Answer:
Mass = 8.46 g
Explanation:
Given data:
Mass of water produced = ?
Mass of glucose = 20 g
Mass of oxygen = 15 g
Solution:
Chemical equation:
C₆H₁₂O₆ + 6O₂ → 6H₂O + 6CO₂
Number of moles of glucose:
Number of moles = mass/molar mass
Number of moles = 20 g/ 180.16 g/mol
Number of moles = 0.11 mol
Number of moles of oxygen:
Number of moles = mass/molar mass
Number of moles = 15 g/ 32 g/mol
Number of moles = 0.47 mol
now we will compare the moles of water with oxygen and glucose.
C₆H₁₂O₆ : H₂O
1 : 6
0.11 : 6/1×0.11 = 0.66
O₂ : H₂O
6 : 6
0.47 : 0.47
Less number of moles of water are produced by oxygen thus it will limit the yield of water and act as limiting reactant.
Mass of water produced:
Mass = number of moles × molar mass
Mass = 0.47 mol ×18 g/mol
Mass = 8.46 g
The correct answer is: [D]: " milk " .
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Choice [A]: "soil" is incorrect; since "soil" is "heterogeneous" {composed of many different "ingredients" .].
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Same with "Choice [B]: "granola" [composed of many different ingredients—clumps of sweetened oats, raisins, coconuts, etc.].
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Same with "Choice [C]: "salad dressing". {Notice how we usually have to "shake the bottle" ? Composed of multiple ingredients, (e.g. oil, vinegar, and spices, or oil and other spices, and more ingredientes).
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Choice: [D]: "milk", as a liquid, is a single, well-mixed, uniform, mixture; as such, it is "homogeneous". Note: "homo-" means "same".
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The correct answer is A, B and C
Answer:
The answer to your question is: SiCl₄
Explanation:
Data
amount of Si 1.71 g
amount of Cl 8.63 g
MW Si = 28 g
MW Cl = 35.5
Process (rule of three)
For Si For Cl
28 g of Si ------------------ 1 mol 35.5 g of Cl --------------- 1 mol
1.71g of Si --------------- x 8.63 g of Cl -------------- x
x = 1.71 x 1 / 28 = 0.06 mol x = 8.63 x 1 / 35.5 = 0.24 mol
Now, divide both results by the lowest of them.
Si = 0.06 mol / 0.06 = 1 molecule of Si Cl = 0.24 / 0.06 = 4 molecules of Cl
Finally
Si₁ Cl₄ or SiCl₄
When E° cell is an electrochemical cell which comprises of two half cells.
So,
when we have the balanced equation of this half cell :
Al3+(aq) + 3e- → Al(s) and E°1 = -1.66 V
and we have also this balanced equation of this half cell :
Ag+(aq) + e- → Ag(s) and E°2 = 0.8 V
so, we can get E° in Al(s) + 3Ag (aq) → Al3+(aq) + 3Ag(s)
when E° = E°2 - E°1
∴E° =0.8 - (-1.66)
= 2.46 V
∴ the correct answer is 2.46 V
