Suppose the gas is ideal gas. According to the ideal gas equation of state, pV = nRT, the pressure is unchanged. So Vi/Ti = nR/p = Vn/Tn. (i stands for initial and n stands for new) So the new volume Vn = 3.67 L
The ideal gas equation is pV = nRT
From that you can derive several equations, depending on which variables are fixed.
1) When n and T are fixed:
pV = nRT = constant
pV = constant => p1 V1 = p2 V2 => p1 / V2 = p2 / V1 ---> Boyle's Law
2) When n and V are constant:
p / T = nR/V = constant
p / T = constant => p1 / T1 = p2 / T2 ----> Gay - Lussac's Law
3) when n and p are constant
V / T = nR/p = constant
V / T = constant => V1 / T1 = V2 / T2 ---> Charles' Law
4) When only n is constant
pV / T = nR = constant
pV / T = constant => p1 V1 / T1 = p2 V2 / T2 ----> Combined gas law.
There you have the four equations that agree with the ideal gas law.
Answer:
15. 2.66 moles .
16. 2.09L.
Explanation:
Molarity of a solution is simply defined as the mole of solute per unit litre of the solvent. Mathematically, it is represented as:
Molarity = mole /Volume.
With the above formula, let us answer the questions given above
15. Data obtained from the question include the following:
Volume of solution = 1.4L
Molarity = 1.9M
Mole of solute =.?
Molarity = mole /Volume
1.9 = mole / 1.4
Cross multiply
Mole = 1.9 x 1.4
Mole = 2.66 moles
Therefore, the mole of the solute present in the solution is 2.66 moles.
16. Data obtained from the question include the following:
Mole of solute = 0.46 mole
Molarity = 0.22M
Volume of solvent (water) =.?
Molarity = mole /Volume
0.22 = 0.46/Volume
Cross multiply
0.22 x Volume = 0.46
Divide both side 0.22
Volume = 0.46/0.22
Volume = 2.09L
Therefore, 2.09L of water is required.
the solid particles take up the intermolecular spaces in the liquid.
In formation of a Type II Binary Compound, the metal atom present is<span>
NOT</span> found in either Group 1 or Group 2 on the periodic table. For the choices, Ba is under Group 2 on the periodic table, which makes it the atom not involved in formation of type II compounds. The answer is B.
