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3 years ago

2. The Density of Water is 1.0000g/cm3. Based on this property of water,

Chemistry

1 answer:

Vanyuwa [196]3 years ago

7 0

Answer:

if the object sank then that object has a greater density then water. if the object floated then its density is lower then water.

Explanation:

lets say object 1 has a density of 24/cm3. the density is greater then water (1.0000g/cm3) so it would sink. now lets say object 2 has a density of 0.79383g/cm3 since it's less then the density of water (1.0000g/cm3) it would float.

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Draw the other possible resonance structure of each organic ion. In each case, draw the structure that minimizes formal charges.

pshichka [43]

Three resonance structures can be drawn for the allyl cation while two resonance structures can be drawn for the amidate ion.

Sometimes, we cannot fully describe the bonding in a chemical specie using a single chemical structure. In such cases, we have to use a number of structures which cooperatively represent the actual bonding in the molecule. These structures are called resonance or canonical structures.

The resonance structures of the allyl cation and the amidate ion are shown in the images attached to this answer. These structures show the different bonding extremes in these organic ions.

Learn more: brainly.com/question/4933048

7 0

3 years ago

Consider a galvanic cell consisting of the following two redox couplesAlB+ 3e Al , Mga + 2e Mg, (ag. E® = 1.676 V ) E° = 2.356 V

malfutka [58]

Answer:

0.68 V

Explanation:

For anode;

3Mg(s) ---->3Mg^2+(aq) + 6e

For cathode;

2Al^3+(aq) + 6e -----> 2Al(s)

Overall balanced reaction equation;

3Mg(s) + 2Al^3+(aq) ----> 3Mg^2+(aq) + 2Al(s)

Since

E°anode = -2.356 V

E°cathode = -1.676 V

E°cell=-1.676 -(-2.356)

E°cell= 0.68 V

4 0

3 years ago

three students made multiple weighings of a copper cylinder each using a different balance. Describe the accuracy and precision

ch4aika [34]

I think that you have put up an incomplete question. However, i am answering the question based on my research and knowledge.

Lissa- accuracy and precision are both low
Lamont- accuracy and precision are definitely high
Leigh Anne- accuracy is low but precision is definitely high.

I hope that this is the answer that you were looking for and the answer has definitely come to your desired help.

3 0

3 years ago

What is a chemical reaction. in your own words but be more detailed about what you say.

harkovskaia [24]

Answer:

A chemical reaction is the change in chemical form rather than physical due to an outside force. This can come from something as simple as a change in temerature to as large as a specific element or compound.

8 0

3 years ago

For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi

natka813 [3]

Answer: The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

For Iron:

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:

$\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol$

For the given chemical reaction:

$2Fe(s)+O_2(g)\rightarrow 2FeO(s)$

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = $\frac{2}{2}\times 0.0771=0.0771mol$ of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:

$0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g$

To calculate the percentage yield of iron (ii) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

$\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%$

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0

3 years ago