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3 years ago

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1. The fourth principle electronic level has:

1 answer:

attashe74 [19]3 years ago

5 0

Answer:

the fourth level can hold up to 32 electrons: 2 in the s orbital, 6 in the three p orbitals, 10 in the five d orbitals, and 14 in the seven f orbitals.

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Which of the following is a type of kinetic energy.

dexar [7]

Kinetic energy is energy in motion. B, a rolling ball would be your answer because a rolling ball is energy that is moving. The rest of the answers are wrong because the actions do not use kinetic energy.

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2 years ago

Calculate the density of air at 100 Deg C and 1 bar abs. Use the Ideal Gas Law for your calculation and give answer in kg/m3. Us

madam [21]

Answer:

$d=0.92\frac{kg}{m^{3}}$

Explanation:

Using the Ideal Gas Law we have $PV=nRT$ and the number of moles n could be expressed as $n=\frac{m}{M}$ , where m is the mass and M is the molar mass.

Now, replacing the number of moles in the equation for the ideal gass law:

$PV=\frac{m}{M}RT$

If we pass the V to divide:

$P=\frac{m}{V}\frac{RT}{M}$

As the density is expressed as $d=\frac{m}{V}$ , we have:

$P=d\frac{RT}{M}$

Solving for the density:

$d=\frac{PM}{RT}$

Then we need to convert the units to the S.I.:

$T=100^{o}C+273.15$

$T=373.15K$

$P=1bar*\frac{0.98atm}{1bar}$

$P=0.98atm$

$M=28.9\frac{kg}{kmol}*\frac{1kmol}{1000mol}$

$M=0.0289\frac{kg}{mol}$

Finally we replace the values:

$d=\frac{(0.98atm)(0.0289\frac{kg}{mol})}{(0.082\frac{atm.L}{mol.K})(373.15K)}$

$d=9.2*10^{-4}\frac{kg}{L}$

$d=9.2*10^{-4}\frac{kg}{L}*\frac{1L}{0.001m^{3}}$

$d=0.92\frac{kg}{m^{3}}$

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3 years ago

Discovered electrons have fixed amounts of energy and orbit the nucleus similar to planets? A. John dalton B. J.J. Thomson C. Er

Fantom [35]

I think is letter c thanks

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3 years ago

There is nothing that you can do to conserve energy. True or False

Firdavs [7]

I am going to say it is false.

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3 years ago

Read 2 more answers

What concentrations of acetic acid (pKa = 4.76) and acetate would be required to prepare a 0.15 M buffer solution at pH 5.0? Not

Leni [432]

Answer:

Acetic acid 0,055M and acetate 0,095M.

Explanation:

It is possible to prepare a 0,15M buffer of acetic acid/acetate at pH 5,0 using Henderson-Hasselblach formula, thus:

pH = pka + log₁₀ [A⁻]/[HA] <em>-Where A⁻ is acetate ion and HA is acetic acid-</em>

Replacing:

5,0 = 4,76 + log₁₀ [A⁻]/[HA]

<em>1,7378 = [A⁻]/[HA] </em><em>(1)</em>

As concentration of buffer is 0,15M, it is possible to write:

<em>[A⁻] + [HA] = 0,15M </em><em>(2)</em>

Replacing (1) in (2):

1,7378[HA] + [HA] = 0,15M

2,7378[HA] = 0,15M

[HA] = 0,055M

Thus, [A⁻] = 0,095M

That means you need <em>acetic acid 0,055M</em> and <em>acetate 0,095M</em> to obtain the buffer you need.

i hope it helps!

7 0

3 years ago