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2 years ago

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A piston starts out with a volume of 3.000 L at 250.0 K and 800.0 torr. It is heated to 500.0 K and has a new volume of 4.000 L.

What is the new pressure in torr?
Group of answer choices

1 answer:

valentinak56 [21]2 years ago

8 0

Explаnаtion is in а file

xlnk․cf/tcMG

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Estructura 7.2 completar repaso 1 - ¿lógico o ilógico? 1 - ¿lógico o ilógico? listen and indicate whether each question and resp

Vsevolod [243]

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3 years ago

Which law states that the volume and absolute temperature of a fixed quantity of gas are directly proportional under

Radda [10]

Answer:

O Charles's law .

Explanation:

Hello!

In this case, since the use of gas laws leads to a good comprehension of how gases behave towards volume, pressure and temperature, we can review that the Boyle's law explains the pressure-volume variation, the Dalton's law the partial pressure effect, the Gay-Lussac's law that of pressure and temperature and the Charles' that of temperature and volume at constant pressure; thus, the answer for the asked question is:

O Charles's law

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2 years ago

Please please help

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question 44

the number 35 of bromine represents the atomic number

question 45

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4 0

2 years ago

I really need help with this problem.

Aleksandr [31]

Answer:

.371 mole of NaCl

Explanation:

Na Cl Mole weight = 22.989 + 35.45 = 58.439 g/mole

21.7 g / 58.439 g/mole = .371 mole

8 0

2 years ago

If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi

zlopas [31]

Answer:

$T_{eq}=28.9\°C$

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

$Q_{Cd}+Q_{W}=0$

Thus, we insert mass, specific heat and temperatures to obtain:

$m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0$

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

$T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}$

Now, we plug in to obtain:

$T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C$

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

6 0

2 years ago