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Scilla [17]
3 years ago
10

What is the molarity of a solution that contains 14.92 grams magnesium oxide (MgO) in 365ml of solution?

Chemistry
2 answers:
AnnZ [28]3 years ago
6 0

Answer:

Option A. 1.01 molar solution

Explanation:

Step 1:

Data obtained from the question. This include the following:

Mass of MgO = 14.92g

Volume of solution = 365mL

Molarity of solution =.?

Step 2:

Determination of the number of mole present in 14.92g of MgO.

This is illustrated below:

Mass of MgO = 14.92g

Molar mass of MgO = 24.3 + 16 = 40.3g/mol

Number of mole MgO =...?

Mole = Mass /Molar Mass

Number of mole of MgO = 14.92/40.3 = 0.370 mole

Step 3:

Determination of the molarity of the solution.

This can be obtained as follow:

Mole of MgO = 0.370 mole

Volume = 365mL = 365/1000 = 0.365L

Molarity =..?

Molarity = mole /Volume

Molarity = 0.370/0.365

Molarity = 1.01M

Yuki888 [10]3 years ago
3 0

Answer:

a. 1.01 molar solution

Explanation:

Now we have to recall the formula;

n= CV

Where

n= number of moles of magnesium oxide

C= concentration of magnesium oxide solution = ???

V= volume of magnesium oxide solution= 365 ml

But;

n= m/M

Where;

m= given mass of magnesium oxide= 14.2 g

M= molar mass of magnesium oxide= 40.3044 g/mol

Substituting values;

14.2g/40.3044 g/mol = C×365/1000

0.3523= 0.365 C

C= 0.3523/0.365

C= 0.965

C= 1.00 molar (approx)

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A chemist weighed out 20.7 g of sodium . Calculate the number of moles of sodium she weighed out
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Two equilibrium reactions of nitrogen with oxygen, with their corresponding equilibrium constants (Kc) at a certain temperature,
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Answer:

Kc = 1.54e - 31 / 2.61e - 24

Explanation:

1 )   N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)  ; Kc = 1.54e - 31

2)   N_{2}(gas) + 1/2O_{2}(gas)\rightarrow N_{2}O(gas)  ; Kc = 2.16e - 24

   upon reversing  ( 2 )  equation

     N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)   Kc = 1/2.16e - 24  

    now adding 1 and reversed equation (2)

       N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)

      N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)

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                  N_{2}O(gas) + 1/2O_{2}(gas)\rightarrow 2NO(gas)  Kc = 1.54e-31 × 1/2.61e - 24

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4 0
2 years ago
Read 2 more answers
what mass of sodium fluoride (FW=42.0 g/mol) must be added to 3.50 x 10^2 mL of water to give a solution with pH = 8.40?
MaRussiya [10]

Answer:

Explanation:

Sodium fluoride, being a salt, dissolves in water completely producing F ⁻ ions. Now  F⁻ is the conjugate base of the weak acid HF, so in water we will have the following equilibrium:

F⁻  +  H₂O ⇆ HF + OH⁻

Given this equilibrium, we need to calculate Kb from the Ka for HF,  the [ OH ⁻] from the given pH, and finally the mass needed to produce that  OH⁻ concentration.  

The equilibrium constant, Kb , can be calculated from Kw = Ka x Kb, where Kw = 10⁻¹⁴ and Ka for HF is  6.6 x 10⁻⁴ from reference tables.

Kb = 10⁻¹⁴ / 6.6 x 10⁻⁴ = 1.5 x 10⁻¹¹

pH + pOH = 14  ⇒ pOH = 14 - 8.40 = 5.60

[ OH⁻ ] = 10^-5.60 = 2.51 x 10⁻⁶

Now we have all the information :

                                   F⁻                    HF                        OH⁻

Equilibrium                 X                  2.51 x 10⁻⁶            2.51 x 10⁻⁶

(2.51 x 10⁻⁶)² / X  =  1.5 x 10⁻¹¹     ⇒  X =  (2.51 x 10⁻⁶)²  / 1.5 x 10⁻¹¹

X = [ F⁻ ] = 0.41 M

For 350 mL ( 0.35 L ) we need to add:

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and finally the mass will be:

0.144 mol NaF *  42.0 g/mol NaF = 6.03 g NaF

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