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Answer:
th relative ability of a solute to devolve into a solvent
The molarity of NaBr = 0.05036M
<h3>
calculation</h3><h3>Step1: calculate the total mass</h3>
0.0460 m NaBr has 0.0460 moles NaBr with 1000g of solvent
therefore the mass of NaBr= 0.0460 moles x 102.89 g/mol =4.73grams
total mass is therefore= 4.733 g +1000= 1004.73 grams
step 2:
use the density to determine the volume
vol=mass/density
=1004.73 g/1.10 g/mol=913.39 ml in liters= 913.39/1000=0.91339 L
now the molarity = moles/vol in liters
=0.0460 m/0.91339=0.05036 M
Molar mass O2 = 31.99 g/mol
Molar mass CO2 = 44.01 g/mol
Moles ratio:
<span>C3H8 + 5 O2 = 3 CO2 + 4 H2O
</span>
5 x 44.01 g O2 ---------------- 3 x 44.01 g CO2
( mass of O2) ------------------ 37.15 g CO2
mass of O2 = 37.15 x 5 x 44.01/ 3 x 44.01
mass of O2 = 8174.8575 / 132.03
mass of O2 = 61.916 g
Therefore:
1 mole O2 ----------------- 31.99 g
moles O2 -------------------- 61.916
moles O2 = 61.916 x 1 / 31.99
moles = 61.916 / 31.99 => 1.935 moles of O2
Answer:
a. 1.728 moles.
b. 262.7g of Cr₂O₃ are required
Explanation:
Based on the reaction:
Cr₂O₃(s) + 3H₂S(g) → Cr₂S₃(s) + 3H₂O(l)
The important thing in the reaction is that 1 mole of Cr₂O₃ produce 1 mole of Cr₂S₃
a. To produce 346g of Cr₂S₃ we must know how many moles of Cr₂S₃ must be produced, and, as 1 mole of Cr₂O₃ produce 1 mole of Cr₂S₃ we can know moles of Cr₂O₃ that are required.
<em>Moles of 346g Cr₂S₃ (Molar mass: 200.19g/mol):</em>
346g Cr₂S₃ * (1mol / 200.19g) = 1.728 moles of Cr₂S₃
Based on the reaction, moles of Cr₂O₃ that are required are
<h3>
1.728 moles of Cr₂O₃</h3>
b. Again, to conver the 1.728 moles of Cr₂O₃ to grams we must use molar mass of Cr₂O₃ (151.99g/mol):
1.728 moles Cr₂O₃ * (151.99g / mol) =
<h3>262.7g of Cr₂O₃ are required</h3>