4 Fe + 6 H₂O + 3 O₂ → 4 Fe(OH)₃

For a principal quantum number n, how many atomic orbitals are possible?

slavikrds [6]

Answer:

The total number of orbitals for a given n value is n2.

Explanation:

For a hydrogen atom with n=1, the electron is in its ground state; if the electron is in the n=2 orbital, it is in an excited state.

5 0

3 years ago

Design an invitation you could use to calculate the density of a penny

Nataliya [291]

Firstly, the density of any substance is represented by the mass (amount of matter) as divided by the volume(amount of space). According to external websites, the mass of a penny is 2.5 grams.However, the volume of a penny is .35cm to the power of 3 (due to the thickness of the penny being extremely minimal.Thus the amount of density is extremely little). Therefore, the density of a penny is 0.875 g/cm cubed (dimensional analysis).As for an invention that could be used, that is possible with the usage of a series of measurements that can both calculate mass and volume and directly allocate that to attain density

7 0

4 years ago

A mixture containing 20 mole % butane, 35 mole % pentane and rest

notka56 [123]

Answer:

2.5 % butane, 42.2 % pentane and 55.3 % hexane

Explanation:

Hello,

In this case, the mass balance for each substance is given by:

$Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB$

Whereas $y$ accounts for the fractions at the outlet distillate and $x$ for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

$0.9=\frac{y_bD}{z_bF}$

So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:

$y_bD=0.9*z_bF=0.9*0.2*100mol=18mol$

The total distillate flow:

$y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol$

And the total bottoms flow:

$F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol$

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

$x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025$

Then, the molar fraction of pentane and hexane:

$x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422$

$x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553$

Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.

NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.

Regards.

8 0

3 years ago

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Sulfur and magnesium both have two valence electrons.<br> a. True<br> b. False

Nezavi [6.7K]

It would be false sulfur has 6

3 0

4 years ago

During an endothermic phase change, what happens to the potential energy and the kinetic energy?

antiseptic1488 [7]

During endothermic phase change, the potential energy of the system always increases while the kinetic energy of the system remains constant. The potential energy of the reaction increases because energy is been added to the system from the external environment.

<u>Explanation</u>:

Those are three distinct methods for demonstrating a specific energy condition of an object. They don't affect one another.

"Potential Energy" is a relative term showing a release of possible energy to the environment. If we accept its pattern as the overall energy state of a compound, at that point, an endothermic phase change would infer an increase in "potential" as energy is being added to the compound by the system.

A phase change will display an increase in the kinetic energy at whatever point the compound is transforming from a high density to a low dense phase. The kinetic energy will decrease at whatever point the compound is transforming from a less dense to high dense phase.

4 0

3 years ago

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