Answer:
There are 0.93 g of glucose in 100 mL of the final solution
Explanation:
In the first solution, the concentration of glucose (in g/L) is:
15.5 g / 0.100 L = 155 g/L
Then a 30.0 mL sample of this solution was taken and diluted to 0.500 L.
- 30.0 mL equals 0.030 L (Because 30.0 mL ÷ 1000 = 0.030 L)
The concentration of the second solution is:
So in 1 L of the second solution there are 9.3 g of glucose, in 100 mL (or 0.1 L) there would be:
1 L --------- 9.3 g
0.1 L--------- Xg
Xg = 9.3 g * 0.1 L / 1 L = 0.93 g
Yes, because the net ionic is equation will yield BaCO3 as a precipitate because it is insoluble in water
Answer:
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Explanation:
<u>1. Balanced molecular equation</u>
<u>2. Mole ratio</u>
<u>3. Moles of HNO₃</u>
- Number of moles = Molarity × Volume in liters
- n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃
<u>4. Moles Ba(OH)₂</u>
- n = 0.700M × 0.0310 liter = 0.0217 mol
<u>5. Limiting reactant</u>
Actual ratio:
Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.
Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.
<u>6. Final molarity of Ba(OH)₂</u>
- Molarity = number of moles / volume in liters
- Molarity = 0.0187 mol / (0.0100 + 0.0031) liter = 0.456M
I'm pretty sure it's 9726 milligrams of iodine. Hope this helps.
Answer:
The answer to your question is 27 g of Al
Explanation:
Data
mass of Al = ?
moles of Al₂O₃ = 0.5
The correct formula for the product is Al₂O₃
Balanced chemical reaction
4Al + 3O₂ ⇒ 2Al₂O₃
Process
1.- Calculate the molar mass of the product
Al₂O₃ = (27 x 2) + (16 x 3)
= 54 + 48
= 102 g
2.- Convert the moles of Al₂O₃ to grams
102 g ---------------- 1 mol
x ---------------- 0.5 moles
x = (0.5 x 102) / 1
x = 51 g of Al₂O₃
3.- Use proportions to calculate the mass of Al
4(27) g of Al --------------- 2(102) g of Al₂O₃
x --------------- 51 g
x = (51 x 4(27)) / 2(102)
x = 5508 / 204
x = 27 g of Al
