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Volgvan
2 years ago
7

If phosphorous has 5 valence electrons, how many does sulfur have?

Chemistry
1 answer:
VARVARA [1.3K]2 years ago
5 0

Answer:

6

Explanation:

The atomic number for phosphorous is 15, meaning that it has 15 electrons (and protons). The first and second shells would be filled up with 2 and 8 electrons respectively, leaving 5 which goes on the third shell, which is also the valence shell, meaning phosphorous has 5 valence electrons.

Since the atomic number of sulfur is 16, the first and second shells are also filled up with 2 and 8 electrons respectively, leaving 6 to be on the third shell, the valence shell. Hence, sulfur has 6 valence electrons.

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At 700 K, the reaction2SO2(g) + O2(g) 2SO3(g)has the equilibrium constant Kc = 4.3 x 106, and the following concentrations are p
dedylja [7]

Explanation:

The given reaction is as follows.

       2SO_{2} + O_{2}(g) \rightarrow 2SO_{3}(g)

Value of equilibrium constant is given as K_{c} = 4.3 \times 10^{6}[/tex].

Concentration of given species is [SO_2] = 0.010 M; [SO_3] = 10.M; [O_2] = 0.010 M.

Formula for experimental value of equilibrium constant (Q) is as follows.

             Q = \frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}

Putting the given concentration as follows.

              Q = \frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}

             Q = \frac{(10)^{2}}{(0.010)^{2}(0.010)}

              Q = 10^{8}

It is known that when Q > K_{eq}, then reaction moves in the backward direction.

When Q < K_{eq}, then reaction moves in the forward direction.

When Q = K_{eq}, then reaction is at equilibrium.

As, for the given reaction Q > K_{eq} then it means reaction moves in the backward direction.

Thus, we can conclude that the reaction is moving in the backward direction, that is, right to left to reach the equilibrium.

5 0
3 years ago
Use the Earth's Energy Budget diagram to answer the question. Based on the diagram, less solar energy that reaches Earth is ____
Ivan
<span>absorbed, radiated
Hope this helps. </span>
5 0
3 years ago
The mass of Carbon is 12 g/mol and the mass of oxygen is 16 g/mol. What is the approximate percent composition by mass of CO2 (f
nikitadnepr [17]

Answer:

the last one

27% by 73%

Explanation:

12/44*100 for C

and

16/44*100 for O

5 0
3 years ago
What can you conclude about these two reactions based on the chemical equations? Reaction A: 6CO2(g) + 6H2O(l) + sunlight ? C6H1
OlgaM077 [116]

Given reactions:

(A)   6CO2(g) + 6H2O(l) + sunlight → C6H12O6(aq) + 6O2(g)

(B)  2H2(g) + O2(g) → 2H2O(g) + energy

Exothermic reactions are those which proceed with the release of heat/energy. In contrast, endothermic reactions proceed with the absorption of energy in the form of heat or light.

Since reaction A required sunlight, it is endothermic. Reaction B releases energy, hence exothermic

Ans: (B)

A is endothermic

B is exothermic


8 0
3 years ago
(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct
Zarrin [17]

Explanation:

The given data is as follows.

   \Lambda^{o}_{m}(NaCl) = 1.264 \times 10^{-2}

   \Lambda^{o}_{m}(H-O=C-ONO) = 1.046 \times 10^{-2}

   \Lambda^{o}_{m}(HCl) = 4.261 \times 10^{-2}

Conductivity of monobasic acid is 5.07 \times 10^{-2} S m^{-1}

     Concentration = 0.01 mol/dm^{3}

Therefore, molar conductivity (\Lambda_{m}) of monobasic acid is calculated as follows.

                 \Lambda_{m} = \frac{conductivity}{concentration}

                                  = \frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}

                                 = \frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}

                                 = 5.07 \times 10^{-3} S m^{2} mol^{-1}

Also, \Lambda^{o}_{m} = \Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}

                            = 4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}

                            = 4.043 \times 10^{-2} S m^{2} mol^{-1}

Relation between degree of dissociation and molar conductivity is as follows.

               \alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}

                             = \frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}

                             = 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

                     K = \frac{c \times \alpha^{2}}{1 - \alpha}

Putting the values into the above formula we get the following.

                     K = \frac{c \times \alpha^{2}}{1 - \alpha}

                        = \frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}

                        = 0.017973 \times 10^{-2}

                       = 1.7973 \times 10^{-4}

Hence, the acid dissociation constant is 1.7973 \times 10^{-4}.

Also, relation between pK_{a} and K_{a} is as follows.

                 pK_{a} = -log K_{a}

                              = -log (1.7973 \times 10^{-4})

                              = 3.7454

Therefore, value of pK_{a} is 3.7454.

                             

3 0
3 years ago
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