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3 years ago

a neutral atom is bombarded with sufficient energy that an electron is liberated. the result is called

Chemistry

1 answer:

timofeeve [1]3 years ago

5 0

Answer:

A cation.

Explanation:

When an electron is liberated, that means it is removed from the atom.

When an electron is lost, the atom is ionized, and becomes a cation.

It becomes a cation because it loses one negative charge, and therefore become positively charged.

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A marine biologist is preparing a deep-sea submersible for a dive. The sub stores breathing air under high pressure in a spheric

Luba_88 [7]

Complete Question

A marine biologist is preparing a deep-sea submersible for a dive. The sub stores breathing air under high pressure in a spherical air tank that measures 74.0 wide. The biologist estimates she will need 2600 L of air for the dive. Calculate the pressure to which this volume of air must be compressed in order to fit into the air tank. Write your answer in atmospheres. Round your answer to significant digits.

Answer:

The pressure required is $P_2= 12.2 \ atm$

Explanation:

Generally the volume of a sphere is mathematically denoted as

$V_s = \frac{4}{3} * \pi r^3$

Substituting $r = \frac{d}{2} = \frac{74}{2} = 37cm$

$V_s = \frac{4}{3} * 3.42 * (37)^2$

$V_s = 2.121746 *10^5 cm^3$

Converting to Liters

$V_s = \frac{2.121746 *10^5}{1000}$

$V_s= 212.1746L$

Assume that the pressure at which the air is given to the diver is 1 atm when the air was occupying a volume of 2600L

From Charles law

$P_1V_1 = P_2 V_s$

Substituting $V_1 =2600 L$ , $P_1 = 1 atm$ , $V_s =212.1746L$ , and making $P_2$ the subject we have

$P_2 = \frac{P_1 * V_1}{V_s}$

$= \frac{1 * 2600}{212.1746}$

$P_2= 12.2 atm$

6 0

4 years ago

A large cylindrical tank contains 0.750 m3 of nitrogen gas at 27°C and 7.50 * 103 Pa (absolute pressure). The tank has a tight‐f

Nuetrik [128]

Answer: The answer is 68142.4 Pa

Explanation:

Given that the initial properties of the cylindrical tank are :

Volume V1= 0.750m3

Temperature T1= 27C

Pressure P1 =7.5*10^3 Pa= 7500Pa

Final properties of the tank after decrease in volume and increase in temperature :

Volume V2 =0.480m3

Temperature T2 = 157C

Pressure P2 =?

Applying the gas law equation (Charles and Boyle's laws combined)

P1V1/T1 = P2V2/T2

(7500 * 0.750)/27 =( P2 * 0.480)/157

P2 =(7500 * 0.750* 157) / (0.480 *27)

P2 = 883125/12.96

P2 = 68142.4Pa

Therefore the pressure of the cylindrical tank after decrease in volume and increase in temperature is 68142.4Pa

8 0

3 years ago

A sample of gas occupies a volume of 67.1 mL . As it expands, it does 135.3 J of work on its surroundings at a constant pressure

Semmy [17]

Answer:

$V_2=1.363x10^{-3}m^3=1363mL$

Explanation:

Hello,

In this case, since the work done at constant pressure as in isobaric process is computed by:

$W= P(V_2-V_1)$

Thus, given the pressure, initial volume and work, the final volume is:

$V_2=V_1+\frac{W}{P}$

Whereas the pressure must be expressed in Pa as the work is given in J (Pa*m³):

$P=783Torr*\frac{101325Pa}{760Torr} =104394Pa$

And the volumes in m³:

$V_1=67.1mL*\frac{1m^3}{1x10^6mL} =6.71x10^{-5}m^3$

Thus, the final volume turns out:

$V_2=6.71x10^{-5}m^3+\frac{135.3Pa*m^3}{104394Pa}\\\\V_2=1.363x10^{-3}m^3=1363mL$

Best regards.

3 0

4 years ago

If 250. ML of water are poured into the measuring cup, the volume reading is 8.45 oz . This indicates that 250. ML and 8.45 oz a

Tamiku [17]

Answer:

$1oz=29.589ml$

Explanation:

From the question we are told that:

Initial Volume $v_1=250ml$

Final Volume $v_2=8.45oz$

Generally the equation for one ounce is mathematically given by

$1oz=\frac{v_1}{v_2}$

Therefore

$1oz=\frac{250}{8.45}$

$1oz=29.589ml$

7 0

3 years ago

The standard free-energy change for this reaction in the direction written is 23.8 kJ/mol. The concentrationsof the three interm

Natali [406]

Answer: The actual free-energy change for the reaction -8.64 kJ/mol.

Explanation:

The given reaction is as follows.

Fructose 1,6-bisphosphate $\rightleftharpoons$ Glyceraldehyde 3-phosphate + DHAP

For the given reaction, $\Delta G^{o}$ is 23.8 kJ/mol.

As we know that,

$\Delta G = \Delta G^{o} + RT ln Q$

Here, R = 8.314 J/mol K, T = $37^{o} C$

= (37 + 273) K

= 310.15 K

Fructose 1,6-bisphosphate = $1.4 \times 10^{-5}$ M

Glyceraldehyde 3-phosphate = $3 \times 10^{-6}$ M

DHAP = $1.6 \times 10^{-5}$ M

Expression for reaction quotient of this reaction is as follows.

Reaction quotient = $\frac{[DHAP][\text{glyceraldehyde 3-phosphate}]}{[/text{Fructose 1,6-bisphosphate}]}$

Q = $\frac{1.6 \times 10^{-5} \times 3 \times 10^{-6}}{1.4 \times 10^{-5}}$

= $3.428 \times 10^{-6}$

Now, we will calculate the value of $\Delta G$ as follows.

$\Delta G = \Delta G^{o} + RT ln Q$

= $23800 + 8.314 \times 310.15 \times ln(3.428 \times 10^{-6})$

= -8647.73 J/mol

= -8.64 kJ/mol

Thus, we can conclude that the actual free-energy change for the reaction -8.64 kJ/mol.

7 0

3 years ago