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vodomira [7]
3 years ago
6

Click this link to view O*NET's Work Context section for Emergency Medical Technicians. Note that common tools

Chemistry
2 answers:
lawyer [7]3 years ago
7 0

Answer:

( B, C, E, F)

Explanation:

Person above is right!

But I hope this helps!!

agasfer [191]3 years ago
4 0

Answer:

2,3,5,6

Explanation:

i got it right

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Give the answer to the problem below using the correct number of significant digits. (1.3 x 103) x (5.724 x 104)
Gemiola [76]
(133.9) x (595.296)= 79,710.1344
Answer with significant digits: 80,000
1.3 has two significant digits so 79,700 rounds up to 80000.
6 0
3 years ago
Read 2 more answers
Write the half-reactions as they occur at each electrode and the net cell reaction for this electrochemical cell containing indi
AlexFokin [52]

Explanation:

The given cell reaction is as follows.

       In(s)| In^{3+}(aq) || Cd^{2+}(aq) | Cd(s)

Hence, reactions taking place at the cathode and anode are as follows.

At anode ; Oxidation-half reaction : In(s) \rightarrow In^{3+}(aq) + 3e^{-} ...... (1)

At cathode; Reduction-half reaction : Cd^{2+}(aq) + 2e^{-} \rightarrow Cd(s) ....... (2)

Hence, balance the half reactions by multiplying equation (1) by 2 and equation (2) by 3.

Therefore, net cell reaction is as follows.

      2In(s) \rightarrow 2In^{3+}(aq) + 6e^{-}

      3Cd^{2+}(aq) + 6e^{-} \rightarrow 3Cd(s)

Net reaction: 2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)

Thus, we can conclude that the overall cell reaction is as follows.

        2In(s) + 3Cd^{2+}(aq) \rightarrow 2In^{3+}(aq) + 3Cd(s)

4 0
3 years ago
Explain why current flowing indicates the presence of ions
seraphim [82]

Explanation:

An electron current, the flow of electrons, contributes to an electric current since the electron 'carries' negative electric charge. ... The flow of ions (either positively or negatively charged) also contributes to an electric current in, for example, the electrolyte of an electrochemical cell.

hope it will work well?!

3 0
3 years ago
Suppose 180 ml of 3.52x10^-4 M NaOH is mixed with 220 mL of 2.47x10^-4 M MgCl2.
velikii [3]
When in water, MgCl2 dissociates into magnesium ions and Cl- ions and NaOH into Na and OH ions. The equation is as follows:

MgCl2 = Mg2+ + 2Cl-
NaOH = Na+ + OH-

The initial concentrations are as follows:

[Mg2+] = .220(<span> 2.47x10^-4) / .220+.180 = 1.36x10^-4 M Mg2+
</span>[OH-] = .180 (3.52x10^-4) / .220+.180 = 1.58x10^-4 M OH-
6 0
3 years ago
What would I write in there? I’m confused
pav-90 [236]
I assume what they are asking you? Sorry if that sound mean
3 0
2 years ago
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