.25 m/s

Explanation:

We are given that Current= 5 m/s in j direction

Velocity= 8 m/s i + 3 m/s j

Now, we have to find Jada's speed with respect to the water.

First we find Jada's velocity with respect to water

v= (8 i + 3 j) - (5 j)

v= 8i - 2 j

To find the speed, we take the magnitude of this velocity vector we have

|v|= $\sqrt{(8)^2+(-2)^2}$

|v|= $\sqrt{68}$ = 8.246 m/s

which comes out to be around = 8.25 m/s

So option d is correct.

5 0

3 years ago

Which of the following is correct in order of lowest to highest energy?

Serggg [28]

It is number 3. Because microwaves are the lowest of the sepectum then visible light the the highest of them all is Gamma Ray's. Hope this helps!!

3 0

3 years ago

Read 2 more answers

How to convert 40m/min to km/h? need step by step explanation

Hitman42 [59]

Answer:

i have no clue i just need brailnly points

Explanation:

7 0

3 years ago

Read 2 more answers

A tank contains one mole of carbon dioxide gas at a pressure of 6.70 atm and a temperature of 23.0°C. The tank (which has a fixe

quester [9]

Answer: 888.45 K or 615.3 °c

Explanation:

According to Gay Lussacs law which states that at constant volume, pressure of an ideal gas is directly proportional to it's absolute temperature.

P/T = Constant

Therefore, P1/T1 = P2/T2

P1 = 6.7 atm

T1= 23°c = 273.15 + 23 = 296.15K

Since P2 is tripled, then,

P2 = 6.7 x 3= 20.1 atm

T2 = (20.1 x 296.15) ÷ 6.7

T2 = 888.45 K

Or in celcius 615.3°c

5 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