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STALIN [3.7K]
2 years ago
11

How many grams of KCl are produced?

Chemistry
1 answer:
ZanzabumX [31]2 years ago
4 0

Answer:

= 74.4 grams / mole. Ernest Z. The reaction will produce 15.3 g of KCl

Explanation:

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expeople1 [14]
<span>Classify is the answer hope this helps :)</span>
3 0
3 years ago
A chemist designs a galvanic cell that uses these two half-reactions: half-reaction standard reduction potential (s)(aq)(aq)(l)
miv72 [106K]

Answer:

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)  

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻        

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

E°cell = 1.10 V

Explanation:

<em>The half-reactions are missing, but I will propose some to show you the general procedure and then you can apply it to your equations.</em>

<em>Suppose we have the following half-reactions.</em>

<em>Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)   E°red = 0.34 V</em>

<em>Zn²⁺(⁺aq) + 2 e⁻ → Zn(s)    E°red = -0.76 V</em>

<em />

To identify how to make a spontaneous cell, we need to consider the standard reduction potentials (E°red). The half-reaction with the higher E°red will occur as a reduction (in the cathode), whereas the one with the lower E°red will occur as an oxidation (in the anode).

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)   E°red = 0.34 V

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻        E°red = -0.76 V

To get the overall equation we add both half-reactions.

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E°cell = E°red, cat - E°red, an

E°cell = 0.34 V - (-0.76 V) = 1.10 V

Since E°cell > 0, the reaction is spontaneous.

5 0
3 years ago
What is kinetic energy related to?<br><br> heat<br><br> position<br><br> motion<br><br> electricity
Anna35 [415]
The Answer Should Be Motion
8 0
3 years ago
Read 2 more answers
Which of the following would be a reasonable synthesis of CH3CH2CH2CH2OH?Group of answer choices
vampirchik [111]

Answer:

B. 1-Butene rightarrow (1) BH3: THF (2)H202, OH-

Explanation:

In the hydroboration of alkenes, an alkene is hydrated to form an alcohol with anti-Markovnikov orientation.

the reagent BH₃:THF is the way that borane is used in organic reactions. The BH₃ adds to the double bond of an alkene to form an alkyl borane. Peroxide hydrogen in basic medium oxidizes the alkyl borane to form an alcohol. Indeed, hydroboration-oxidation converts alkenes to alcohols by adding water through the double bond, with anti-Markovnikov orientation.

8 0
3 years ago
Predict the products of La(s) + O2(aq) -&gt;
mina [271]

Answer:

i. La (s) + O2 (g) => LaO2 (s)

ii. La (s) + O2 (g) => La2O (s)

III. La (s) + O2 (g) => La2O3 (s)

Explanation:

<em>Hello </em><em>there!</em>

When you are given such a problem for completing the chemical equations, what you have to understand is that metals are found in groups I, II and III. While Oxygen is a group VI element.

From the above question I have considered that my La(s), solid is either Sodium (Na) - group I, Magnesium - group II and Aluminum - group III.

In a reaction, there is exchange of electrons given by their oxidation numbers (I, II and III - for our metals above)

The chemical equations are thus;

i. Na (s) + O2 (g) => NaO (s)

ii. Mg (s) + O2 (g) => Mg2O (s)

iii. Al (s) + O2 (g) => Al2O3 (s)

Relate this to the problem and it will be;

i. La (s) + O2 (g) => LaO2 (s)

ii. La (s) + O2 (g) => La2O (s)

III. La (s) + O2 (g) => La2O3 (s)

<em>I hope this </em><em>helps </em><em>you</em><em> </em><em>to </em><em>understand</em><em> </em><em>better</em><em>.</em><em> </em><em>Enjoy </em><em>your</em><em> </em><em>studies</em>

3 0
2 years ago
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