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Helga [31]
3 years ago
9

If the only chemical you can find is labeled with the wrong concentration:_______

Chemistry
1 answer:
Softa [21]3 years ago
3 0

Answer:inform the lab instructor and get instructions

Explanation:

If you come across a chemical in the laboratory which has been wrongly labelled, do not be quick to dilute it or take any further action. The laboratory instructor who may have prepared the reagent himself or has better knowledge about the reagent should be contacted immediately so that he/she can give you instructions about what to do with the wrongly labelled reagent.

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Identify the type of the bond​
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That is phosphorus trichloride and a COVALENT BOND

Cl needs one electron and P has metallic character, it donates electron easy
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What might happen if an endocrine hormone such as thyroid hormone was controlled by positive instead of negative feedback?​
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4 0
2 years ago
12. An electrolysis reaction is
tia_tia [17]

Answer: D.) non-spontaneous.

Explanation:

6 0
2 years ago
The potential difference across the terminals of a battery of e.m.f. 12 V and internal resistance 2 ohm drops to 10 V when it is
lesantik [10]

Explanation:

The given data is as follows.

      E.m.f = 12 V,        Voltage = 10 V,        Resistance = 2 ohm

Hence, calculate the current as follows.

          I = \frac{E - V_{t}}{r_{int}}

Putting the given values into the above formula as follows.

         I = \frac{E - V_{t}}{r_{int}}

           = \frac{12 - 10}{2}

           = 1 A

Atomic weight of copper is 63.54 g/mol. Therefore, equivalent weight of copper is \frac{M}{2}.

That is,          \frac{M}{2}

                = \frac{63.54 g/mol}{2}

Hence, electrochemical equivalent of copper is as follows.

                   Z = (\frac{E}{96500}) g/C

                       = (\frac{63.54 g/mol}{2 \times 96500}) g/C

                       = 3.29 \times 10^{-4} g/C

Therefore, charge delivered from the battery in half-hour is calculated as follows.

                     It = Q

                       = 1 \times \frac{1}{2} \times 60 times 60  

                       = 1800 C

So, copper deposited at the cathode in half-an-hour is as follows.

                     M = ZQ

                = 3.29 \times 10^{-4} g/C \times 1800 C

                = 0.5927 g

Thus, we can conclude that 0.5927 g of copper is deposited at the cathode in half an hour.

8 0
3 years ago
TRUE or FALSE: Units in the Sl system include feet, pounds, and<br> degrees.<br><br> True<br> False
luda_lava [24]

Answer:

true true true true true

4 0
2 years ago
Read 2 more answers
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