Answer:
Yes, it is possible. Let us consider an example of two solutions, that is, solution A having 20 percent mass RbCl (rubidium chloride) and solution B is having 15 percent by mass NaCl or sodium chloride.
It is found that solution A is having more concentration in comparison to solution B in terms of mass percent. The formula for mass percent is,
% by mass = mass of solute/mass of solution * 100
Now the formula for molality is,
Molality = weight of solute/molecular weight of solute * 1000/ weight of solvent in grams
Now molality of solution A is,
m = 20/121 * 1000/80 (molecular weight of RbCl is 121 grams per mole)
m = 2.07
Now the molality of solution B is,
m = 15/58.5 * 1000/85
m = 3.02
Therefore, in terms of molality, the solution B is having greater concentration (3.02) in comparison to solution A (2.07).
Answer:
The answer to your question is 242 ml
Explanation:
Data
HI 0.211 M Volume = x
KMnO₄ 0.354 M Volume = 24 ml
Balanced Chemical reaction
12HI + 2KMnO₄ + 2H₂SO₄ → 6I₂ + Mn₂SO₄ + K₂SO₄ + 8H₂O
Process
1.- Calculate the moles of KMnO₄ 0.354 M in 24 ml
Molarity = moles / volume (L)
moles = Molarity x volume (L)
moles = 0.354 x 0.024
moles = 0.0085
2.- From the balanced chemical reaction we know that HI and KMnO₄ react in the proportion 12 to 2. Then,
12 moles of HI --------------- 2 moles of KMnO₄
x --------------- 0.0085 moles of KMnO₄
x = (0.0085 x 12)/2
x = 0.051 moles of HI
3.- Calculate the milliliters of HI 0.211 M
Molarity = moles/volume
Volume = moles/molarity
Volume = 0.051/0.211
Volume = 0.242 L or Volume = 242 ml
Hey there,
Answer:
4 valence electrons.
Hope this helps :D
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Answer:
false. acids do not neutralize bases.
Explanation:
Answer:
ExplanatioFirst let us compute for the molar mass of Mg(OH)2.
molar mass = 24.3 + 2 (16) + 2(1.0)
molar mass = 58.3 g/mol
So the mass is then:
mass = 3.2 mol * (58.3 g/mol)
mass = 186.56 gramsn:
