M(dextrose) = 50 g.
V(solution) = 1 L.
n(dextrose) = 50 g ÷ 180 g/mol.
n(dextrose) = 0,27 mol.
Osmotic concentration (osmolarity)<span> is a measure of how many </span><span>osmoles of particles of solute</span><span> it contains </span>per liter.
The osmolarity = n(dextrose) ÷ V(solution).
The osmolarity = 0,27 mol ÷ 1 L.
The osmolarity = 0,27 mol/L · 1000 mmol/m.
The osmolarity (dextrose) = 270 mosm/L.
The osmolarity (dextrose monohydrate) = 50 g÷197 g/mol·1000 =254mosm/L
Answer:
Metal
Explanation:
Na or sodium though soft... it is a metal...
Answer:
7.90×10²¹ formula units
Explanation:
From the question given above, the following data were obtained:
Mass of Cu(NO₃)₂ = 2.46 g
Formula units of Cu(NO₃)₂ =?
From Avogadro's hypothesis,
1 mole of Cu(NO₃)₂ = 6.02×10²³ formula units
Next, we shall determine the mass of 1 mole of Cu(NO₃)₂. This can be obtained as follow:
1 mole of Cu(NO₃)₂ = 63.5 + 2[14 + (3×16)]
= 63.5 + 2[14 + 48]
= 63.5 + 2[62]
= 63.5 + 124
= 187.5 g
Thus,
187.5 g of Cu(NO₃)₂ = 6.02×10²³ formula units
Finally, we shall determine the formula units contained in 2.46 g of Cu(NO₃)₂. This can be obtained as follow:
187.5 g of Cu(NO₃)₂ = 6.02×10²³ formula units.
Therefore,
2.46 g of Cu(NO₃)₂ =
(2.46 × 6.02×10²³)/187.5
= 7.90×10²¹ formula units
Thus, 2.46 g of Cu(NO₃)₂ contains 7.90×10²¹ formula units
In order to balance reactions, the number of each element must be the same on the left and right side. If I have 3 of element X on one side, I must have 3 of element X on the other side. That being said:
Zn + Fe(NO3)2 —> ZnNO3 + Fe
I have two NO3 groups on the left, so I will add a coefficient of 2 in front of the compound which has an NO3 group on the right side.
Zn + Fe(NO3)2 —> 2ZnNO3 + Fe
Since I now have 2 Zn’s on the right, and only 1 on the left, I will add a coefficient of 2 in front of Zn on the left.
2Zn + Fe(NO3)2 —> 2ZnNO3 + Fe
Lets check:
Zn: 2 right, 2 left (correct)
Fe: 1 right, 1 left (correct)
NO3 groups: 2 right, 2 left
We have successfully balanced the equation. :)
Answer:
Excess reagent is the ZnS. After the reaction is complete, 1.67 moles of sulfide remain.
Explanation:
This is the reaction to work with:
2ZnS + 3O₂ → 2ZnO + 2SO₂
Limiting reagent is the oxygen. We confirm it by a rule of three
2 moles of sulfide can react to 3 moles of O₂
Therefore 3 moles of ZnS will react to (3 . 3) / 2 = 4.5 moles (we need 4.5 moles of O₂ and we only have 2 moles, that's why the O₂ is the limiting)
Excess reagent is the zinc (II) sulfide
3 moles of oxygen react to 2 moles of ZnS
2 moles of O₂ will react to (2 . 2) / 3 = 1.33 moles of ZnS (it is ok, be cause we have 3 moles, and we need only 1.33)
After the reaction is complete ( 3 - 1.33) = 1.67 moles of sulfide remain.