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Irina-Kira [14]
3 years ago
12

What are the signs given to the electrodes in both the voltaic and electrolytic cells? What reaction (oxidation or reduction) is

occurring at each?
Chemistry
1 answer:
Alex3 years ago
4 0
In a voltaic cell, electrons flow from the anode to the cathode. Anode is negative while cathode is positive.

In an electrolytic cell, electron flows from the cathode to the anode. Anode is positive and cathode is negative.

For both, Oxidation occurs at the anode and Reduction occurs at the cathode.
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The type of social bonds present in modern societies that are based on difference, interdependence, and individual rights is cal
Goryan [66]

Answer:

Organic Solidarity

Explanation:

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GR 11 CHEM M3
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<em>A:  When burning Sulfur, Sulfur Dioxide is released. Having more Oxygen available provides more reactive potential for the burning Sulfur, making it burn much more fiercely. In water, the Sulfur Dioxide forms Sulfurous acid. Added: 12 years ago.</em>

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3 years ago
PLEASE HELP ME BABES AND ASAP PLEASE!!
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What special structures are needed for green plants?
<span> A.chloroplasts and chlorophyll

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5 0
3 years ago
Read 2 more answers
NaHCO3 (s) + HC2H3O2 (aq) = NaC2H3O2 (aq) + H2O (I) + CO2 (g)
Misha Larkins [42]

Moles=volume*concentration   
         =0.1*.83
         =.083 Moles of HC2H3O2
Mole ratio between HC2H3O2 and CO2 is 1:1
This means .083 Moles of CO2

Mass =Moles*Rfm of CO2
         =.083*(12+16+16)
         =3.7grams
8 0
4 years ago
4. Magnesium and oxygen undergo a chemical reaction to
erastovalidia [21]

Answer:

About 16.1 grams of oxygen gas.

Explanation:

The reaction between magnesium and oxygen can be described by the equation:
\displaystyle 2\text{Mg} + \text{O$_2$} \longrightarrow 2\text{MgO}

24.4 grams of Mg reacted with O₂ to produce 40.5 grams of MgO. We want to determine the mass of O₂ in the chemical change.

Compute using stoichiometry. From the equation, we know that two moles of MgO is produced from every one mole of O₂. Therefore, we can:

  1. Convert grams of MgO to moles of MgO.
  2. Moles of MgO to moles of O₂
  3. And moles of O₂ to grams of O₂.

The molecular weights of MgO and O₂ are 40.31 g/mol and 32.00 g/mol, respectively.

Dimensional analysis:

\displaystyle 40.5\text{ g MgO} \cdot \frac{1\text{ mol MgO}}{40.31\text{ g MgO}} \cdot \frac{1\text{ mol O$_2$}}{2\text{ mol MgO}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}

In conclusion, about 16.1 grams of oxygen gas was reacted.

You will obtain the same result if you compute with the 24.4 grams of Mg instead:

\displaystyle 24.4\text{ g Mg}\cdot \frac{1\text{ mol Mg}}{24.31\text{ g Mg}} \cdot \frac{1\text{ mol O$_2$}}{1\text{ mol Mg}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}

3 0
2 years ago
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