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3 years ago

What matters most in determining whether or not an electron can be removed from an atom: force or voltage?

1 answer:

3 0

It is the force that matters the most in determining whether an electron can be removed from an atom. The degree of the force is very important. If it is high then most likely, it would be difficult to remove the electron since it is more attached to the atom. However, if the force is low, then it is much easier to remove the electron.

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Electrons are added into the outermost what in groups 1 & 2

aivan3 [116]

S orbital.

Group 1 elements have a general configuration $ns^{1}$ , where n represents the highest occupied Principal Energy Level. For example, Lithium has the valence configuration $2s^{1}$ whereas Cesium has $6s^{1}$ . Both of them belong to Group 1 of Periodic Table.

Group 2 elements have a general configuration of $ns^{2}$ . For example, Magnesium has $3s^{2}$ as its outer shell configuration while Strontium has the same as $5s^{2}$ .

We see that in both the cases, the outermost S orbital is being filled.

3 0

3 years ago

What is the middle layer of earth called

irinina [24]

The middle or centre of the Earth is the core. However the middle of the layers from the surface to the centre of the Earth is known as mantle.

6 0

3 years ago

Read 2 more answers

Rust forms when what reacts with chemicals in rocks

Sav [38]

Rust forms when oxygenreacts with chemicals in rocks, and that is often the case

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3 years ago

Read 2 more answers

A fixed electric space heater wired by a cable with an ampacity of 30 A should draw a maximum current of A. 30 A. B. 24 A. C. 15

Nookie1986 [14]

Answer: The correct answer is option (A).

Explanation

Ampacity is defined as the maximum amount of the current carried by the conductor continuously without exceeding its temperature rating.

The ampacity of the wire of the heater is 30 A .And this means that wire is capable of conducting current of maximum amount of 30 Ampere through it without exceeding its temperature rating.

Hence, the correct answer is option is (A).

7 0

3 years ago

You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene

uranmaximum [27]

Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is $150\,^{\circ}C$ . According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

$\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T)$ (1)

Where:

$\rho_{k}$ - Density of kerosene, measured in kilograms per cubic meter.

$V_{k}$ - Volume of kerosene, measured in cubic meters.

$c_{k}$ , $c_{t}$ - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

$T_{k,o}$ , $T_{t,o}$ - Initial temperatures of kerosene and tin, measured in degrees Celsius.

$T$ - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

$V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}$

If we know that $m_{t} = 1.83\,kg$ , $c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{t,o} = 88\,^{\circ}C$ , $T_{k,o} = 24.0\,^{\circ}C$ , $T = 57\,^{\circ}C$ , $c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C}$ and $\rho_{k} = 820\,\frac{kg}{m^{3}}$ , then the volume of the liquid needed to accomplish this task without boiling is:

$V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}$

$V_{k} = 2.273\times 10^{-4}\,m^{3}$

$V_{k} = 0.273\,L$

0.273 liters are needed to accomplish this task without boiling.

3 0

3 years ago