Which of these claims about the way energy is transferred in Earth’s interior is supported by evidence ?

it takes 513 kj to remove a mole of electrons from the atoms at the surface of a piece of metal. how much energy does it take to

schepotkina [342]

Answer:

The right solution is " $8.5\times 10^{-19} \ joule$ ".

Explanation:

As we know,

1 mole electron = $6.023\times 10^{23} \ no. \ of \ electrons$

Total energy = $513 \ KJ$

= $513\times 1000 \ joule$

For single electron,

The amount of energy will be:

= $\frac{513\times 1000}{(6.023\times 10^{23})}$

= $8.5\times 10^{-19} \ joule$

8 0

1 year ago

Consider the following system at equilibrium where H° = -87.9 kJ, and Kc = 83.3, at 500 K. PCl3(g) + Cl2(g) PCl5(g) If the VOLUM

7nadin3 [17]

Answer:

The value of Kc C. remains the same.

The value of Qc C. is less than Kc.

The reaction must: A. run in the forward direction to reestablish equilibrium

The number of moles of Cl2 will B. decrease.

Explanation:

Le Chatelier's Principle states that if a system in equilibrium undergoes a change in conditions, it will move to a new position in order to counteract the effect that disturbed it and recover the state of equilibrium.

A decrease in volume causes the system to evolve in the direction in which there is less volume, that is, where the number of gaseous moles is less.

But temperature is the only variable that, in addition to influencing equilibrium, modifies the value of the constant Kc. So if the volume of the equilibrium system is suddenly decreased at constant temperature: The value of Kc remains the same.

As mentioned, if the volume of an equilibrium gas system decreases, the system moves to where there are fewer moles. In this case, being:

PCl₃(g) + Cl₂(g) ⇔ PCl₅(g)

The equilibrium in this case then shifts to the right because there is 1 mole in the term on the right, compared to the two moles on the left. So, The reaction must: A. run in the forward direction to reestablish equilibrium.

By decreasing the volume, and so that Kc remains constant, being:

$Kc=\frac{[PCl_{5} ]}{[PCl_{3}]*[Cl_{2} ]}=\frac{\frac{nPCl_{5} }{Volume} }{\frac{nPCl_{3}}{Volume}*\frac{nCl_{2} }{Volume} } =\frac{nPCl_{5}}{nPCl_{3}*nCl_{2}} *Volume$

where nPCl₅, nPCl₃ and nCl₂ are the moles in equilibrium of PCl₅, PCl₃ and Cl₂

so, the number of moles of Cl₂ should decrease.The number of moles of Cl2 will B. decrease.

If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system will evolve to the right, the direct reaction prevailing, to increase the concentration of products. So in this case, if the reaction moves to the right, the value of Qc C. is less than Kc.

3 0

1 year ago

How much heat must be removed from 25.0g of steam at 118.0C in order to form ice at 15C

NemiM [27]

Answer:

-10778.95 J heat must be removed in order to form the ice at 15 °C.

Explanation:

Given data:

mass of steam = 25 g

Initial temperature = 118 °C

Final temperature = 15 °C

Heat released = ?

Solution:

Formula:

q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 15 °C - 118 °C

ΔT = -103 °C

now we will put the values in formula

q = m . c . ΔT

q = 25 g × 4.186 J/g.°C × -103 °C

q = -10778.95 J

so, -10778.95 J heat must be removed in order to form the ice at 15 °C.

3 0

1 year ago

Gino made a table to describe parts of the electromagnetic spectrum. A 3-column table with 4 rows. The first column labeled wave

IgorC [24]

Answer:

D. Ultraviolet light should have a short wavelength, not a long wavelength.

Explanation:

just took the quiz on Ed

7 0

2 years ago

Read 2 more answers

THESE QUESTIONS ARE ELECTRICIAN REALATED

Vladimir [108]

1. A.

It is a technique typically engaged in electrical power and electronic devices, wherein the devices are run at less than their rated maximum power degeneracy, taking into account the case or body temperature, the ambient temperature and the type of cooling mechanism used.

2. B.

In order to answer the question, you need to know the area of a circle.
Area = pi * r^2
Area = 3.14 * (100/2) ^ 2
Area = 3.14 * 50^2
Area = 7850 mils^2 = 0.00785 in^2

3. A.

V / 15 A = 0.213 Ohm 
0.213 Ohm / 125 ft * 1000 ft = 1.71 Ohm / 1000 ft
Closest gauge is #12 AWG (solid wire)
#12 is 1.588 Ohm / ft, which is a 2.9775 V drop @ 15A

4. A.

RHW cale insulation can be used up to 167 degrees F.

5. D.

This is a type of electrical connector used to fasten two or more low-voltage (or extra-low-voltage) electrical conductors.

6. C. the ease with which a material allows electricity to move is called CONDUCTIVITY

7. D.

Stranded conductors are not acceptable in the pressure terminals of a duplex receptacle. Aluminum conductors of suitable gauge for a 15 ampere duplex outlet will not fit in the pressure terminal holes.

8. D.

A 3/0 AWG THHN insulated Copper is rated at 225 amps and it falls under the 90 degree C table.

9. D.

A #12 copper conductor with an insulation factor of 90 degrees C is rated at 20 amps. A #12 aluminum conductor with an insulation rating of 90 degrees C is rated at 15 amps. These conductors ratings only applies to three conductors in a rated at 15 amps. These conductors’ ratings only applies to three conductors in a raceway. From 7 to 24 conductors in a raceway, both aluminum and copper conductor's ratings have to be reduced by .70, so 15 amps x .7 = 10.5 amps and 20 amps x .7 = 14 amps respectively.

10. A.

NEC 310.15(B)(1) and (2), table 310.15(B)(2)(a), and table 310.15(B)(17).
Convert 158°F to C result is 70°C. 50 amps x .33 correction factor = 16.5 amps

11. A.

12. B.

Ideally, nm or armored cable should be installed in a wall stud at least 1 1/4 inches from the front edge of the stud.

13. C.

Always use flux for electrical connections and never use acid flux on electrical connections

14. D.

In wire gauge, the lower the number, the larger the diameter. So a 6 gauge is larger than 12 gauge, but smaller than 4 gauge.

15. A.

16. D.

17. C.

Table 310-16 NEC

18. A 
MCM means 1000's (M) of (C)ircular (M)ils

19. D.

Rubber tape is used to round sharp edges

20. B.

The larger the diameter, the larger the ampacity, or the ability to carry current.

4 0

2 years ago

Read 2 more answers