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vichka [17]
3 years ago
8

Graduated cylinder is filled to 41.5mL with water, and a price of granite is placed in the cylinder displacing the level to 47.6

mL. What is the volume of the granite piece in cubic centimeters
Chemistry
1 answer:
Y_Kistochka [10]3 years ago
8 0

Answer:

6.1 cm³

Explanation:

To solve this problem we first need to keep in mind <em>Archimedes' principle</em>:

  • The volume of water (or any fluid) displaced by a submerged object is equal to the object's volume.

With that in mind we <u>calculate the volume of the granite piece in mililiters</u>:

  • Volume displaced = 47.6 mL - 41.5 mL = 6.1 mL
  • Volume of the granite piece = 6.1 mL

Given that one cubic centimeter is equal to one mililiter, the volume of the granite piece in cm³ is 6.1 cm³.

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No

Explanation:

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It's the oesophagus.

Explanation:

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8 0
1 year ago
Consider a galvanic cell in which Al 3 + Al3+ is reduced to elemental aluminum and magnesium metal is oxidized to Mg 2 + Mg2+ .
olga nikolaevna [1]

Answer:

Anode:

3Mg(s) ----------> 3Mg2+(aq) + 6e

Cathode:

2Al3+(aq) +6e ---------> 2Al(s)

Explanation:

Anode:

3Mg(s) ----------> 3Mg2+(aq) + 6e

Cathode:

2Al3+(aq) +6e ---------> 2Al(s)

Magnesium is more electro positive than aluminum hence it functions as the anode. Six electrons are lost/gained in the redox process as shown in the oxidation and reduction half reaction equations above. Magnesium is oxidized to magnesium ion while aluminum is reduced to elemental aluminum.

5 0
3 years ago
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This question involves two calculations. The answer to the first part will be
ozzi

Answer :

(a) The mass of Al_2O_3 produced is, 15.2 grams.

(b) The percent yield of the reaction is, 72.5 %

Explanation :

Part (a) :

Given,

Mass of Al = 85.1 g

Molar mass of Al = 27 g/mol

First we have to calculate the moles of Al

\text{Moles of }Al=\frac{\text{Given mass }Al}{\text{Molar mass }Al}=\frac{85.1g}{27g/mol}=3.15mol

Now we have to calculate the moles of Al_2O_3

The balanced chemical equation is:

4Al+3O_2\rightarrow 2Al_2O_3

From the reaction, we conclude that

As, 4 moles of Al react to give 2 moles of Al_2O_3

So, 3.15 moles of Al react to give \frac{2}{4}\times 3.15=1.58 mole of Al_2O_3

Now we have to calculate the mass of Al_2O_3

\text{ Mass of }Al_2O_3=\text{ Moles of }Al_2O_3\times \text{ Molar mass of }Al_2O_3

Molar mass of Al_2O_3 = 102 g/mole

\text{ Mass of }Al_2O_3=(1.58moles)\times (102g/mole)=161.2g

Therefore, the mass of Al_2O_3 produced is, 161.2 grams.

Part (b) :

Now we have to calculate the percent yield of the reaction.

\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield = 116.9 g

Theoretical yield = 161.2 g

Now put all the given values in this formula, we get:

\text{Percent yield}=\frac{116.9g}{161.2g}\times 100=72.5\%

Therefore, the percent yield of the reaction is, 72.5 %

8 0
3 years ago
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