Answer:
0.209 mol/L
Explanation:
Given data
- Mass of copper(lI) sulfate (solute): 11.7 g
- Volume of solution: 350 mL = 0.350 L
The molar mass of copper(Il) sulfate is 159.61 g/mol. The moles corresponding to 11.7 grams are:
11.7 g × (1 mol/159.61 g) = 0.0733 mol
The molarity of copper(Il) sulfate is:
M = moles of solute / liters of solution
M = 0.0733 mol / 0.350 L
M = 0.209 mol/L
Answer:
Sodium (Na): (.5 point)
22.99+35.453=58.433. 22.99/58.433= 39%
Chlorine (Cl): (.5 point)
22.99+35.453=58.433. 35.453/58.433= 61%
Explanation:
Molar mass:
O2 = 16 x 2 = 32.0 g/mol Mg = 24 g/mol
<span>2 Mg(s) + O2(g) --->2 MgO(s)
</span>
2 x 24.0 g Mg -------------> 32 g O2
5.00 g Mg -----------------> ( mass of O2)
mass of O2 = 5.00 x 32 / 2 x 24.0
mass of O2 = 160 / 48
= 3.33 g of O2
hope this helps!
Answer:
The answer is 1.15m.
Since molality is defined as moles of solute divided by kg of solvent, we need to calculated the moles of H2SO4 and the mass of the solvent, which I presume is water.
We can find the number of H2SO4 moles by using its molarity
C=nV→nH2SO4=C⋅VH2SO4=6.00molesL⋅48.0⋅10−3L=0.288
Since water has a density of 1.00kgL, the mass of solvent is
m=ρ⋅Vwater=1.00kgL⋅0.250L=0.250 kg
Therefore, molality is
m=nmass.solvent=0.288moles0.250kg=1.15m
