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o-na [289]
3 years ago
12

Which of the following is not part of the proper protocol for using acids and bases?

Chemistry
1 answer:
Anni [7]3 years ago
8 0

Answer:

B.Add acid to water,not water to acid

Explanation:

they should not be mixed

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The answer is silicon.

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Write balanced equations for each of the following by inserting the correct coefficients in the blanks:
Aleksandr-060686 [28]

Balanced equation :

Cu(NO₃)₂(aq) + 2KOH(aq) → Cu(OH)₂(s)  + 2KNO₃(aq)

Balancing a chemical equation :

A chemical equation shows us the substances involved in a chemical reaction - the substances that react (reactants) and the substances that are produced (products). In general, a chemical equation looks like this:

                                 Reactant →Product

According to the law of conservation of mass, when a chemical reaction occurs, the mass of the products should be equal to the mass of the reactants. Therefore, the amount of the atoms in each element does not change in the chemical reaction. As a result, the chemical equation that shows the chemical reaction needs to be balanced. A balanced chemical equation occurs when the number of the atoms involved in the reactants side is equal to the number of atoms in the products side.

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3 0
1 year ago
Which factors influence the salinity of ocean surface water?
In-s [12.5K]
C and D

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4 0
3 years ago
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The melting point of h2o is 0 degrees celsius. this is the same as its:
professor190 [17]
This is the same as its freezing point

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6 0
3 years ago
Using the standard reduction potentials, Pb 2+(aq) + 2e– => Pb(s), E° = –0.13 V Fe 2+(aq) + 2e– => Fe(s), E° = –0.44 V Zn
mojhsa [17]

Answer:

Pb(s), Fe(s) and Zn(s) will reduce Mn^{3+} to Mn^{2+}

Fe(s) and Zn(s) will reduce Cr^{3+} to Cr

Explanation:

Standard reduction potential denotes ability to consume electrons from another species.

Hence, higher the standard reduction potential, higher will be the ability to oxidize another species.

Metal with E_{red}^{0} value lower than 1.51 V will donate electron to Mn^{3+} and thus reduces Mn^{3+} to Mn^{2+}.

So, Pb(s), Fe(s) and Zn(s) will reduce Mn^{3+} to Mn^{2+}.

Metal with E_{red}^{0} value lower than -0.40 V will donate electron to Cr^{3+} and thus reduces Cr^{3+} to Cr.

So, Fe(s) and Zn(s) will reduce Cr^{3+} to Cr.

6 0
3 years ago
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