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3 years ago

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In a healthy pond, the temperature is 16°C (61°F). What is the most likely pH of this pond?

1 answer:

masha68 [24]3 years ago

7 0

Answer:

A . 6.3 In a healthy pond, the temperature is 16°C (61°F). What is the most likely pH of this pond

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A reaction has a forward rate constant of 2.3 × 106 s-1 and an equilibrium constant of 4.0 × 108. what is the rate constant for

avanturin [10]

Answer is: 5,75·10⁻¹.
Kf = 2,3·10⁶ 1/s.
K = 4,0·10⁸ 1/s.
Kr = ?
Kf - <span>forward rate constant.
K - </span><span>equilibrium constant.
Kr - </span><span>reverse rate constant.
</span>Since both Kf and Kr are constants at a given temperature, their ratio is also a constant that is equal to the equilibrium constant K.<span>
K = Kf/Kr.
Kr = Kf/K = </span>2,3·10⁶ 1/s ÷ 4,0·10⁸ 1/s = 5,75·10⁻¹.

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3 years ago

Figuring out how to make a better type of plastic is which type of research

quester [9]

Answer:applied chemistry

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3 years ago

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The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

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3 years ago

Convert 1.5 km to millimeters, and express the result in scientific notation.

Grace [21]

The answer is 1.5e+6. Hope this helped!

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What does saturated mean in science?<br>

Sergeu [11.5K]

A saturated is one in which the atoms are linked by single bonds. :)

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