When a solute is completely dissolved in a solvent, a solution is formed.
Answer:
68133080.02 g
Explanation:
I believe that the question is to find the mass of air in the room and not the molar mass of air since the molar mass of air was already given in the question as 28.97 g/mol.
Now, if 1 mole of a gas occupies 22.4 L
x moles of air occupies 52,681,428.8 Liters
x = 1 * 52,681,428.8 /22.4
x = 2351849.5 moles of air
Now, number of moles = mass/ molar mass
but molar mass = 28.97 g/mol
2351849.5 = mass/28.97
mass = 2351849.5 * 28.97
mass = 68133080.02 g
Diffusion, facilitated diffusion, filtration, and osmosis
Answer:
M of HI = 5.4 M.
Explanation:
- We have the rule: at neutralization, the no. of millimoles of acid is equal to the no. of millimoles of the base.
<em>(XMV) acid = (XMV) base.</em>
where, X is the no. of (H) or (OH) reproducible in acid or base, respectively.
M is the molarity of the acid or base.
V is the volume of the acid or base.
<em>(XMV) HI = (XMV) Ca(OH)₂.</em>
For HI; X = 1, M = ??? M, V = 25.0 mL.
For Ca(OH)₂, X = 2, M = 1.5 M, V = 45.0 mL.
<em>∴ M of HI = (XMV) Ca(OH)₂ / (XV) HI</em> = (2)(1.5 M)(45.0 mL) / (1)(25.0 mL) = <em>5.4 M.</em>
