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Nikolay [14]
3 years ago
5

Discord anyone? RaTqUeEn2#8785

Chemistry
2 answers:
riadik2000 [5.3K]3 years ago
8 0
The answer is A I’m answering because I need points
GarryVolchara [31]3 years ago
6 0
The answer would be letter A
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5.1169 mol of Ne is held at 0.9148 atm and 911 K. What is the volume of its container in liters?
sveticcg [70]

By applying the Boyle's equation and substituting our given data the volume of the container was found to be 418.14 Litres

<h3>Boyle's  Law</h3>

Given Data

  • number of moles of Ne = 5.1169 mol
  • Pressure = 0.9148 atm
  • Temperature = 911 K

We know that the relationship between pressure and temperature is given as

PV = nRT

R = 0.08206

Making the volume subject of formula we have

V= nRT/P

Substituting our given data to find the volume we have

V = 5.1169*0.08206*911/0.9148

V = 382.522353554/0.9148

V = 418.14 L

Learn more about Boyle's law here:

brainly.com/question/469270

5 0
2 years ago
What is a substance?
shusha [124]

I think the answer is C. An element or compound that cannot be physically separated. Sorry if im wrong.

7 0
4 years ago
Toothpaste is an alkali. How could you use the toothpaste to show that red cabbage is an indicator?
Alecsey [184]

Answer:

PUT THE TOOTHPAST ON CABBAGE AND CHECK IT OUT AFTER A DAY

3 0
3 years ago
How is electrolysis most commonly used to produce an energy source?
elena-s [515]

Answer:

Splitting water molecules produces hydrogen gas, which is used to power machines through hydrogen fuel cells. ( B-)

4 0
3 years ago
What is the vapor pressure of CS2CS2 in mmHgmmHg at 26.5 ∘C∘C? Carbon disulfide, CS2CS2, has PvapPvap = 100 mmHgmmHg at −−5.1 ∘C
Digiron [165]

Answer: 26.5 mm Hg

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1= initial pressure at 26.5^oC = ?

P_2 = final pressure at -5.1^oC = 100 mm Hg

= enthalpy of vaporisation = 28.0 kJ/mol =28000 J/mol

R = gas constant = 8.314 J/mole.K

T_1= initial temperature = 26.5^oC=273+26.5=299.5K

T_2 = final temperature =-5.1^oC=273+(-5.1)=267.9K

Now put all the given values in this formula, we get

\log (\frac{P_1}{100})=\frac{28000}{2.303\times 8.314J/mole.K}[\frac{1}{299.5}-\frac{1}{267.9}]

\log  (\frac{P_1}{100})=-0.576

\frac{P_1}{100}=0.265

P_1=26.5mmHg

Thus the vapor pressure of CS_2CS_2 in mmHg at 26.5 ∘C is 26.5

7 0
4 years ago
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