Ask question

Ask question

All categories

2 years ago

12

Why was Roemer's measurement of the speed of light inaccurate?

1 answer:

DanielleElmas [232]2 years ago

7 0

C bc you have to have the accurate diameter of earths orbit

You might be interested in

Phosphate never enters the<br> O atmosphere<br> O ground<br> O water<br> O ocean

gizmo_the_mogwai [7]

I think you can google this because I really don’t know the answer I’m so sorry

7 0

3 years ago

The speed of a wave is 2ms, and its wavelength 0.4 meters. What is the period of the wave?

AnnyKZ [126]

Answer:

5

Explanation:

6 0

2 years ago

Read 2 more answers

The strength of an electromagnet can be altered by

denpristay [2]

The strength of an electromagnet can be altered by increasing the number of coils around the core. The more times the coil is wrapped, the stronger the electromagnet is.

Your answer is: B) Increasing the number of coils around the core

Have an amazing day and stay hopeful!

3 0

2 years ago

Read 2 more answers

How many atoms of oxygen in the chemical formula 2 Ca(ClO2)2?

alexgriva [62]

Answer:

8

Explanation:

Ca(ClO2)2 - 2*2 = 4 Oxygen atoms

2 Ca(ClO2)2 - 2*4 = 8 Oxygen atoms

8 0

2 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