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Alex Ar [27]
3 years ago
10

Doing an experiment about electroplating, you attempt to coat silver in gold using a basic electroplating set-up. You take the m

ass of the silver before and after the experiment. What can you say about the mass of the silver/gold product at the end of the experiment?
a. it’s mass was greater than when it started

b. it’s mass was lower than when it started
Chemistry
1 answer:
Elan Coil [88]3 years ago
3 0

Answer:

it’s mass was greater than when it started

Explanation:

When a metal is coated with another metal, the plating metal deposits on the plated metal. Usually, the plating metal functions as the anode while the plated metal functions on the cathode. The anode metal is oxidized and reduced at the cathode and become deposited on the cathode material. This increases the mass of the cathode. Hence the mass of the silver/gold product is greater than the mass of silver at the beginning of the electroplating process.

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As a chemist for an agricultural products company, you have just developed a new herbicide,"Herbigon," that you think has the po
Ganezh [65]

Answer:

pH = 2.03

Explanation:

The pH can be calculated using the following equation:

pH = -log [H_{3}O^{+}]  (1)

The concentration of H₃O⁺ is calculated using the dissociation constant of the next reaction:

CH₃COOH + H₂O ⇄  CH₃COO⁻ + H₃O⁺    

   1.00 M    

K_{a} = \frac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]}

Solving the above equation for H₃O⁺, we have:    

[H_{3}O^{+}] = \frac{Ka*[CH_{3}COOH]}{[CH_{3}COO^{-}]}    (2)    

The dissociation constant is equal to:    

pKa = -log(Ka) \rightarrow Ka = 10^{-pKa} = 10^{-4.76} = 1.74 \cdot 10^{-5}    

Now, by solving the equation of the solubility product for Herbigon, we can find [CH₃COO⁻]:

CH₃COOX  ⇄  CH₃COO⁻ +  X⁺  

                                             5.00x10⁻³ M

K_{sp} = [CH_{3}COO^{-}][X^{+}]

[CH_{3}COO^{-}] = \frac{K_{sp}}{[X^{+}]} = \frac{9.40 \cdot 10^{-6}}{5.00 \cdot 10^{-3}} = 1.88 \cdot 10^{-3} M

By entering the values of [CH₃COO⁻] and Ka, into equation (2) we can calculate [H₃O⁺]:

[H_{3}O^{+}] = \frac{1.74 \cdot 10^{-5}*[1.00]}{[1.88 \cdot 10^{-3}]} = 9.26 \cdot 10^{-3} M

Hence, the pH is:

pH = -log [H_{3}O^{+}] = -log [9.26 \cdot 10^{-3}] = 2.03

Therefore, the pH must be 2.03 to yield a solution in which the concentration of X⁺ is 5.00x10⁻³M.

I hope it helps you!  

6 0
4 years ago
In his pamphlet Common Sense, Thomas Paine
kati45 [8]
In his pamphlet Common Sense, Thomas Paine urged American colonists to establish their own nation. The correct option among all the options that are given in the question is option "1". The pamphlet by Thomas Paine actually urged the colonists to denounce the British rule. He was of the opinion that it was not fair that an island would rule over a total continent.<span>He also wrote that being a part of Britain would unnecessarily drag America into unwanted wars with European countries.</span>


6 0
3 years ago
How do you determine the theoretical number of electrons in an energy level?
Makovka662 [10]
I'm pretty sure the formula is 2n(/\) 2
3 0
3 years ago
Most caves form in: *<br> O Igneous rock<br> O Compacted soil<br> O Limestone<br> Sandstone
KonstantinChe [14]
Igneous Rock (I believe)
6 0
3 years ago
Match the words in the below to the appropriate blanks in the sentences.
12345 [234]

Answer:

Sn and Ge

Explanation:

To obtain the more metallic element, we compare the group in which both elements are. Generally the element with the lower ionzation energy is he more metallic one.

Ionization energy increases fro left to right across a period. Ionization energy decreases down the group.

1. When comparing the two elements A s and S n , the more metallic element is ______based on periodic trends alone.

Sn has a lower ionization energy so it is the more metallic one.

2. When comparing the two elements G e and S b , the more metallic element is ________ based on periodic trends alone.

Ge has a lower ionizaiton energy compared to Sb. So it is more metallic element than Sb.

6 0
4 years ago
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