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Readme [11.4K]
3 years ago
8

Someone please help?!?

Chemistry
1 answer:
ss7ja [257]3 years ago
4 0

Answer: to late sorry maybe next time

Explanation:

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Good day please help me thank you
NARA [144]
I think c I’m not positive about the question but I think c
6 0
3 years ago
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How many grams of KCl are produced?
ZanzabumX [31]

Answer:

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

Explanation:

4 0
2 years ago
The following data was collected when a reaction was performed experimentally in the laboratory.
REY [17]

Answer:

The answer to your question is 3 moles of AlCl₃

Explanation:

Process

1.- Write and balance the equation

                  Al(NO₃)₃ + 3NaCl   ⇒   3NaNO₃  +  AlCl₃

2.- Determine the limiting reactant

Theoretical proportion     1 mol Al(NO₃)₃ :  3 moles of NaCl                  

Experimental proportion   4 moles Al(NO₃)₃ : 9 moles NaCl

From these values, we determine that the limiting reactant is NaCl because the number of moles increases three times and the number of moles of Al(NO₃)₃  increases four times.

3.- Determine the amount of AlCl₃ using proportions

                       3 moles of NaCl --------------- 1 mol of AlCl₃

                       9 moles of NaCl ----------------  x

                       x = (9 x 1) / 3

                       x = 9 /3

                       x = 3 moles

                     

5 0
3 years ago
For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia
Fittoniya [83]

Answer : The value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction:  Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction:  Cu^{2+}+2e^-\rightarrow Cu

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu

To calculate the E^o_{(Cu^{2+}/Cu)} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}

Putting values in above equation, we get:

1.10V=E^o_{(Cu^{2+}/Cu)}-(-0.76V)

E^o_{(Cu^{2+}/Cu)}=0.34V

Hence, the value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

8 0
3 years ago
Which of the following instruments is a combination of a percussion and a stringed instrument?
murzikaleks [220]

Answer:B

Explanation:

3 0
3 years ago
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