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2 years ago

1 mole of Cl2 will produce how many atoms of KCl?

Chemistry

1 answer:

olchik [2.2K]2 years ago

6 0

Answer:

1.1 mole

Explanation:

The mass of potassium chloride is 3.8 g . Look at the mole ratio you have between potassium chlorate and potassium chloride; what you will see is that every 2 moles of potassium chlorate will produce 2 moles of potassium chloride, i.e. you have a 1:1 mole ratio between the two compounds

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What does your data indicate about the optimum ph level for this enzyme-catalyzed reaction?

mestny [16]

At an optimum pH of 7.0, there are more molecules per minute in all amounts of substrate thus this pH is ideal for maximum growth. 5. Enzymes function most efficiently at the temperature of a typical cell, which is 37 degrees Celsius. Increases or decreases in temperature can significantly lower the reaction rate.

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4 years ago

15.24. Calcium hydroxide, also known as slaked lime, is used in industrial processes in which low concentrations of base are req

Ivahew [28]

The concentration of the hydroxide ions, OH¯ in 250 mL of the solution containing the maximum amount of dissolved calcium hydroxide is 0.01728 M

We'll begin by calculating the number of mole of in 0.16 g of Ca(OH)₂. This can be obtained as follow:

Mass of Ca(OH)₂ = 0.16 g

Molar mass of Ca(OH)₂ = 40 + 2[16 + 1] = 74 g/mol

Mole = mass / molar mass

Mole of Ca(OH)₂ = 0.16 / 74

Next, we shall determine the molarity of the stock solution of Ca(OH)₂.

Mole of Ca(OH)₂ = 0.00216 mole

Volume = 100 mL = 100 / 1000 = 0.1 L

<h3>Molarity of Ca(OH)₂ =? </h3>

Molarity = mole / Volume

Molarity of Ca(OH)₂ = 0.00216 / 0.1

<h3>Molarity of Ca(OH)₂ = 0.0216 M</h3>

Next, we shall determine the molarity of the diluted solution. This can be obtained as follow:

Volume of stock solution (V₁) = 100 mL

Molarity of stock solution (M₁) = 0.0216 M

Volume of diluted solution (V₂) = 250 mL

<h3>Molarity of diluted solution (M₂) =?</h3>

0.0216 × 100 = M₂ × 250

2.16 = M₂ × 250

Divide both side by 250

M₂ = 2.16 / 250

Thus, the molarity of the diluted solution is 0.00864 M

Finally, we shall determine the concentration of the hydroxide ions, OH¯ in the diluted solution. This can be obtained as follow:

Ca(OH)₂(aq) —> Ca²⁺(aq) + 2OH¯(aq)

From the balanced equation above,

1 mole of Ca(OH)₂ contains 2 moles of OH¯

Therefore,

0.00864 M Ca(OH)₂ will contain = 2 × 0.00864 = 0.01728 M OH¯

Thus, the concentration of the hydroxide ions, OH¯ in 250 mL of the solution containing the maximum amount of dissolved calcium hydroxide is 0.01728 M

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7 0

3 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

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2 years ago

21 mL It required 42.35 mL of H2SO4 to neutralize 21.17 mL of 0.5000 M NaOH. Calculate the concentration of H2SO4

bija089 [108]

The balanced equation for the above reaction is
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of NaOH moles required-0.5000 M / 1000 mL/L x 21.17 mL = 0.010585 mol
According to stoichiometry, acid moles required are 1/2 of the base moles reacted
Therefore number of H₂SO₄ moles reacted - 0.010585 /2 mol
Number of moles in 42.35 mL of H₂SO₄ - 0.010585 /2 mol
Therefore in 1 L solution - (0.010585) /2 / 42.35 mL x 1000 mL/L = 0.125 M
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3 years ago

The freezing point of benzene C6H6 is 5.50°C at 1 atmosphere. A nonvolatile, nonelectrolyte that dissolves in benzene is DDT . H

Vlad1618 [11]

Answer: 0.028 grams

Explanation:

Depression in freezing point :

Formula used for lowering in freezing point is,

$\Delta T_f=k_f\times m$