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GenaCL600 [577]
4 years ago
5

How many grams of sodium oxide can be synthesized from 17.4 g of sodium? assume that more than enough oxygen is present. the bal

anced equation is: 4na(s) + o2(g) → 2na2o(s)?
Chemistry
1 answer:
Tatiana [17]4 years ago
7 0
Equation is as follow,

<span>                          4 Na (s)  +  O</span>₂ <span>(g)    →    2Na</span>₂<span>O (s)

According to equation,

      91.92 g (4 moles) of Na produces  =  123.92 g (2 moles) of Na</span>₂O
So,
                    17.4 g of Na will produce  =  X g of Na₂O

Solving for X,
                                   X  =  (17.4 g × 123.92 g) ÷ 91.92 g

                                   X  =  23.45 g of Na₂O
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102. What is the Pauli exclusion principle? What is Hund's rule?
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<em> Pauli Exclusion Principle is about quantum numbers of an atom. Hund rule is about how electrons are filled to the orbitals of an atom.</em>

5 0
2 years ago
A sample of gas occupies a volume of 72.0 mL . As it expands, it does 141.2 J of work on its surroundings at a constant pressure
bagirrra123 [75]

The final volume of the gas is  73.359 mL

<h3 />

Given :

A sample gas has an initial volume of 72.0 mL

The work done = 141.2 J

Pressure = 783 torr

The objective is to determine the final volume of the gas.

Since, the process does 141.2 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.

Converting the external pressure to atm; we have

External Pressure Pext:

P ext = 783 torr × \frac{1 atm}{760 torr}

Pext = 1.03 atm

The work done W =  Pext V

The change in volume ΔV= \frac{W}{Pext}

ΔV = \frac{141.2 J \frac{1 L atm}{101.325 J} }{1.03 atm}

ΔV = \frac{0.0014}{1.03}

ΔV = 0.001359 L

ΔV = 1.359 mL

The initial  volume = 72.0 mL

The change in volume V is ΔV = V₂ - V₁

-  V₂ = - ΔV - V₁

multiply both sides by (-), we have:

V₂ = ΔV + V₁

     =  1.359 mL + 72.0 mL

     = 73.359 mL

Therefore, the final volume of the gas is  73.359 mL .

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3 0
2 years ago
Water absorbs energy when it undergoes___
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Water absorbs energy D. Melting
5 0
3 years ago
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3. What is the atomic mass of phosphorous if phosphorous-29 has a percent abundance of 35.5%, phosphorous-30 has a percent abund
daser333 [38]

Answer:

The atomic mass of phosphorus is 29.864 amu.

Explanation:

Given data:

Atomic mass of phosphorus = ?

Percent abundance of P-29 = 35.5%

percent abundance of P-30 = 42.6%

Percent abundance of P-31 = 21.9%

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass  / 100

Average atomic mass  = (29×35.5)+(30×42.6) + (31×21.9) /100

Average atomic mass =  1029.5 + 1278 + 678.9/ 100

Average atomic mass  = 2986.4 / 100

Average atomic mass = 29.864 amu.

The atomic mass of phosphorus is 29.864 amu.

5 0
3 years ago
What is the molar solubility of aucl3 (ksp = 3. 2 x 10-23) in a 0. 013 m solution of magnesium chloride (a soluble salt)?
Inessa [10]

The molar solubility of AuCl₃ in a 0.013 M solution of magnesium chloride is 1.81×10⁻²⁸M.

<h3>What is Ksp?</h3>

The solubility product constant, Ksp, is the equilibrium constant for a solid substance dissolving in an aqueous solution. And for the AuCl₃, Ksp will be written as: Ksp = [Au³⁺][Cl⁻]³

Let the solubility of the AuCl₃ in 0.013M solution of magnesium chloride is x, of Au³⁺ is x and of Cl⁻ is 3x. But we know that MgCl₂ is a strong electrolyte and it completely dissociates into its ions and will produce 2 moles of chloride ions. For this solution let we consider the volume is 1 liter then the concentration of chloride ions in MgCl₂ is 2(0.013)=0.026M.

So, in MgCl₂ solution concentration of Cl⁻ becomes = 3x + 0.026.

Value of Ksp for AuCl₃ = 3.2 × 10⁻²³

On putting all values on the Ksp equation, we get

Ksp = (x)(3x + 0.026)³

Value of 3x is negligible as compared to the 0.026, so the equation becomes

3.2 × 10⁻²³ = (x)(0.026)³

x = 3.2×10⁻²³ / (0.026)³

x = 1.81×10⁻²⁸M

Hence the molar solubility of AuCl₃ in 0.010M MgCl₂ is 1.81×10⁻²⁸M.

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