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Eduardwww [97]
2 years ago
15

A toothpaste contains sodium fluoride (NaF) What percentage of Fluoride is present.(4cs)

Chemistry
1 answer:
MariettaO [177]2 years ago
7 0

Answer:

45.2%

Explanation:

To calculate the percent of an element in a compound we divide the molar mass of the element by the compound and multiply that by 100

First lets find the molar mass of Fluoride

Looking at the periodic table Fluoride has a molecular mass of 18.998 g

Now we need to find the molecular mass of NaF

Looking at a periodic table, Sodium (Na) has a molecular mass of 22.990g and Fluoride has a molecular mass of 18.998 so NaF has a molecular mass of 22.990(1) + 18.998(1) = 41.988g

Now we divide the mass of fluoride by the mass of sodium fluoride and multiply that by 100 to find the percentage of fluoride that is present in NaF

Mass of Fluoride = 18.998g

Mass of Sodium Fluoride = 41.988g

Percentage of fluoride present in NaF = (18.998g / 41.988g) * 100 = 45.2%

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What could be the fourth quantum number of a 1s^2 electron be?
Bumek [7]

The forth quantum number is Spin quantum number and the value of Spin Quantum Number of 1s² will be +1/2. Therefore, the correct option is (B) option

<h3>What is Spin Quantum Number ?</h3>

It describes the orientation of the electron spin (rotation) in space.

In the given case of 1s² configuration,

since, the Quantum numbers are known as:

  • Principal quantum number (n) which cannot be zero.

       The allowed values of n are therefore 1, 2, 3, 4, and so on.

  • Angular quantum number (l) is able to be any integer between 0  

       and n-1.

       Therefore, for example, If n = 1, l will be 0,

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  • Magnetic quantum number (m) is able to be any integer between -l and +l.    

       Therefore, for example, If l = 3, m can be either -3 to +3.

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Thus, since the fourth quantum number is Ms, spin, we infer that the answer is (B) option ie, Ms = +1/2 because it represents the positive spin on 1s².

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brainly.com/question/15325270

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6 0
2 years ago
Consider the reaction below. At 500 K, the reaction is at equilibrium with the following concentrations. [PCI5]= 0.0095 M [PCI3]
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Concentration of [Cl_2] =  0.020 M

The expression of the equilibrium constant is given as:

K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}=\frac{0.020 M\times 0.020 M}{0.0095 M}

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The equilibrium constant for the given reaction is 0.0421.

6 0
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