You swing a 2.2 kg stone in a circle with radius 75 cm. At what speed do you need to swing it so its

Liono4ka [1.6K]

A compound is a substance that consists of two or more elements, which is chemically combined, meaning that it could only be split by chemical means.

Examples:-

- NaCl is salt, chemically combined of Sodium and Chloride.

- H2O is water, chemically combined of Hydrogen and oxygen.

8 0

3 years ago

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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

Will an atom emit light if all the atoms electrons are in the ground state

Oliga [24]

No,

To emit light an electron has to jump down to a lower energy level but in an electron is unable to go any lower in ground state.

6 0

3 years ago

A train at rest emits a sound at a frequency of 1057 Hz. An observer in a car travels away from the sound source at a speed of 2

il63 [147K]

Answer:

993.52 Hz

Explanation:

The frequency of sound emitted by the stationery train is 1057 Hz.

The car travels away from the train at 20.6 m/s.

The frequency the observer hears is given by the formula:

$f_o = \frac{v - v_o}{v}f$

where v = velocity of sound = 343 m/s

vo = velocity of observer

f = frequency from source

This phenomenon is known as Doppler's effect.

Therefore:

$f_o = \frac{343 - 20.6}{343} * 1057\\ \\f_o = 322.4 / 343 * 1057\\\\f_o = 993.52 Hz$

The frequency heard by the observer is 993.52 Hz.

4 0

3 years ago

Where is the sun's energy most concentrated

Maru [420]

B. At the equator

Explanation:

The energy coming from the Sun hits the Earth's surface at different angles, depending on the latitude of the place. The more perpendicular the ray of lights hit the surface, the more the energy transmitted to the Earth's surface, the warmer the location.

The angle at which the ray of lights hit the Earth is related to the latitude: in particular, the ray of lights arrive perpendicular at the equator ( $0^{\circ}$ ), they arrive at larger angle in the United States (which is located at intermediate latitudes) and they arrive at the largest angles at the poles. For this reason, the sun's most energy is concentrated at the equator.

5 0

2 years ago