The name given to these electrons are that they are valence electrons or binding electrons as these are directly involved in chemical Bonding and allow for different compounds to be made.

5 0

3 years ago

The town of Natrium, West Virginia, derives its name from the sodium produced in the electrolysis of molten sodium chloride (NaC

Alona [7]

Explanation:

The reaction equation will be as follows.

$Na^{+} + e^{-} \rightarrow Na(s)$

Hence, moles of Na = moles of electron used

Therefore, calculate the number of moles of sodium as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{4500 g}{23 g/mol}$ (as 1 kg = 1000 g)

= 195.65 mol

As, Q = $n \times F$ where F = Faraday's constant

= $195.65 mol \times 96500 C$

= $1.88 \times 10^{7}$ mol C

Relation between electrical energy and Q is as follows.

E = $Q \times V$

Hence, putting the given values into the above formula and then calculate the value of electricity as follows.

E = $Q \times V$

= $1.88 \times 10^{7} \times 5$

= $9.4 \times 10^{7} J$

As 1 J = $2.77 \times 10^{-7}$ kWh

Hence, $\frac{9.4 \times 10^{7}}{2.77 \times 10^{-7}}$ kWh

= 3.39 kWh

Thus, we can conclude that 3.39 kilowatt-hours of electricity is required in the given situation.

7 0

3 years ago

Read 2 more answers

1,645.6 expressed in scientific notation equals?

12345 [234]

1.6456 x 10^3 (ten to the third power)

8 0

3 years ago

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Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water

madreJ [45]

Explanation:

The given data is as follows.

Temperature of dry bulb of air = $25^{o}C$

Dew point = $10^{o}C$ = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

Partial pressure of water vapor $(P_{a})$ = vapor pressure of water at a given temperature $(P^{s}_{a})$

Using Antoine's equation we get the following.

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}$

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}$

= 0.17079

$P^{s}_{a}$ = 1.18624 kPa

As total pressure $(P_{t})$ = atmospheric pressure = 760 mm Hg

= 101..325 kPa

The absolute humidity of inlet air = $\frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

$\frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

= 0.00735 kg $H_{2}O$ / kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg $H_{2}O$ / kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

0.07 kg - 0.00735 kg

= 0.06265 kg $H_{2}O$ for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

$0.06265 kg \times 100$

= 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

3 0

3 years ago