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storchak [24]
2 years ago
12

Elements within the same group of the periodic table behave similarly because they have the same number of ?

Chemistry
1 answer:
sammy [17]2 years ago
4 0
The elements in each group have the same number of electrons in the outer orbital. Or also called valence electrons. Khan academy has a great video online explaining why this happens. (It only happens for main group elements). Here is a link (sorry you can’t click it in Brainly) https://www.khanacademy.org/science/chemistry/periodic-table/copy-of-periodic-table-of-elements/v/periodic-table-valence-electrons. Feel free to message me for a better explanation, I would explain now but I’m not sure how much you know about this. If you know how to write an electron configuration you can see how all the electron configurations for the same group (not the transitional metals only the main groups) have the same number of valence electrons. I hope that helped, sorry I was vague about the explanation :)
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Calculate the molar mass of Cr4(P2O7)
Norma-Jean [14]

Hey there!

Cr₄(P₂O₇)₃

Cr: 4 x 51.996 = 207.984

P: 6 x 30.97 = 185.82

O: 21 x 16 = 336

-------------------------------------

                      729.804 g/mol

The molar mass of Cr₄(P₂O₇)₃ is 729.804 g/mol.

Hope this helps!

7 0
3 years ago
How does carbon dioxide behave differently (its properties) as a gas compared with as a solid?
MaRussiya [10]
<span> is made up of a carbon atom covalently double bonded to two oxygen atoms. Carbon dioxide exists in Earth's atmosphere as a trace gas at a concentration of about 0.04 percent by volume.</span>
5 0
3 years ago
Which of the following is an example of a chemical change?
wolverine [178]

Answer:

C

Explanation:

6 0
2 years ago
Read 2 more answers
Consider the following reaction NHAHS(s)NH3(g) + H2S(g) If a flask maintained at 302 K contains 0.196 moles of NH4HS(s) in equil
quester [9]

Answer:

Kc = 3.72 × 10⁶

Explanation:

Let's consider the following reaction:

NH₄HS(g) ⇄ NH₃(g) + H₂S(g)

At equilibrium, we have the following concentrations:

[NH₄HS] = 0.196 M (assuming a 1 L flask)

[NH₃] = 9.56 × 10² M

[H₂S] = 7.62 × 10² M

We can replace this data in the Kc expression.

Kc=\frac{[NH_{3}] \times [H_{2}S] }{[NH_{4}HS]} =\frac{9.56 \times 10^{2}  \times 7.62  \times 10^{2}}{0.196} =3.72 \times 10^{6}

7 0
3 years ago
Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2O(g) (Delta
shusha [124]

Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

Explanation:

The balanced chemical reaction is,

2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

The expression for enthalpy change is,

\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]

Putting the values we get :

\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]

\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]

\Delta H=-5314.8kJ

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat = \frac{5314.8}{2}\times 1=2657.4kJ

Thus enthalpy of combustion per mole of butane is -2657.4 kJ

3 0
3 years ago
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