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OlgaM077 [116]
2 years ago
5

If I have 4 moles of a gas at a pressure of 5.6 atm and a volume of 12 liters, what is the temperature? (2005 K)

Chemistry
1 answer:
jolli1 [7]2 years ago
8 0

Answer:

205K

Explanation:

The following were obtained from the question:

n = 4moles

P = 5.6atm

V = 12L

R = 0.082atm.L/Kmol

T =?

PV = nRT

T = PV/nR

T = (5.6 x 12)/ (4 x 0.082)

T = 205K

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Please help me with this.<br><br><br>And show all work as well ASAP!!​
qaws [65]

Answer: The partial pressure of oxygen is 187 torr.

Explanation:

According to Raoult's law, the partial pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the total pressure.

p_1=x_1p_{total}  

where, x = mole fraction  

p_{total} = total pressure  

x_{oxygen}=\frac{\text {moles of oxygen}}{\text {total moles}},  

x_{oxygen}=\frac{3.0}{12.33}=0.243,  

p_{oxygen}=0.243\times 770torr=187torr  

Thus the partial pressure of oxygen is 187 torr.

6 0
2 years ago
Allena and Charissa were discussing whether Cl2 is an element or a compound. Allena said that it is an element because it has on
kobusy [5.1K]
Answer is: Allena is correct. It is an element because it is only made of chlorine atoms.
A chemical element bonded to an identical chemical element is not a chemical compound since it is made from only one element and not from two different elements. Chlorine is molecule, but not compound.
8 0
3 years ago
each element is designated by its __________ which is usually from the first letters of the elements name
klemol [59]

Answer:

each element is designated by its <u>Chemical Symbol </u>which is usually from the first letters of the elements name

Explanation:

7 0
2 years ago
Suppose you are a food chemist working for a company that makes and manufactures soda. Your job is to create a new soft drink wi
Mekhanik [1.2K]

Answer:

The answer to your question is given after the questions so I just explain how to get it.

Explanation:

a)

Get the molecular weight of Phosphoric acid

        H₃PO₄ =  (3 x 1) + (31 x 1) + (16 x 4)

                    = 3 + 31 + 64

                    = 98 g

         98 g -----------------  1 mol

      0.045 g ---------------   x

          x = (0.045 x 1) / 98

          x = 0.045 / 98

          x = 0.00046 moles or 4.6 x 10 ⁻⁴

b)

Molarity = \frac{moles}{volume}

Molarity = \frac{0.00046}{0.35}

Molarity = 0.0013 or 1.31 x 10⁻³

c)

Formula            C₁V₁ = C₂V₂

                              V₁ = C₂V₂ / C₁

Substitution

                              V₁ = (0.0013)(1) / 0.01

Simplification and result

                              V₁ = 0.0013 / 0.1

                              V₁ = 0.13 l = 130 ml            

7 0
3 years ago
Need help !!!!! ASAP
Ksivusya [100]
<h2>Hello!</h2>

The answer is:

We have that there were produced 0.120 moles of CO_{2}

n=0.120mol

<h2>Why?</h2>

We are asked to calculate the number of moles of the given gas, also, we  are given the volume, the temperature and the pressure of the gas, we can calculate the approximate volume using The Ideal Gas Law.

The Ideal Gas Law is based on Boyle's Law, Gay-Lussac's Law, Charles's Law, and Avogadro's Law, and it's described by the following equation:

PV=nRT

Where,

P is the pressure of the gas.

V is the volume of the gas.

n is the number of moles of the gas.

T is the absolute temperature of the gas (Kelvin).

R is the ideal gas constant (to work with pressure in mmHg), which is equal to:

R=62.363\frac{mmHg.L}{mol.K}

We must remember that the The Ideal Gas Law equation works with absolute temperatures (K), so, if we are given relative temperatures such as Celsius degrees or Fahrenheit degrees, we need to convert it to Kelvin before we proceed to work with the equation.

We can convert from Celsius degrees to Kelvin using the following formula:

Temperature(K)=Temperature(C\°) + 273K

So, we are given the following information:

Pressure=760mmHg\\Volume=2.965L\\Temperature=25.5C\°=25.5+273K=298.5K

Now, isolating the number of moles, and substituting the given information, we have:

PV=nRT

n=\frac{PV}{RT}

n=\frac{PV}{RT}

n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}

n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}\\\\n=\frac{2242mmHg.L}{18615.355\frac{mmHg.L}{mol.}}\\\\n=0.120mole

Hence, we have that there were produced 0.120 moles of CO_{2}

n=0.120mol

Have a nice day!

7 0
3 years ago
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