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stiv31 [10]
3 years ago
14

Based on your observations of the laboratory assignment(s) that produced gases, can you conclude

Chemistry
1 answer:
Alja [10]3 years ago
5 0

Answer:

With the use of same chemicals, same gas will be produced while the use of different chemicals, different gases will be produced.

Explanation:

The same gas was produced in each assignment if the same chemicals are mixed with each other while different gases were produced in each assignment when different chemicals are allowed to mix. If every time, same chemicals and same concentration of chemicals are mixed together, the result will be the same means production of same gas but if different chemicals are allowed to mix, then the product will be the production of different gases.

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1-<br> the<br> Nalt and Gil-<br> Cu2+<br> ad out- <br> what are the spectator ions in this reaction
Mrac [35]

Answer:

cu2 is 5he correct answer

3 0
2 years ago
Help me in my this plzzz ​
eduard

Answer:

a is oxidation

b is reduction

c is reduction

d is oxidation

hope it helps you

8 0
3 years ago
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A certain metal M forms a soluble sulfate salt MSO, Suppose the left half cell of a galvanic cell apparatus is filled with a 3.0
dimaraw [331]

Explanation:

It is known that for high concentration of M^{2+}, reduction will take place. As, cathode has a positive charge and it will be placed on left hand side.

Now, E^{o}_{cell} = 0 and the general reaction equation is as follows.

         M^{2+} + M \rightarrow M + M^{2+}

                3.00 M        n = 2       30 mM

         E = 0 - \frac{0.0591}{2} log \frac{50 \times 10^{-3}}{1}

            = -\frac{0.0591}{2} log (5 \times 10^{-2})

            = 0.038 V

Therefore, we can conclude that voltage shown by the voltmeter is 0.038 V.

5 0
3 years ago
Số hợp chất tố đa của nguyên tố này với nguyên tố khác theo hóa trị của nó
SVETLANKA909090 [29]

Answer:

Carbon

Explanation:

Carbon

5 0
2 years ago
Calculate the solubility of mn(oh)2 in grams per liter when buffered at ph=7.0. assume that buffer capacity is not exhausted
Vanyuwa [196]
When PH + POH = 14 
∴ POH = 14 -7 = 7

when POH = -㏒[OH-]

          7    = -㏒ [OH-]
∴[OH-] = 10^-7

by using ICE table:

           Mn(OH)2(s) ⇄  Mn2+ (aq)  + 2OH-(aq)
initial                              0                     10^-7
change                           +X                      +2X
Equ                                 X                  (10^-7 + 2X)

when Ksp = [Mn2+][OH-]^2

when Ksp of Mn(OH)2 = 4.6 x 10^-14

by substitution:

4.6 x 10^-14 = X*(10^-7+2X)^2  by solving this equation for X

∴ X =2.3 x 10-5 M

∴ The solubility of Mn(OH)2 in grams per liter (when the molar mass of Mn(OH)2 = 88.953 g/mol
= 2.3 x10^-5 moles/L * 88.953 g/mol

= 0.002 g/ L
3 0
3 years ago
Read 2 more answers
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