All categories

2 years ago

5. Chlorine gas was formed at the ... *

1 answer:

5 0

Clorine gas was formed at the positive electrode. 

You might be interested in

Be sure to answer all parts. what are the concentrations of hso4−, so42−, and h in a 0. 31 m khso4 solution? (hint: h2so4 is a s

disa [49]

At equilibrium the concentrations of:

[HSO₄⁻] = 0.10 M;

[SO₄²⁻] = 0.037 M;

[H⁺] = 0.037 M;

There is initially very little H+ and no SO₄²⁻ in the solution. A salt is KHSO₄⁻. All KHSO₄⁻ will split apart into K⁺ and HSO₄⁻ ions. [HSO₄⁻] will initially be present at a concentration of 0.14 M.

HSO₄⁻ will not gain H⁺ to produce H₂SO₄ since H₂SO₄ is a strong acid. HSO₄⁻ may act as an acid and lose H⁺ to form SO₄²⁻. Let the final H⁺ concentration be x M. Construct a RICE table for the dissociation of HSO₄²⁻.

R $HSO_4^-$ ⇄ $H^+ + SO_4^2^-$

I $0.14$

C $- x$ $+x$ $+x$

E $0.14-x$ $x$ $x$

$K_a = 1.3$ × $10^-^2$ for $HSO^-_4$ . As a result,

$\frac{[H^+]. [SO_4^2^-]}{HSO_4^-} = K_a$

$K_a$ is large. It is no longer valid to approximate that $[HSO^-_4]$ at equilibrium is the same as its initial value.

$\frac{x^2}{0.14-y} = 1.3 * 10^-^2$

$x^2+1.3*10^-^2x - 0.14$ × $1.3$ × $10^-^2= 0$

Solving the quadratic equation for $x , x \geq 0$ since $x$ represents a concentration;

$x=0.0366538$

Then, round the results to 2 significant figure;

$[SO_4^2^-] = x = 0.037 mol. L ^-^1$
$[H^+] = x = 0.037 mol. L ^-^1$
$[HSO_4^-] = 0.14 - x = 0.10 mol. L ^-^1$

Learn more about concentration here:

brainly.com/question/14469428

#SPJ4

3 0

1 year ago

A buffer is a solution that is a mixture of either a weak acid and its conjugate base or a weak base and its conjugate acid. Whe

babymother [125]

Answer:

b. The weak base of an alkaline buffer will accept hydrogen protons when a strong acid is added to the solution

d.The conjugate acid of an alkaline buffer will donate hydrogen protons when a strong base is added to the solution.

Explanation:

A buffer is a solution that resist pH change, it shows minimal change upon addition of small amount of strong acid or strong base. An alkaline buffer will have a basic pH, above 7. It is made by mixing a weak base and its salt with a strong acid. An example of an alkaline buffer is carbonate-bicarbonate buffer that is prepared using varying amount of anhydrous sodium carbonate and volume of solution of sodium bicarbonate to get pH range between 9.2 to 10.7

Within the buffer,the salt is completely ionized while the weak base is partly ionized. on addition of acid, the released protons will be removed by the bicarbonate ion to form sodium carbonate; on addition of base, the hydroxide ion released by the base will be removed by the hydrogen ions to form water and the pH remains relatively the same

6 0

2 years ago

(-5)+(+7)-(-4) + (+12) A 8 B 10 C 18 d 20

Vladimir79 [104]

Answer:

C-18

Explanation:

Step one follow order of operations

Add and subtract from left to right(-5)+(7)=2-(-4)+(12)

STEP 2

Apply negative Rule -(-4)=+4=2+4+(12)

then add 2+4+12=18

5 0

2 years ago

A liquid that evaporates at a slow rate exhibits __________.

Marizza181 [45]

Strong internolecurar forces (A) hope it helps

3 0

3 years ago

The equation represents the combustion of sucrose. C12H22O11 + 12O2 12CO2 + 11H2O If there are 10.0 g of sucrose and 8.0 g of ox

Vaselesa [24]

0.0292 moles of sucrose are available. First, lookup the atomic weights of all involved elements Atomic weight Carbon = 12.0107 Atomic weight Hydrogen = 1.00794 Atomic weight Oxygen = 15.999 Now calculate the molar mass of sucrose 12 * 12.0107 + 22 * 1.00794 + 11 * 15.999 = 342.29208 g/mol Divide the mass of sucrose by its molar mass 10.0 g / 342.29208 g/mol = 0.029214816 mol Finally, round the result to 3 significant figures, giving 0.0292 moles

7 0

2 years ago