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3 years ago

What mass of Fe(OH)3 will be obtained when 100. mL of 0.240 M FeCl3 is mixed with 200. mL of 0.182 M NaOH?

Chemistry

1 answer:

ratelena [41]3 years ago

7 0

Answer:

648.68 mg

Explanation:

The reaction that takes place is:

FeCl₃ + 3NaOH → Fe(OH)₃ + 3NaCl

First we calculate how many moles of each reactant were added, using the given volumes and concentrations:

FeCl₃ ⇒ 100 mL * 0.240 M = 24 mmol FeCl₃

NaOH ⇒ 100 mL * 0.182 M = 18.2 mmol NaOH

24 mmol of FeCl₃ would react completely with (24 * 3) 72 mmol of NaOH. There are not as many NaOH mmoles, so NaOH is the limiting reactant.

Now we calculate how many moles of Fe(OH)₃ are formed, using the moles of the limiting reactant:

18.2 mmol NaOH * $\frac{1mmolFe(OH)_3}{3mmolNaOH}$ = 6.07 mmol Fe(OH)₃

Finally we convert 6.07 mmol Fe(OH)₃ to grams, using its molar mass:

6.07 mmol Fe(OH)₃ * 106.867 mg/mmol = 648.68 mg

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Identify the solute with the highest van't Hoff factor. And how do you determine which one is highest?

german

Answer : The correct option is, (A) $AlCl_3$

Explanation :

Van't Hoff factor : It is defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property.

We can determine the van't Hoff factor by the association and dissociation of the compound.

(A) $AlCl_3$

It is an electrolyte that dissociates to give aluminum ion and chloride ion.

The dissociation of $AlCl_3$ will be,

$AlCl_3\rightarrow Al^{3+}+3Cl^{-}$

So, Van't Hoff factor = Number of solute particles = $Al^{3+}+3Cl^{-}$ = 1 + 3 = 4

(B) $KI$

It is an electrolyte that dissociates to give potassium ion and iodide ion.

The dissociation of $KI$ will be,

$KI\rightarrow K^{+}+I^{-}$

So, Van't Hoff factor = Number of solute particles = $K^{+}+I^{-}$ = 1 + 1 = 2

(C) $CaCl_2$

It is an electrolyte that dissociates to give calcium ion and chloride ion.

The dissociation of $CaCl_2$ will be,

$CaCl_2\rightarrow Ca^{2+}+2Cl^{-}$

So, Van't Hoff factor = Number of solute particles = $Ca^{2+}+2Cl^{-}$ = 1 + 2 = 3

(D) $MgSO_4$

It is an electrolyte that dissociates to give magnesium ion and sulfate ion.

The dissociation of $MgSO_4$ will be,

$MgSO_4\rightarrow Mg^{2+}+SO_4^{2-}$

So, Van't Hoff factor = Number of solute particles = $Mg^{2+}+SO_4^{2-}$ = 1 + 1 = 2

(E) Non-electrolyte

The dissociation non-electrolyte is not possible. So, the Van't Hoff factor will always be, 1.

Hence, the highest van't Hoff factor of solute is, $AlCl_3$

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3 years ago

By replacing hardware safety features with software controls in the therac-25 machine, aecl was able to?

Ipatiy [6.2K]

Therac-25, AECL was able to reduce costs by replacing hardware safety features with software controls.

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The Therac-25 machine was a state-of-the-art linear accelerator developed by using the company Atomic strength Canada restricted (AECL) and a French company CGR to provide radiation treatment to cancer sufferers. The Therac-25 become the most computerized and complicated radiation therapy device state its time.

learn more about cancer here brainly.com/question/11710623

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2 years ago

Which of the following transitions represent the emission of a photon with the largest energy?

s2008m [1.1K]

Answer:

n=6 to n=3 (B)

Explanation:

Energy of an electron present in the $n^{th}$ orbit is directly proportional to $n^{2}$ .Hence a transistion from one orbit to another orbit emits an energy proportional to the difference of their squares of the orbits. that is if an electron travels from orbit n1 to orbit n2 then it emits an energy corresponding to $n_{1} ^{2} -n_{2} ^{2}$ .So in the above question the highest energy emission occurs when an electron moves from n=6 to n=3.(Highest difference of energy levels).

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3 years ago

Indicate which solution in each pair has the lower pH. Your response should be a four letter "word". The first letter should be

JulijaS [17]

Answer:

bcfh

Explanation:

HClO₄ reacts with water thus:

HClO₄ + H₂O → H₃O⁺ + ClO₄⁻

That means HClO₄ produce H₃O⁺ that decreases pH. That means the higher concentration of HClO₄ decreases pH. Thus, lower pH will be:

b) 0.2 M HClO4

The reaction of NaClO₄ is:

NaClO₄ + H₂O → OH⁻ + HClO₄ + Na⁺

The higher concentration of NaClO₄ the higher production of OH⁻ that increase pH, that means the lower concentration of NaClO₄ the lower pH, thus, the answer is:

c) 0.1 M NaClO or

HF reacts with water thus;

HF ⇄ H⁺ + F⁻

The equilibrium constant is:

k = [H⁺] [F⁻] / [HF] = 3,5x10⁻⁴

For HNO₂ equilibrium is:

HNO₂ ⇄ H⁺ + NO₂⁻

k = [H⁺] [NO₂⁻] / [HNO₂] = 4,5x10⁻⁴

As k value is higher for HNO₂, the concentration of H⁺ will be higher in this system doing the HNO₂ with the lower pH.

f) 0.1 M HNO2

NaOH is a strong base that produce OH⁻ that increase pH, pure water is neutral, thus, the lowe pH is:

h) pure water