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timofeeve [1]
3 years ago
15

What will be the pressure of 2.00 mol of an ideal gas at a temperature of 20.5 degrees Celsius and a volume of 62.3L?

Chemistry
1 answer:
erastova [34]3 years ago
7 0

.774atm

First, look at what you have and look at the equations you can use to solve this problem. The best equation would be PV=nRT.

P being pressure, V being volume, n being moles, R being the gas constant, and T being temperature.

Before you start doing any of the math, make sure of two things. Since you're looking for pressure, you'll need a gas constant. When I did the problem, I used the gas constant of atm or atmospheres which is .0821.

Also! Remember to always convert celsius into kelvin, to do this, add 273 to the given celsius degree. After this is all set and done, your equation should look like this:

P = \frac{2 x .0821 x 293.5}{62.3}

The reason that the equation is divided by the volume is due to the fact that you need to isolate the variable or pressure.

Multiply everything on the top and divide by the bottom and you should receive the final answer of .774atm.

Hope this helps!

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Answer:

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<h2>250 g</h2>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

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volume of object = 25 mL

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The mass of the object is

mass = 25 × 10

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A mixture of calcium carbonate, CaCO3, and barium carbonate, BaCO3, weighing 5.40 g reacts fully with hydrochloric acid, HCl(aq)
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Answer:

CaCO₃ = 85.18%

BaCO₃ = 14.82%

Explanation:

The acid will react with the salts, the partial reactions are:

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

BaCO₃ + 2HCl → BaCl₂ + CO₂ + H₂O

So, the total amount of CaCO₃ BaCO₃ will form the CO₂.

Using the ideal gas law to calculate the number of moles of CO₂:

PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.082 atm*L/mol*K), and T is the temperature (50ºC + 273 = 323 K).

0.904*1.39 = n*0.082*323

26.486n = 1.25656

n = 0.05 mol

So, the number of moles of the mixture is 0.05 mol.

The molar masses of the components are:

CaCO₃ = 40 g/mol of Ca + 12 g/mol of C + 3*16 g/mol of O = 100 g/mol

BaCO₃ = 137.3 g/mol of Ba + 12 g/mol of C + 3*16 g/mol of O = 197.3 g/mol

Let's call x the number of moles of CaCO₃ and y the number of moles of BaCO₃, so:

100x + 197.3y = 5.4

x + y = 0.05 mol

y = 0.05 - x

100x + 197.3*(0.05 - x) = 5.4

100x - 197.3x = 5.4 - 9.865

97.3x = 4.465

x = 0.046 mol of CaCO₃

y = 0.004 mol of BaCO₃

So, the masses are:

CaCO₃ = 100* 0.046 = 4.60 g

BaCO₃ = 137.3*0.004 = 0.80 g

The percentages in the mixture are:

CaCO₃ = (4.60/5.40)*100% = 85.18%

BaCO₃ = (0.80/5.40)*100% = 14.82%

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