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3 years ago

The second ionization energy of lithium is much greater than the first ionization energy

1 answer:

7 0

Mg: [Ne] 3s2. The second ionization energy of Mg is larger than the first because it always takes more energy to remove an electron from a positively charged ion than from a neutral atom. The third ionization energy of magnesium is enormous, however, because the Mg2+ ion has a filled-shell electron configuration.

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Most full scuba tanks have a pressure of between 1,800 p.s.i. and 3,000 p.s.i. (pounds per square inch). Approximately 70 to 85

ArbitrLikvidat [17]

Answer: The Diver's air tank is filled with oxygen gas.

The filling of air tank while keeping in water can be explained in reference to the Charles' Law.

Charles' Law: It states that the volume of the ideal gas is directly proportional to the absolute temperature at constant pressure.

$V\propto T$ (at constant Pressure)

The Scuba tank contains cold water and is present at high pressure. The gas which is put in the tank will be compact and would not expand. As the temperature is low inside the tank, hence by using Charles' Law, the volume will also decrease inside the tank.

When the air tank is filled, it is always kept in a tub of water because the pressure is high and hence, more oxygen gas can be filled inside the air tank.

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3 years ago

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PLEASE ANSWER THIS QUESTION ASAP.

Natali5045456 [20]

Equal is the answer

Hope this helps

6 0

3 years ago

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9.11 x 10^3 in standard notation?

Irina18 [472]

Answer:

the right answer is 9110

3 0

3 years ago

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A rigid vessel contains 3.98 kg of refrigerant-134a at 700 kPa and 60°C. Determine the volume of the vessel and the total intern

Arada [10]

<u>Answer:</u> The volume of the vessel is $0.1542m^3$ and total internal energy is 162.0 kJ.

<u>Explanation:</u>

To calculate the volume of water, we use the equation given by ideal gas, which is:

$PV=nRT$

or,

$PV=\frac{m}{M}RT$

where,

P = pressure of container = 700 kPa

V = volume of container = ? L

m = Given mass of R-134a = 3.98 kg = 3980 g (Conversion factor: 1kg = 1000 g)

M = Molar mass of R-134a = 102.03 g/mol

R = Gas constant = $8.31\text{L kPa }mol^{-1}K^{-1}$

T = temperature of container = $60^oC=[60+273]K=333K$

Putting values in above equation, we get:

$700kPa\times V=\frac{3980g}{102.03g/mol}\times 8.31\text{L kPa }\times 333K\\\\V=154.21L$

Converting this value into $m^3$ , we use the conversion factor:

$1m^3=1000L$

So, $\Rightarrow (\frac{1m^3}{1000L})\times 154.21L$

$\Rightarrow 0.1542m^3$

To calculate the internal energy, we use the equation:

$U=\frac{3}{2}nRT$

or,

$U=\frac{3}{2}\frac{m}{M}RT$

where,

U = total internal energy

m = given mass of R-134a = 3.98 kg = 3980 g (Conversion factor: 1kg = 1000g)

M = molar mass of R-134a = 102.03 g/mol

R = Gas constant = $8.314J/K.mol$

T = temperature = $60^oC=[60+273]K=333K$

Putting values in above equation, we get:

$U=\frac{3}{2}\times \frac{3980g}{102.03g/mol}\times 8.314J/K.mol\times 333K\\\\U=161994.6J$

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, 161994.6 J = 162.0 kJ

Hence, the volume of the vessel is $0.1542m^3$ and total internal energy is 162.0 kJ.

6 0

3 years ago

What what is the equation for percent error

Stella [2.4K]

When you calculate results that are aiming for known values, the percent error formula is useful tool for determining the precision of your calculations. The formula is given by: The experimental value is your calculated value, and the theoretical value is your known value.

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3 years ago