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Gennadij [26K]
3 years ago
6

Read the given expression.

Chemistry
1 answer:
Jlenok [28]3 years ago
6 0

Answer:

Choice number two. The value of "X" in this equation should be constant for all elements across a period.

Explanation:

Electrons are negative while protons are positive. Electrons are attracted to the proton but repel each other.

Consider an atom where electrons occupy more than one energy level. Consider the Bohr Model for that atom. Protons in the nucleus attract the electrons towards the center of the atom. However, at the same time, electrons in the inner shell will repel the valence electrons. That creates an outward force that pushes the valence electrons away from the atom.

The two forces mostly balance each other, but the attraction is slightly stronger. As a result, the overall force on the valence electrons is attractive. The effective nuclear charge gives the number of protons required to produce an attraction of that strength if there was no repulsion at all.

The value of effective nuclear charge is approximately the same as atomic number minus the number of inner-shell electrons. Apparently, the "X" in this question stands for the number of inner-shell electrons.

By the Aufbau Principle, all spots in the inner shell must be filled before more electrons can be added. Additionally, atoms in the same period have the same number of inner shells. As a result, the number of inner-shell electrons will be the same for all atoms in each period. Hence, the value of "X" should stay (approximately) the same across each period.

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Which postulate of Dalton's atomic theory was later proven wrong?
ruslelena [56]

Answer:

Option c and d

Explanation:

John Dalton. In 1808, John Dalton proposed a theory known as Dalton’s Atomic Theory. The theory was published in a paper titled “A New Chemical Philosophy”. This theory was new to that era

The 5 postulates of Daltons' atomic theory are:

1. All the matters are made of atoms.

2. Atoms of different elements combine to form compounds

3. Compounds contain atoms in small whole-number ratios

4. Atoms can neither be created nor destroyed . (This was later proven wrong )

5. All atoms of an element are identical and have the same properties (This was later proven wrong as atoms of same element may be different in case of elements having isotopes )

Therefore, options c and d are the answer.

6 0
3 years ago
If 3 moles of a compound use 12 J of energy in a reaction, what is the Hreaction in kJ/mol
igomit [66]

Answer:

\Delta _RH=4x10^{-3}\frac{kJ}{mol}

Explanation:

Hello,

In this case, the molar enthalpy of reaction is obtained by dividing the involved energy by the reacting moles:

\Delta _RH=\frac{12J}{3mol} =4\frac{J}{mol}

Thus, it is important to notice that the compound "uses" the energy, it means that it absorbs the energy, for that reason the sign is positive. Moreover, computing the result in kJ/mol we finally obtain:

\Delta _RH=4\frac{J}{mol}*\frac{1kJ}{1000J} =4x10^{-3}\frac{kJ}{mol}

Best regards.

5 0
3 years ago
Which phrase best describes activation energy? (1 point)
iren [92.7K]

Answer: the minimum amount of energy required to break bonds and start a chemical reaction

Explanation: got a 100% on the quick check

8 0
2 years ago
Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 3.50 moles of magnesium
Firdavs [7]

Answer:

n_{Mg}=3.50molMg\\\\ n_{Cl}=7.00molCl\\\\n_O=28.0molO

Explanation:

Hello there!

In this case, according to the given information it turns out possible for us to realize that one mole of the given compound, Mg(ClO₄)₂, has one mole of Mg, two moles of Cl and eight moles of O; thus, we proceed as follows:

n_{Mg}=3.50molMg(ClO_4)_2*\frac{1molMg}{1molMg(ClO_4)_2}=3.50molMg\\\\ n_{Cl}=3.50molMg(ClO_4)_2*\frac{2molCl}{1molMg(ClO_4)_2}=7.00molCl\\\\n_O=3.50molMg(ClO_4)_2*\frac{8molO}{1molMg(ClO_4)_2}=28.0molO

Best regards!

3 0
3 years ago
The mineral enargite is 48.41% cu, 19.02% as, and 32.57% s by mass. what is the empirical formula of enargite?
LuckyWell [14K]
Empirical formula is the simplest ratio of whole numbers of components in a compound. 
Assuming for 100 g of the compound 
                                Cu                                 As                             S
mass                      48.41 g                          19.02 g                      32.57 g
number of moles    48.41 / 63.5 g/mol      19.02 / 75 g/mol        32.57 / 32 g/mol 
                                = 0.762 mol                = 0.2536 mol            = 1.018 mol 
divide by the least number of moles 
                               0.762 / 0.2536             0.2536 / 0.2536         1.018 / 0.2536
                               = 3.00                          = 1.00                         = 4.01
once they are rounded off 
Cu - 3
As - 1
S - 4
therefore empirical formula is Cu₃AsS₄
8 0
3 years ago
Read 2 more answers
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