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velikii [3]
3 years ago
6

You might need:

Chemistry
1 answer:
nikitadnepr [17]3 years ago
8 0

Answer:

Mass = 14.3 g

Explanation:

Given data:

Mass of Mg(OH)₂  = 16.0 g

Mass of HCl = 11.0 g

Mass of MgCl₂ = ?

Solution:

Chemical equation:

Mg(OH)₂ + 2HCl    →   MgCl₂ + 2H₂O

Number of moles of Mg(OH)₂ :

Number of moles = mass/ molar mass

Number of moles = 16.0 g/ 58.3 g/mol

Number of moles = 0.274 mol

Number of moles of HCl :

Number of moles = mass/ molar mass

Number of moles = 11.0 g/ 36.5 g/mol

Number of moles = 0.301 mol

Now we will compare the moles of Mg(OH)₂  and HCl with MgCl₂.

                           Mg(OH)₂          :           MgCl₂

                                 1                 :               1

                                 0.274        :          0.274

                               HCl             :              MgCl₂

                                  2              :               1

                                0.301         :           1/2×0.301 = 0.150

The  number of moles of MgCl₂ produced by HCl are less so it will limiting reactant.

Mass of MgCl₂:

Mass = number of moles × molar mass

Mass = 0.150 ×  95 g/mol

Mass = 14.3 g

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Answer:

eg=linear, mg=linear

Explanation:

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6 0
3 years ago
47.0ml of a HBr solution were titrated with 37.5ml of a 0.215M LiOH solution to reach the equivalence point. what is the molarit
faltersainse [42]

Hello!

The molarity of the HBr solution is 0,172 M.

Why?

The neutralization reaction between LiOH and HBr is the following:

HBr(aq) + LiOH(aq) → LiBr(aq) + H₂O(l)

To solve this exercise, we are going to apply the common titration equation:

M1*V1=M2*V2

M1=\frac{M2*V2}{V1}= \frac{0,215 M * 37,5 mL}{47 mL}=0,172 M

Have a nice day!

4 0
3 years ago
A buffer solution contains 0.479 M NaHCO3 and 0.342 M Na2CO3. Determine the pH change when 0.091 mol HNO3 is added to 1.00 L of
pshichka [43]

Answer:

ΔpH = 0.20

Explanation:

The buffer of HCO₃⁻ + CO₃²⁻ has a pka of 10.2

HCO₃⁻ ⇄ H⁺ + CO₃²⁻

There are 0.479moles of NaHCO₃ and 0.342moles of Na₂CO₃.

Using Henderson-Hasselbalch formula:

pH = pka + log [Base] / [Acid]

pH = 10.2 + log 0.342mol / 0.479mol

<em>pH = 10.05</em>

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0.091 moles of NaOH produce the same moles of CO₃²⁻ and consume HCO₃⁻. Moles of these species are:

CO₃²⁻: 0.342mol + 0.091mol: 0.433mol

HCO₃⁻: 0.479mol - 0.091 mol: 0.388mol

Using Henderson-Hasselbalch formula:

pH = pka + log [Base] / [Acid]

pH = 10.2 + log 0.433mol / 0.388mol

pH = 10.25

That means change of pH, ΔpH is:

ΔpH = 10.25 - 10.05 = <em>0.20</em>

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I hope it helps!

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