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2 years ago

Which of the following is a limitation of Bohr's model?

Chemistry

1 answer:

Murljashka [212]2 years ago

6 0

Answer:

The answer is:

The first option

Explanation:

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1) You are asked to make 10 miles of iron (Fe) from iron oxide (Fe2O3) and excess carbon monoxide (CO). Fe2O3(s) + 3CO(g)—> 2

Inessa05 [86]

For both of them, used the balanced equation and it’s mole ratio to convert whatever you need to into moles. See the attacked work.

1) D 5 mols

2) A 0.55 mols

8 0

2 years ago

Can someone please help as soon as possible

Kobotan [32]

Answer:

Explanation:

#1 2HCL and 2H20

4 0

2 years ago

What observation did Rutherford make from his gold-foil experiment which

JulsSmile [24]

Answer:

A piece of gold foil was hit with alpha particles, which have a positive charge. Most alpha particles went right through. This showed that the gold atoms were mostly empty space. Some particles had their paths bent at large angles. A few even bounced backward. The only way this would happen was if the atom had a small, heavy region of positive charge inside it.

8 0

2 years ago

In methane combustion, the following reaction pair is important: At 1500 K, the equilibrium constant Kp has a value of 0.003691

andre [41]

Explanation:

Let us assume that the value of $K_{r}$ = $2.82 \times 10^{5} \times e^({\frac{-9835}{T}}) m^{6}/mol^{2}s$

Also at 1500 K, $K_{r}$ = $2.82 \times 10^{5} \times e^({\frac{-9835}{1500}}) m^{6}/mol^{2}s$

$K_{r} = 400.613 m^{6}/mol^{2}s$

Relation between $K_{p}$ and $K_{c}$ is as follows.

$K_{p} = K_{c}RT$

Putting the given values into the above formula as follows.

$K_{p} = K_{c}RT$

$0.003691 = K_{c} \times 8.314 \times 1500$

$K_{c} = 2.9 \times 10^{-7}$

Also, $K_{c} = \frac{K_{f}}{K_{r}}$

or, $K_{f} = K_{c} \times K_{r}$

= $2.9 \times 10^{-7} \times 400.613$

= $1.1617 \times 10^{-4} m^{6}/mol^{2}s$

Thus, we can conclude that the value of $K_{f}$ is $1.1617 \times 10^{-4} m^{6}/mol^{2}s$ .

6 0

2 years ago

Enter the correct ground-state (or lowest energy) configuration based on the number of electrons: 1s^4 2s^4 2p^12.

Blababa [14]

Answer:

The ground state configuration is the lowest energy, most stable arrangement. An excited state configuration is a higher energy arrangement (it requires energy input to create an excited state). Valence electrons are the electrons utilised for bonding.

or the

FIGURE 5.9 The arrow shows a second way of remembering the order in which sublevels fill. Table 5.2 shows the electron configurations of the elements with atomic numbers 1 through 18.

Element Atomic number Electron configuration

sulfur 16 1s22s22p63s23p4

chlorine 17 1s22s22p63s23p5

argon 18 1s22s22p63s23p6

or the

Two electrons

Two electrons fill the 1s orbital, and the third electron then fills the 2s orbital. Its electron configuration is 1s22s1.

Explanation:

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6 0

2 years ago