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zhannawk [14.2K]
3 years ago
12

How many moles are present in a 24.5 gram sample of K2Cr2O7?

Chemistry
1 answer:
attashe74 [19]3 years ago
6 0
The answer to this is solved through stochiometry: the answer is  this: 0.0833mol

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What is an example of an interest leading to a career choice
Leno4ka [110]

Answer:

An example of these would be "I have always been interested in cars so I became a auto mechanic." or "I have always been interested in cars so I decided to work at a car dealership."

Explanation:

A big interest that you have that leds you to pick a job related to that interest.

4 0
3 years ago
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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
True or false. (Pls help)
grigory [225]
I believe they’re both true.
5 0
2 years ago
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A spectrophotometer measures the transmittance or the absorbance, or both, of a particular wavelength of light after it has pass
Cerrena [4.2K]

Answer:Answer: The step that is NOT necessary to complete before a cuvette is placed into the spectrophotometer is option B (Write, in ink, either sample or blank on the side of the cuvette to keep track of them)

Explanation: spectrophotometer is an instrument used to measure the light intensity absorbed after being passed through a solution. Before the absorbance of the sample solution, a solvent solution called blank is used for the calibration of the machine and this blank solvent is placed in a cuvette. The procedure usually comes first before the main sample is processed. Therefore there is no need to

Write, in ink, either sample or blank on the side of the cuvette to keep track of them. This is so since sample and blank is not absorbed at the same time by the machine.

7 0
3 years ago
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A buffer contains 0.18 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. What is the pH
Yuki888 [10]

Answer:

1) pH = 5.05

2) pH = 5.13

3) pH = 4.97

Explanation:

Step 1: Data given

Number of moles of propionic acid = 0.18 moles

Number of moles sodium propionate = 0.26 moles

Volume = 1.20 L

Ka = 1.3 * 10^-5    → pKa = 4.989

Step 2: Calculate concentrations

Concentration = moles / volume

[acid]= 0.18/ 1.2 =0.150 M

[salt]= 0.26/ 1.3 = 0.217 M

pH = 4.89 + log(0.217/0.150)=<u>5.05</u>

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What is the pH of the buffer after the addition of 0.02 mol of NaOH?

moles acid = 0.18 - 0.02 = 0.16

[acid]= 0.16/ 1.2=0.133 M

moles salt = 0.26 + 0.02 = 0.28

[salt]= 0.28/ 12=0.233

pH = 4.89 + log 0.233/ 0.133 = 5.13

What is the pH of the buffer after the addition of 0.02 mol of HI?

moles acid = 0.18+ 0.02 = 0.20 moles

[acid]= 0.20/ 1.2 = 0.167 M

[salt]= 0.26 - 0.02= 0.24 moles

[salt]= 0.24/ 1.2 = 0.20 M

pH = 4.89 + log 0.20/ 0.167= 4.97

8 0
3 years ago
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