I am a metalloid. I have 7 valence electrons. I have 6 energy levels.

Chemistry

1 answer:

Alexxx [7]3 years ago

3 0

1 electron Bc of the electron you have

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Which three of the following are steps of aerobic cellular respiration?

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A.
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3 years ago

You need a 20% alcohol solution. on hand, you have a 240 ml of a 15% alcohol mixture. you also have 40% alcohol mixture. how muc

juin [17]

0.15*240=36 ml of alcohol in <span>240 ml of a 15% alcohol mixture
0.4x = </span>ml of alcohol in x ml of a 40% alcohol mixture
0.2(x+240)= ml of alcohol in (x+240) ml of a 20% alcohol mixture

0.15*240 + 0.4x = 0.2(x+240)
36+0.4x=0.2x+48
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Calculate the molar mass of the acid.

JulijaS [17]

Citric acid has the molecular formula C6H8O7 so you can add the molar masses of the elements from the periodic table. C has a molar mass of 12.01 g/mol, H has 1.01 g/mol and O has 15.999 g/mol. Now you calculate the total molar mass= (6*12.01 + 8*1.01 + 7*15.999). This yields a molar weight of 192.124 g/mol (anhydrous)

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Oxygen and hydrogen are kept under the same temperature and pressure. Oxygen molecule is 16 times more massive than hydrogen one

lozanna [386]

Explanation:

When we increase the temperature of a substance then there will occur an increase in the kinetic energy of its molecules.

Also, K.E = $\frac{3}{2}kT$

So, kinetic energy is directly proportional to the temperature.

Hence, when temperature and pressure are kept the same for both oxygen and hydrogen gas then values of their kinetic energy will be the same irrespective of their masses.

Thus, we can conclude that kinetic energy of oxygen molecule is the same as compared to hydrogen molecule.

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8. Estimate the maximum volume percent of Methanol vapor that can exist at standard conditions. Vapor pressure = 88.5 mm Hg in a

sashaice [31]

Explanation:

Vapor pressure is defined as the pressure exerted by vapors or gas on the surface of a liquid.

It is known that at standard condition, vapor pressure is 760 mm Hg.

And, it is given that methanol vapor pressure in air is 88.5 mm Hg.

Hence, calculate the volume percentage as follows.

Volume percentage = $\frac{\text{given vapor pressure}}{\text{standard vapor pressure}} \times 100$

= $\frac{88.5}{760} \times 100$

= 11.65%

Thus, we can conclude that the maximum volume percent of Methanol vapor that can exist at standard conditions is 11.65%.

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3 years ago