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jekas [21]
1 year ago
10

if 3.95 mol of an ideal gas has a pressure of 2.33 atm and a volume of 60.87 l, what is the temperature of the sample in degrees

celsius? g
Chemistry
1 answer:
eduard1 year ago
4 0

If 3.95 mol of an ideal gas has a pressure of 2.33 atm and a volume of 60.87 l, 164.65 temperature of the sample in degrees Celsius.

T=PV/nR

T= 2.33*60.87/3.95*0.082 atms

T=437.8

Convert the temperature of sample from K to С = k-273.15

C=437.8-273.15

⁰C=164.6

The temperature of the sample is 164.6⁰С

The relationship between temperature and pressure for a given volume of gas is straightforward. Pressure increases in systems when temperature increases and vice versa. The Gay-law Lussac's specifies the link between a gas's pressure and temperature. Air pressure is the result of molecules colliding and air pressing against its surroundings. The difference between temperature and air density is how many molecules there are in a given volume of air.

Learn more about Pressure here:

brainly.com/question/1357378

#SPJ4

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Answer:

B

Explanation:

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7 0
3 years ago
. Write the following isotope in nuclide notation (e.g., “ ”): copper-70
vekshin1

Answer:

{\boxed{\text{$_{29}^{70}${Cu}}}

Explanation:

The atomic number (Z) of copper is 29 and this isotope has an atomic mass (A) of 70.

The general symbol for an isotope E is _{Z}^{A}\text{E}.

The atomic number is a left subscript, and the atomic mass is a left superscript.

\rm {\text{The nuclide notation for copper-70 is }}{\boxed{\textbf{$_{29}^{70}${Cu}}}

4 0
3 years ago
Let P and V represent the pressure and volume of the Xe(g) in the container in diagram 3. If a piston is used to reduce the volu
Reil [10]

Answer:

2p

Explanation:

To solve this question, we can use Boyle's Law, which states that:

"For a fixed mass of an ideal gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume"

Mathematically:

p\propto \frac{1}{V}

where

p is the pressure of the gas

V is its volume

The equation can be rewritten as

p_1 V_1 = p_2 V_2

where in this problem we have:

p_1 = p is the initial pressure of the Xe(g) gas

V_1=V is the initial volume of the Xe(g) gas

V_2=\frac{V}{2}  is the final volume of the Xe(g) gas

Solving for p2, we find the final pressure of the gas:

p_2=\frac{p_1 V_1}{V_2}=\frac{pV}{V/2}=2p

So, the final pressure is twice the initial pressure.

7 0
3 years ago
With 143.6 grams of acetylene and an excess amount of oxygen gas, what is the percent yeild of carbon dioxide if the actual yiel
nata0808 [166]

Answer:

93.28%

Explanation:

To solve the percent yield we need to find theoretical yield:

<em>Percent yield = Actual Yield (452.78g) / Theoretical yield * 100</em>

<em />

Theoretical yield is obtained converting the mass of acetylene to moles and using the balanced equation determine the moles of CO₂ produced assuming a 100% of reaction:

<em>Moles acetylene (Molar mass: 26.04g/mol)</em>

143.6g C₂H₂ * (1mol / 26.04g) = 5.515 moles C₂H₂

<em>Moles CO₂:</em>

5.515 moles C₂H₂ * (4 moles CO₂ / 2mol C₂H₂) = 11.029moles CO₂

<em>Mass CO₂ (Molar mass: 44.01g/mol):</em>

11.029moles CO₂  * (44.01g / mol) = 485.39g CO₂ is theoretical yield

Percent yield is:

Percent yield = 452.78g / 485.39g * 100

= 93.28%

8 0
3 years ago
You measure water in two containers:a 10-mL graduated cylinder with marks at every mL, and a 1-mL pipetmarked at every 0.1 mL. I
Reptile [31]

Answer:

B. 0.1 mL

Explanation:

The solution explanation has been attached.

6 0
3 years ago
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