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babunello [35]
2 years ago
15

How to solve 5gNaCl= MolE NaCl

Chemistry
1 answer:
valentinak56 [21]2 years ago
4 0

0.086mole

Explanation:

The problem is very simple. It involves conversion from mass to number of moles of NaCl

What is mole?

 A mole is a unit of measuring chemical substances. It is the amount that contains the avogadro's number of particles.

How is it related and can be converted from mass;

 

Number of moles of a substance = \frac{mass}{molar mass}

Since we know the mass, we can find the molar mass and put into this equation;

Molar mass of NaCl = 23 + 35.5 = 58.5g/mol

Number of moles of NaCl = \frac{5}{58.5}  = 0.086mole

learn more:

Number of moles brainly.com/question/1841136

#learnwithBrainly

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Lady bird [3.3K]

<u>Answer</u>:-

Name of element = Lead

Symbol ♾ of Lead = Pb

Atomic no. = 82

Atomic mass = 207.2 amu

No. of protons = 82

No. of electrons = 82

Yes, lead(Pb) is a metallic element and certainly it has 6 electron shells which means 6 energy level.

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Calculate the hydronium ion concentration in an aqueous solution with a ph of 9.85 at 25°c.
UkoKoshka [18]

Answer:

1.41 × 10⁻¹⁰ M

Explanation:

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pH = -log [H⁺]

[H⁺] = antilog -pH

[H⁺] = antilog -9.85

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At equilibrium, the concentrations in this system were found to be [ N 2 ] = [ O 2 ] = 0.200 M and [ NO ] = 0.600 M . N 2 ( g )
alexandr1967 [171]

Answer:

0.78 M

Explanation:

First, we need to know which is the value of Kc of this reaction. In order to know this, we should take the innitial values of N2, O2 and NO and write the equilibrium constant expression according to the reaction. Doing this we have the following:

N2(g) + O2(g) <------> 2NO(g)   Kc = ?

Writting Kc:

Kc = [NO]² / [N2] * [O2]

Replacing the given values we have then:

Kc = (0.6)² / (0.2)*(0.2)

Kc = 9

Now that we have the Kc, let's see what happens next.

We add more NO, until it's concentration is 0.9 M, this means that we are actually altering the reaction to get more reactants than product, which means that the equilibrium is being affected. If this is true, in the reaction when is re established the equilibrium, we'll see a loss in the concentration of NO and a gaining in concentrations of the reactants. This can be easily watched by doing an ICE chart:

      N2(g) + O2(g) <------> 2NO(g)

I:      0.2        0.2                 0.9

C:     +x         +x                   -2x

E:    0.2+x    0.2+x             0.9-2x

Replacing in the Kc expression we have:

Kc =  [NO]² / [N2] * [O2]

9 = (0.9-2x)² / (0.2+x)*(0.2+x)   ----> (this can be expressed as 0.2+x)²

Here, we solve for x:

9 = (0.9-2x)² / (0.2+x)²

√9 = (0.9-2x) / (0.2+x)

3(0.2+x) = 0.9-2x

0.6 + 3x = 0.9 - 2x

3x + 2x = 0.9 - 0.6

5x = 0.3

x = 0.06 M

This means that the final concentration of NO will be:

[NO] = 0.9 - (2*0.06)

[NO] = 0.78 M

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Answer:

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