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alukav5142 [94]
3 years ago
15

Which does not describe the relationship between the nucleus and the electron cloud of an atom?

Chemistry
2 answers:
trapecia [35]3 years ago
5 0

Answer: b. the masses of the nucleus and electron cloud are balanced.

Explanation:

1) From J.J. Thompson plum pudding model of the atom, and thereafter, it is know the the mass of the electrons are much, much smaller than the mass of an atom.

2) Also, from Ernest Rutherford gold foil experiment, it was determined that the atom consisted of a small and heavy positvely charged nucleus surrounded by electrons.

3) As per the scientific evidence, the current atom model (quantum model) we know that:

a. It is true that the electron cloud takes up much more space than the nucleus: the electrons are in regions called orbitals which are around the tiny nucleus.

b. It is not true that the masses of the nucleus and electron cloud are balanced: the relative masses of protons, neutrons and electrons are:

1 : 1 : 1 / 1840.

Which means that the nucleous contains almost all the mass of the atom.

c. It is true that the electrons move much faster than the particles in the nucleus: the particles in the nucleous stand in a very reduced space whose motion is only vibrational, while the electrons move rapidly i n the orbitals.

d. It is true that the charges of the nucleus and electron cloud are balanced: the atom is neutral because the positive charges of the nuclous equals the negative charges of the electrons.

rusak2 [61]3 years ago
4 0
I think the correct answer among the choices is option B. The statement which does not describe the relationship between the nucleus and the electron cloud of an atom will be that the <span> masses of the nucleus and electron cloud are balanced.</span>
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Solid sodium iodide is slowly added to a solution that is 0.0050 M Pb 2+ and 0.0050 M Ag +. [K sp (PbI 2) = 1.4 × 10 –8; K sp (A
UkoKoshka [18]

Answer:

[Ag⁺] = 5.0x10⁻¹⁴M

Explanation:

The product solubility constant, Ksp, of the insoluble salts PbI₂ and AgI is defined as follows:

Ksp(PbI₂) = [Pb²⁺] [I⁻]² = 1.4x10⁻⁸

Ksp(AgI) = [Ag⁺] [I⁻] = 8.3x10⁻¹⁷

The PbI₂ <em>just begin to precipitate when the product  [Pb²⁺] [I⁻]² = 1.4x10⁻⁸</em>

<em />

As the initial [Pb²⁺] = 0.0050M:

[Pb²⁺] [I⁻]² = 1.4x10⁻⁸

[0.0050] [I⁻]² = 1.4x10⁻⁸

[I⁻]² = 1.4x10⁻⁸ / 0.0050

[I⁻]² = 2.8x10⁻⁶

<h3>[I⁻] = 1.67x10⁻³</h3><h3 />

So, as the [I⁻] concentration is also in the expression of Ksp of AgI and you know concentration in solution of I⁻ = 1.67x10⁻³M:

[Ag⁺] [I⁻] = 8.3x10⁻¹⁷

[Ag⁺] [1.67x10⁻³] = 8.3x10⁻¹⁷

<h3>[Ag⁺] = 5.0x10⁻¹⁴M</h3>

6 0
3 years ago
How many liters of hydrogen gas are formed from the complete reaction of 1.04 mol of C? Assume that the hydrogen gas is collecte
Alisiya [41]

The volume is 2.23 liters of hydrogen gas.

<u>Explanation</u>:

                  moles of C = grams / molecular mass of C

                                     = 1.04 g / 12.011 g/mol.

                                     = 0.086

        The ratio between C and H2 is 1 : 1

                 moles H2 = 0.086

                               V = nRT / p

                                  = 0.086 x 0.08206 x 316 K / 1.0 atm

                              V  = 2.23 L.

8 0
3 years ago
Calculate the vapor pressure in torr of a solution containng 24.5 g of glycerin (C3H8O3) in 135 mL water at 30.0* C; the vapor p
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2 years ago
An unknown compound contains 38.7 % calcium, 19.9 % phosphorus and 41.2 % oxygen. what is the empirical formula of this compound
Temka [501]

The amount per 100 g is:

38.7 % calcium = 38.7g Ca / 100g compound = 38.7g

19.9 % phosphorus = 19.9g P / 100g compound = 19.9g

41.2 % oxygen = 41.2g O / 100g compound = 41.2g

The molar amounts of calcium, phosphorus and oxygen in 100g sample are calculated by dividing each element’s mass by its molar mass:

Ca = 38.7/40.078 = 0.96

P = 19.9/30.97 = 0.64

O = 41.2/15.99 = 2.57

C0efficients for the tentative empirical formula are derived by dividing each molar amount by the lesser value that is 0.64 and in this case, after that multiply wih 2.

Ca = 0.96 / 0.64 = 1.5=1.5 x 2 = 3

P = 0.64 / 0.64 = 1 = 1x2= 2

O = 2.57 / 0.64 = 4= 4x2= 8

Since, the resulting ratio is calcium 3, phosphorus 2 and oxygen 8

<span>So, the empirical formula of the compound is Ca</span>₃(PO₄)₂

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