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Blizzard [7]
3 years ago
12

In the example in the lesson, 0.10 mole of sodium chloride or magnesium chloride or aluminum chloride was added to one liter of

water. How many moles of each chloride are in one milliliter of the respective solutions?
Chemistry
1 answer:
Elza [17]3 years ago
4 0

Answer:

1x10⁻⁴ moles of Cl⁻ in NaCl solution

2x10⁻⁴ moles of Cl⁻ in MgCl₂ solution

3x10⁻⁴ moles of Cl⁻ in AlCl₃ solution

Explanation:

Let's produce the reactions of each salt

NaCl  → Na⁺ +  Cl⁻

0.1 m                 0.1m

MgCl₂ → Mg²⁺  +  2Cl⁻

0.1m                      0.2m

AlCl₃ → Al³⁺  +  3Cl⁻

0.1 m                 0.3m

0.1, 0.2 and 0.3 are the moles of each chloride in each solution and this moles are added to one liter of solvent.

1 L = 1000mL

Let's prepare the rule of three for each

1000 mL ____ 0.1 m ____ 0.2 m _____ 0.3m

1 mL______ 1x10⁻⁴ m ___ 2x10⁻⁴ m _____ 3x10⁻⁴ m

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trapecia [35]

Answer:

1.12M

Explanation:

Given parameters:

Volume of solution  = 2.5L

Mass of Calcium phosphate  = 600g

Unknown:

Concentration  = ?

Solution:

Concentration is the number of moles of solute in a particular solution.

Now, we find the number of moles of the calcium phosphate from the given mass;

        Formula of calcium phosphate  = Ca₃PO₄

         molar mass = 3(40) + 31 + 4(16) = 215g/mol  

Number of moles of  Ca₃PO₄  = \frac{600}{215}   = 2.79moles

   Now;

  Concentration  = \frac{Number of moles }{volume }  

 Concentration  = \frac{2.79}{2.5}   = 1.12M

7 0
3 years ago
The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W
Leokris [45]

Answer:

0.0010 mol·L⁻¹s⁻¹  

Explanation:

Assume the rate law is  

rate = k[A][B]²

If you are comparing two rates,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}

You are cutting each concentration in half, so

\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}

Then,

\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}

8 0
2 years ago
What is the significant digit in 23.45
klio [65]
If you mean the number of Significant Figures/Digits in 23.45, it would be 4. This is because every single non-zero digit is counted as a significant figure
7 0
2 years ago
Why is an apple an example of matter​
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8 0
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6 0
3 years ago
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