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shepuryov [24]
3 years ago
11

Given the balanced ionic equation representing the reaction in an operating voltaic cell: zn(s) + cu2+(aq) → zn2+(aq) + cu(s) th

e flow of electrons through the external circuit in this cell is from the
Chemistry
1 answer:
soldi70 [24.7K]3 years ago
3 0

Answer: from the Zn anode to the Cu cathode


Justification:


1) The reaction given is: Zn(s) + Cu₂⁺ (aq) -> Zn²⁺ (aq) +Cu(s)


2) From that, you can see the Zn(s) is losing electrons, since it is being oxidized (from 0 to 2⁺), while Cu²⁺, is gaining electrons, since it is being reduced (from 2⁺ to 0).


3) Then, you can already tell that electrons go from Zn to Cu.


4) The plate where oxidation occurs is called anode, and the plate where reduction occus is called cathode.


So you get that the electrons flow from the anode (Zn) to the cathode (Cu).


Always oxidation occurs at the anode, and reduction occurs at the cathode.

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In Nuclear chemistry, the key requirement for a chain reaction is that...a. each event must produce more than one particle capab
ahrayia [7]

Cccccccccccccccccccccccc

5 0
3 years ago
A compound contains only change and n combustion of 35.0mg of the compound produces 33.5mg co2 and 41.1mg h2o. What is the empir
Viefleur [7K]

Answer:

The empirical formula is CH6N2

Explanation:

A compound containing only C, H, and N yields the following data. Complete combustion of 35.0 mg of the compound produced 33.5 mg of CO2 and 41.1 mg of H2O. What is the empirical formula of the compound

Step 1: Data given

Mass of the compound = 35.0 mg = 0.035 grams

Mass of CO2 = 33.5 mg = 0.0335 grams

Mass of H2O = 41.1 mg = 0.0411 grams

Molar mass CO2 = 44.01 g/mol

Molar mass H2O = 18.02 g/mol

Molar mass C = 12.01 g/mol

Molar mass O = 16.0 g/mol

Molar mass H = 1.01 g/mol

Step 2: Calculate moles CO2

Moles CO2 = 0.0335 grams / 44.01 g/mol

Moles CO2 = 7.61 *10^-4 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 7.61 *10^-4 moles CO2 we have 7.61 *10^-4 moles C

Step 4: Calculate mass C

Mass C = 7.61 *10^-4 moles * 12.01 g/mol

Mass C = 0.00914 grams = 9.14 mg

Step 5: Calculate moles H2O

Moles H2O = 0.0411 grams / 18.02 g/mol

Moles H2O = 0.00228 moles

Step 6: Calculate moles H

For 1 mol H2O we have 2 moles H

For 0.00228 moles H2O we have 2* 0.00228 = 0.00456 moles H

Step 7: Calculate mass H

Mass H = 0.00456 moles * 1.01 g/mol

Mass H = 0.00461 grams = 4.61 mg

Step 8: Calculate mass N

Mass N = 35.0 mg - 9.14 - 4.61 = 21.25 mg = 0.02125 grams

Step 9: Calculate moles N

Moles N = 0.02125 grams / 14.0 g/mol

Moles N = 0.00152 moles

Step 10: Calculate the mol ratio

We divide by the smallest amount of moes

C: 0.000761 moles / 0.000761 moles= 1

H:  0.00456 moles / 0.000761 moles = 6

N: 0.00152 moles  / 0.000761 moles = 2

For every C atom we have 6 H atoms and 2 N atoms

The empirical formula is CH6N2

5 0
3 years ago
The amount ofcalcium present in milk can be determined by adding oxalate to asample and measuring the massof calcium oxalate pre
Sauron [17]

<u>Answer:</u> The mass percent of calcium in milk is 0.107 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

Given mass of calcium oxalate = 0.429 g

Molar mass of calcium oxalate = 128.1 g/mol

Putting values in equation 1, we get:

\text{Moles of calcium oxalate}=\frac{0.429g}{128.1g/mol}=0.0033mol

The given chemical equation follows:

Na_2C_2O_4(aq.)+Ca^{2+}(aq.)\rightarrow CaC_2O_4(s)+2Na^+(aq.)

Sodium oxalate is present in excess. So, it is considered as an excess reagent. And, calcium ion is a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of calcium oxalate is produced from 1 mole of calcium ion

So, 0.0033 moles of calcium oxalate is produced from = \frac{1}{1}\times 0.0033=0.0033mol of calcium ions

  • Now, calculating the mass of calcium ions by using equation 1, we get:

Moles of calcium ions = 0.0033 moles  

Molar mass of calcium ions = 40 g/mol

Putting values in equation 1, we get:

0.0033mol=\frac{\text{Mass of calcium ions}}{40g/mol}\\\\\text{Mass of calcium ions}=(0.0033mol\times 40g/mol)=0.132g

  • To calculate the mass percentage of calcium ions in milk, we use the equation:

\text{Mass percent of calcium ions}=\frac{\text{Mass of calcium ions}}{\text{Mass of milk}}\times 100

Mass of milk = 125 g

Mass of calcium ions = 0.132 g

Putting values in above equation, we get:

\text{Mass percent of calcium ions}=\frac{0.132g}{125g}\times 100=0.107\%

Hence, the mass percent of calcium in milk is 0.107 %

7 0
3 years ago
How many oxygen molecules are needed to make 10 carbon dioxide molecules according to the following balanced chemical equation?
11Alexandr11 [23.1K]
<h2 /><h2 /><h2>answer.</h2>

five oxygen molecules

step by step explanation.

according to the equation,one molecule of oxygen is enough to react with two carbon molecules thus 10 carbon molecules need 5oxygen molecules

6 0
3 years ago
How much heat, in calories, is given off when 1.25 grams of silver is cooled from 100.0 degrees Celsius to 80 degrees Celsius? (
lapo4ka [179]
The heat required to raise the temperature of a certain mass of sample to a specific temperature change, we use the formula mCpΔT where m is mass, Cp is the specific heat of the substance and ΔT is the temperature change. In this case, we substitute and form 1.25 g x 0.057 cal/g C *20 C equal to 1.425 calories.
7 0
3 years ago
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