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AURORKA [14]
3 years ago
7

Trimix 10/50 is a gas mixture that contians 10% oxygen and 50% helium, and the rest is nitrogen. If a tank of trimix 10/50 has a

total pressure of 2.07 x 104 kPa, then what is the partial pressure of helium?
Chemistry
1 answer:
Nataly_w [17]3 years ago
5 0

Answer: 1.61 x 10⁴ kPa

Dalton's law <u>states that the sum of the partial pressures of each gas equals the total pressure of the gas mixture.</u> According to this law,

Pi = xi P

where Pi is the partial pressure of the gas i, xi is the mole fraction of the gas i in the gas mixture and P is the total pressure.

The mole fraction <u>is defined as the quotient between the moles of solute (ni) and the total moles of the mixture (nt)</u>, which is calculated by adding the moles of all its components:

xi = \frac{n_{i} }{n_{t} }

In the Trimix 10/50 mix you have 10% oxygen, 50% helium and 40% nitrogen.

To calculate the total number of moles of the mixture and thus determine the molar fraction of helium, we consider 100 g and calculate the number of moles that represent 10 g of O₂ (n₁), 50 g of He (n₂) and 40 g of N₂ (n₃):

n₁ =  10 g x \frac{1 mol}{31.998 g} = 0.313 mol

n₂ =  50 g x \frac{1 mol}{8.005 g} = 6.246 mol

n₃ =  40 g x \frac{1 mol}{28.013 g} = 1.428 mol

Then the total number of moles (nt) will be:

nt = n₁ + n₂ + n₃ = 0.313 mol + 6.246 mol +1.428 mol

nt = 7,987 mol

Then, the mole fraction of helium (x₂) in the mixture will be,

x₂ =  \frac{6.246 mol}{7.987 mol} = 0.78

and the partial pressure of helium in the mixture, according to Dalton's law, will be:

P₂ = x₂ P = 0.78 x 2.07 x 10⁴ kPa

P₂= 1.61 x 10⁴ kPa

So, <u>the partial pressure of helium if a tank of trimix 10/50 has a total pressure of 2.07 x 104 kPa is  1.61 x 10⁴ kPa</u>

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lutik1710 [3]

Answer:

C_{metal}=126.6\frac{J}{g\°C}

Explanation:

Hello!

In this case, when two substances at different temperature are put in contact and an equilibrium temperature is attained, we can evidence that the heat lost by the hot substance (metal) is gained by the cold substance (water) and we can write:

Q_{metal}=-Q_{water}

Which can be also written as:

m_{metal}C_{metal}(T_{EQ}-T_{metal})=-m_{water}C_{water}(T_{EQ}-T_{water})

Thus, since we need the specific heat of the metal, we solve for it as shown below:

C_{metal}=\frac{m_{water}C_{water}(T_{EQ}-T_{water})}{-m_{metal}(T_{EQ}-T_{metal})} \\\\C_{metal}=\frac{250.0g*4.184\frac{J}{g\°C}(29.0\°C-98.0\°C)}{95.0g(29.0\°C-23.0\°C)} \\\\C_{metal}=126.6\frac{J}{g\°C}

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7 0
2 years ago
Determine the empirical formula of the following compound if a sample contains 0.104 molK, 0.052 molC, and 0.156 molO;?
faltersainse [42]

Answer:

K₂CO₃    

Explanation:

Given parameters:

Number of moles of K = 0.104mol

Number of moles of C = 0.052mol

Number of moles of O = 0.156mol

Method

From the given parameters, to calculate the empirical formula of the elements K, C and O, we reduce the given moles to the simplest fraction.

Empirical formula is the simplest formula of a compound and it differs from the molecular formula which is the actual formula of a compound.

  • Divide the given moles through by the smallest which is C, 0.052mol.
  • Then approximate values obtained to the nearest whole number of multiply by a factor to give a whole number ratio.
  • This is the empirical formula

Solution

Elements                             K                       C                    O

Number of moles            0.104                0.052            0.156

Dividing by the

smallest                       0.104/0.052     0.052/0.052  0.156/0.052

                                            2                           1                     3

The empirical formula is K₂CO₃      

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