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satela [25.4K]
3 years ago
12

A gas has density 2.41 g/liter at 25°C and 770 mm Hg. Calculate it's molecular mass (R = 0.0821 L atm.mol-1K-1 ​

Chemistry
1 answer:
Evgesh-ka [11]3 years ago
8 0

Answer:

Explanation:

Given : Density - 2.41 g/liter  

Temperature - 25° C  

Pressure : 770 mm Hg  

R = 0.0821 L atm mol-¹K-¹

 

Find : Molecular mass of gas

Solution : Ideal gas equation with respect to density will be : PM = dRT. In the formula, P is pressure, M is molecular mass, d is density, R is gas constant and T is temperature.

Keeping the values in equation-  

Pressure : 770 mm Hg = 1 atm  

Temperature : 273 + 25 = 298 K

 

M = dRT/P  

M = (2.41*0.0821*298)/1  

M = 58.96 gram/mol

Thus, the molecular mass of gas is 58.96 gram/mol.

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How many milliliters of a 0.266 m lino3 solution are required to make 150.0 ml of 0.075 m lino3 solution?
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We need an equation that would relate the concentration of the original solution to that of the desired solution. To solve this we use the equation expressed as follows, 

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Therefore, we need about 42.29 mL of the 0.266 M of lithium nitrate solution to make 150.0 mL of the 0.075 M lithium nitrate solution.

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What quantity of heat is required to raise the temperature of 460g of aluminum from 15C to 85C?
Hoochie [10]

Answer:

Q = 28.9 kJ

Explanation:

Given that,

Mass of Aluminium, m = 460 g

Initial temperature, T_i=15^{\circ} C

Final temperature, T_f=85^{\circ}

We know that the specific heat of Aluminium is 0.9 J/g°C. The heat required to raise the temperature is given by :

Q=mc\Delta T\\\\Q=mc(T_f-T_i)\\\\Q=460\ g\times 0.9\ J/g^{\circ} C\times (85-15)^{\circ} C\\\\Q=28980\ J\\\\\text{or}\\\\Q=28.9\ kJ

So, 28.9 kJ of heat is required to raise the temperature.

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The partial pressure of CO2 gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO2 gas (in g) will be released f
frutty [35]

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. We can calculate the concentration of CO₂ using Henry's law.

C = k \times P = \frac{1.65 \times 10^{-3} M }{atm}  \times 4.60 atm = 7.59 \times 10^{-3} M

We can calculate the mass of CO₂ in 1.1 L considering its molar mass is 44.01 g/mol.

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Now, we will repeat the same procedure for a partial pressure of 1.28 atm.

C = k \times P = \frac{1.65 \times 10^{-3} M }{atm}  \times 1.28 atm = 2.11 \times 10^{-3} M

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The mass of CO₂ released will be equal to the difference in the masses at the different pressures.

m = 0.367 g - 0.102 g = 0.265 g

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

Learn more: brainly.com/question/18987224

<em>The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO₂ gas (in g) will be released from 1.1 L of the carbonated water when the partial pressure of CO2 is lowered to 1.28 atm? At 25 ºC, the Henry’s law constant for CO₂ dissolved in water is 1.65 x 10⁻³ M/atm, and the density of water is 1.0 g/cm³.</em>

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