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lutik1710 [3]
3 years ago
6

PLS HELP ASAP WITH SCIENCE HW

Chemistry
1 answer:
polet [3.4K]3 years ago
6 0

Answer:

Concave

Explanation:

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In the galvanic cell Al(s) ǀ Al3+(aq, 1 M) ǀǀ Cu2+(aq, 1 M) ǀ Cu(s) which of the following changes will increase the cell potent
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The cathode electrode is copper. This means that the concentration of copper must be increased to increase the cell potential. This can be done by diluting the Al3+ solution. Increasing the surface area will not change the current but will increase the voltage. The answer is (D) I and III only.
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A 0.199 M weak acid solution has a pH of 3.99. Find Ka for the acid.
Alex17521 [72]

Answer:

Ka = ( About ) 5 x 10^ - 8

Explanation:

Let us first identify the dissociation equation for this weak acid,

HA ⇌ ( H+ ) + A¯

Knowing this, we can tell what the equilibrium expression is, respectively,

Ka = ( [ H+ ] [ A¯ ] ) / [ HA ]

_________________________________________________

Now let us use the given pH 3.99 to calculate [ H+ ], knowing that pH = −log [H+],

3.99 = - log[H+],

[H+] = 10 ^ - 3.99,

[H+] = ( About ) 1 * 10^-4 M

Substitute known values into the equilibrium expression,

Ka = [( 1 x 10^ - 4 ) ( 1 x 10^ ¯4 )] / 0.199,

Ka = ( About ) 5 x 10^ - 8

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3 years ago
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