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3 years ago

The reaction that takes place when iron metal combines with aqueous hydrochloric acid.

1 answer:

8 0

The reaction gives out a yellow showery sparks and produces a black solid. Iron reacts with dilute hydrocloric acid to give iron chloride and hydrogen gas.

2Fe+2Hcl3 = 2FeCl3+H2

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A 14.1% by mass HClO3 solution has a density of 1.1690 g/mL. What is the concentration of this solution in molarity?

Scilla [17]

Answer:

[ HClO₃] = 1.93M

Explanation:

X % by mass, means that in 100 g of solution, we have X g of solute.

In this case, 14.1 g of HClO₃ are contained in 100 g of solution.

Density always referrs to solution. Let's calculate the volume of solution.

Solution density = Solution mass / Solution volume

1.1690 g/mL = 100 g / Solution volume

Solution volume = 100 g /1.1690 g/mL → 86.2 mL.

For molarity we must get moles of solute and volume of solution (L), because molarity is mol/L

Let's convert the mL of solution in L

86.2 mL . 1L / 1000mL = 0.0862 L

Now, let's determine the moles of solute. (Mass / Molar mass)

14.1 g / 84.45 g/mol = 0.167 moles

Molarity is mol/L → 0.167 moles / 0.0862 L = 1.93M

6 0

4 years ago

12. A voltaic cell consists of a chromium electrode dipped in a 1.20 M chromium (III) nitrate

Lilit [14]

For a voltaic cell consisting of chromium, an electrode dipped in a 1.20 M chromium (III) nitrate solution and a tin electrode dipped in a 0.400 M tin (II) nitrate solution, the cell potential at 298 K is mathematically given as

Ecell = 0.577 V

<h3 /><h3>What is the cell potential at 298 K?</h3>

Generally, the equation for the Oxidation and Reduction is mathematically given as

Cr(s) ------------------ Cr+3(aq) + 3e- ] x 2 ...O

Sn+2(aq) + 2e- ------------ Sn(s) ] x 3 ...R

Reaction

2 Cr(s) + 3 Sn+2(aq) --------------- 2 Cr+3(aq) + 3 Sn(s)

Therefore

Eicell = - 0.14 - ( - 0.74)

Eicell = 0.60

In conclusion

$Ecell= E0cell - \frac{0.0591}{n} * \frac{log[Cr+3]^2}{ [ Sn+2]^3}$

$Ecell = 0.60 - \frac{0.0591 }{6} \frac{log( 1.20)^2}{ ( 0.200)^3}$

Ecell = 0.577 V

Read more about Temperature

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5 0

3 years ago

PLEASE HELP...............................................

Serjik [45]

Umm its not a i think... lemme see lol

5 0

4 years ago

WILL GIVE BRAINLIEST!!!

Rus_ich [418]

Answer:

The answer is B

Explanation:

The answer is B because representative particles can only be atoms.

3 0

3 years ago

A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 23.0 mL of

Georgia [21]

NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa

NaOH + CH3COOH → CH3COONa + H2O

Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH

Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH

These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L

Molarity of CH3COOH = 0.0106/0.071 = 0.1493M

CH3COONa = 0.0076 / 0.071 = 0.1070M

pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74.

pH using Henderson - Hasselbalch equation:

pH = pKa + log ([salt]/[acid])

pH = 4.74 + log ( 0.1070/0.1493)

pH = 4.74 + log 0.717

pH = 4.74 + (-0.14)

pH = 4.60.

7 0

3 years ago