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dsp73
4 years ago
13

Marshall determines that a gas has a gage pressure of 276 kPa what's the absolute pressure of this gas

Chemistry
2 answers:
defon4 years ago
7 0

And object has a mass of 120 KG’s on the moon what is the force of gravity acting on the object of the moon

Wittaler [7]4 years ago
3 0
To determine the absolute pressure of this gas, all you need to do is to add the value of atmospheric pressure and the value of gage pressure.

Atmospheric pressure is equivalent to 100 kPa. 
Gage pressure is 276 kPa.

Then, we add both values. 

N = 100 kPa + 276 kPa
N =  376 kPa

The absolute pressure of this gas is 376 kPa.

Hope this helps :)
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4 0
3 years ago
Metals have low ionization energies and readily share their ________ or outer electrons with each other to form an electron ____
Dmitry [639]

Explanation:

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But when we move down a group then there occurs an increase in atomic size of the atoms due to addition of number of electrons in the atoms. Hence, ionization energy decreases along a group.

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What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91
Vikentia [17]

Answer:  A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L

Temperature = 91^{o}C = (91 + 273) K = 364 K

As molar mass of O_2 is 32 g/mol. Hence, the number of moles of O_2 are calculated as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

PV = nRT\\P  \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen (O_2) if it occupies 31.6 mL at 91^{o}C.

4 0
3 years ago
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