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2 years ago

5

A student pours 800.0 mL of a 3.000 molar solution of sodium hydroxide into a 2.00 liter volumetric flask and fills the flask up

with water. What is the new molarity of the solution?
A) 12.00 M
B) 1.20 M
C) 14.00 M
D) 0.0750 M

1 answer:

Rus_ich [418]2 years ago

8 0

Dont mind this but dont click on the lino

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How many formula units are contained in 2.35mol of Iron (III) Oxide

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Remember, 1 mole= 6.022x10^23 atoms, molecules, or formula units.

Answer is 1.42x10^24

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2 years ago

Match each element with the element that is most similar to it. If you don’t know the answer, use this periodic table to help yo

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3 years ago

At STP how many molecules of nitrogen gas are in 224 L?

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Answer:

6.022×10²² molecules

Explanation:

Given data:

Volume of nitrogen = 224 L

Pressure = standard = 1 atm

Temperature = standard = 273 K

Number of molecules = ?

Solution:

PV = nRT

1 atm × 224 L = n × 0.0821 atm.L/mol.K × 273 K

224 atm.L = n ×22.41 atm.L/mol

n = 224 atm.L/22.41 atm.L/mol

n = 10 mol

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2 years ago

Which gas is least likely to obey the ideal gas laws at very high pressures and very low temperatures?

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3 years ago

One mole of an ideal gas is contained in a cylinder with a movable piston. The temperature is constant at 77°C. Weights are remo

frozen [14]

Answer:

The total work is 4957.45J

Explanation:

For an ideal gas, at constant temperature the definition of work (W) is

$W = - P.dV = - n.R.T \int\limits^i_f {\frac{dV}{V} \\W = - n.R.T. Ln (\frac{V_{f}}{V_{I}})\\W = - n.R.T. Ln (\frac{P_{i}}{P_{f}})$

where P is the pressure, V the volume, n the moles number, T the temperature and R the gas constant.

To solve the problem is necessary to replace the two steps in the equation

Stape 1: n = 1 mol, R = 0.082atm.L/K.mol, T = 77ºC = 350K, Pi = 5.50atm and Pf = 2.43atm.

$W_{1} = - 1molx0.082\frac{atm.L}{K.mol}x350KxLn (\frac{5.50atm}{2.43atm}) = 23.44atm.Lx101.325\frac{J}{atm.L} =2375.44J$

Stape 2: n = 1 mol, R = 0.082atm.L/K.mol, T = 77ºC = 350K, Pi = 2.43atm and Pf = 1.00atm.

$W_{2} = - 1molx0.082\frac{atm.L}{K.mol}x350KxLn (\frac{2.43atm}{1.00atm}) = 25.48atm.Lx101.325\frac{J}{atm.L} =2582.01J$

The total work is the sum of the two steps

$W = W_{1} + W_{2} = 2375.44J + 2582.01J = 4957.45J$

4 0

3 years ago