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noname [10]
3 years ago
14

In metallic bonds, the mobile electrons surrounding the positive ions are called a(n)?

Chemistry
1 answer:
Anastaziya [24]3 years ago
6 0

Answer: In metallic bonds, the mobile electrons surrounding the positive ions are called <u><em>dipole</em></u>.

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Which method of separation would be most appropriate for separating a mixture of water and alcohol?
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Answer:C. distillation

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Compounds having  different boiling points when  mixed can be separated by Fractional distillation.

Now when there is a mixture of  water and alcohol, Upon heat, we would see that the alcohol will boil at a lower temperature than water  for eg, ethanol will boil at 78.5 degrees Celsius, vaporizes and  distills into the fractionating columns leaving the  water which we know boils at 100 degrees  Celsius.

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2 years ago
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3 years ago
A light year measure the distance a lightbeam can travel in one year one light year is equivalent to 9500000000000 km why do sci
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8 0
3 years ago
Consider the balanced equation for the following reaction:
Zanzabum

<u>Answer:</u> The amount of carbon dioxide formed in the reaction is 5.663 grams

<u>Explanation:</u>

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

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

Given mass of oxygen gas = 8 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{8g}{32g/mol}=0.25mol

For the given chemical equation:

7O_2(g)+2C_2H_6(g)\rightarrow 4CO_2(g)+6H_2O(l)

By Stoichiometry of the reaction:

7 moles of oxygen gas produces 4 moles of carbon dioxide

So, 0.25 moles of oxygen gas will produce = \frac{4}{7}\times 0.25=0.143mol of carbon dioxide

Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide = 0.143 moles

Putting values in equation 1, we get:

0.143mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(0.143mol\times 44g/mol)=6.292g

To calculate the experimental yield of carbon dioxide, we use the equation:

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

Percentage yield of carbon dioxide = 90 %

Theoretical yield of carbon dioxide = 6.292 g

Putting values in above equation, we get:

90=\frac{\text{Experimental yield of carbon dioxide}}{6.292g}\times 100\\\\\text{Experimental yield of carbon dioxide}=\frac{90\times 6.292}{100}=5.663g

Hence, the amount of carbon dioxide formed in the reaction is 5.663 grams

7 0
3 years ago
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