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3 years ago

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Check and double-click on the red polygon for the Henry Mountains laccolith complex in the folder labeled Problem 1. This comple

x consists of many blister-like intrusions. Note the size difference between this intrusion and the Sierra Nevada batholith. Estimate how much smaller in length the Henry Mountains laccolith complex is relative to the batholith. Select one:a. ~1/2b. ~1/4c. ~1/6d. ~1/17

2 answers:

Kobotan [32]3 years ago

8 0

Answer:

d

Explanation:

The batholith is roughly 720km, while the laccolith is roughly 42km. Take 42/720 and it gives you roughly 1/17/

MArishka [77]3 years ago

4 0

Answer: ghg

Explanation:

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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

What is the speed of a wave that has a frequency of 125 Hz and a wavelength of 1.25 meters? Express your answer to the nearest w

shtirl [24]

156 is the answer. so 156.25 is almost the same thing, you just round. It's not hard. Thank you!!!

3 0

3 years ago

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A sample of n2 gas occupies a volume of 746 ml at stp. What volume would n2 gas occupy at 155 ◦c at a pressure of 368 torr?

musickatia [10]

Answer:

2.41 L

Explanation:

We can solve the problem by using the ideal gas equation, which can be rewritten as:

$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}$

where we have:

$p_1 = 1.01\cdot 10^5 Pa$ (initial pressure is stp pressure)

$V_1 = 746 mL = 0.746 L = 7.46\cdot 10^{-4}m^3$ is the initial volume

$T_1 = 0^{\circ}=273 K$ is the initial temperature (stp temperature)

$p_2 = 368 torr = 4.9\cdot 10^4 Pa$ is the final pressure

$V_2 = ?$ is the final volume

$T=155^{\circ}=428 K$ is the final temperature

By substituting the numbers inside the formula and solving for V2, we find the final volume:

$V_2 = \frac{p_1 V_1 T_2}{T_1 p_2}=\frac{(1.01\cdot 10^5 Pa)(7.46\cdot 10^{-4} m^3)(428 K)}{(273 K)(4.9\cdot 10^4 Pa)}=2.41\cdot 10^{-3} m^3$

which corresponds to 2.41 L.

7 0

3 years ago

Why is temperature a good criterion for searching for Earthlike exoplanets?

grandymaker [24]

Because it gives us the abilitity to find planets that have decent temperatures relative to earths temp so we can determine if the planet even has a possiblilty to sustain life hope this helps

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3 years ago

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Please help me and thank you

fredd [130]

Answer:

536.56 m/s

Explanation:

We'll begin by calculating the momentum of the Porsche. This can be obtained as follow:

Mass (m) of Porsche = 1361 kg

Velocity (v) of Porsche = 26.82 m/s

Momentum of Porsche =?

Momentum = mass × velocity

Momentum = 1361 × 26.82

Momentum of Porsche = 36502.02 Kgm/s

Finally, we shall determine the velocity you need to be running with in order to have the same momentum as the Porsche. This can be obtained as follow:

Your Mass = 68.03 kg

Your Momentum = Momentum of Porsche = 36502.02 Kgm/s

Your velocity =?

Momentum = mass × velocity

36502.02 = 68.03 × velocity

Divide both side by 68.03

Velocity = 36502.02 / 68.03

Velocity = 536.56 m/s

Thus you must be running with a speed of 536.56 m/s in order to have the same momentum as Porsche.

8 0

2 years ago