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zimovet [89]
1 year ago
12

Are the following instructions for diluting a 10.0 M solution to a 1.00 M solution correct: ""Take 100.0 mL of the 10.0 M soluti

on and add 900.0 mL water""? Explain.
Chemistry
1 answer:
Molodets [167]1 year ago
7 0

The following directions for diluting a 10.0 M answer to a 1.00 M solution is no longer correct.

<h3>.What is the rule of dilution?</h3>

A popular rule to use in calculating the attention of solutions in a sequence is to multiply the unique concentration by means of the first dilution factor, this by the second dilution factor, this by the 1/3 dilution factor, and so on till the ultimate attention is known.

Example: A 5M answer of HCl is diluted 1/5.

<h3>How do you do a 1 in 10 dilution?</h3>

For example, to make a 1:10 dilution of a 1M NaCl solution, you would combine one "part" of the 1M solution with 9 "parts" of solvent (probably water), for a total of ten "parts." Therefore, 1:10 dilution ability 1 part + 9 components of water (or different diluent).

Learn more about dilution here:

<h3>brainly.com/question/25292980</h3><h3 /><h3>#SPJ4</h3>
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Suppose you stir a little baking soda into water until the water looks clear again how could you prove to someone the material i
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3 years ago
Consider the following chemical reaction: CO (g) + 2H2(g) ↔ CH3OH(g) At equilibrium in a particular experiment, the concentratio
boyakko [2]

Answer:

The concentration of CH₃OH in equilibrium is [CH₃OH] = <em>2,8x10⁻¹ M</em>

Explanation:

For the equilibrium:

CO (g) + 2H₂(g) ⇄ CH₃OH(g) keq= 14,5

Thus:

14,5 = \frac{[CH_{3}OH]}{[CO][H_{2}]^2}

In equilibrium, as [CO] is 0,15M and [H₂] is 0,36M:

14,5 = \frac{[CH_{3}OH]}{[0,15][0,36]^2}

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I hope it helps!

8 0
3 years ago
Iron(II) chloride is formed from the reaction between iron and copper(II) chloride. Fe + CuCl2 FeCl2 + Cu If the reactants have
sertanlavr [38]

Answer: see figure attached and explanation below.

Explanation:

1) Chemical equation (given):

Fe + CuCl₂ → Cu + FeCl₂

2) ΔHf reactants: -256 kJ/mol (given)

3) ΔHf products: - 321 kJ/mol (given)

4) ΔH reaction = ΔHf products - ΔHf reactants = - 321 kJ/mol - (- 256 kJ/mol) = - 65 kJ/mol

5) Conclusion:

i) Since ΔHf of products is less (more negative) than ΔHf of reactants, the reaction is exhotermic: the reaction released energy, which is the reason why the products content less potential energy than the reactants.

ii) Then, the energy diagram is the typical one of an exothermic reaction: the products start a certain potential energy level, the energy incrases until reaching the activation energy (the energy barrier to form the activated complex) and then energy decreases until a level below the energy of the reactants.

iii) See the attached figure with such kind of diagram showing the products at a lower level than the reactans

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