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MariettaO [177]
2 years ago
7

Which half-reaction is most easily oxidized?

Chemistry
2 answers:
timama [110]2 years ago
8 0

Which half-reaction is most easily oxidized?

Answer: Na^+ + e → Na= -2.71 V (D for plato users)


alisha [4.7K]2 years ago
7 0

Answer:

Na+ + e- Na = -2.71 V

Explanation:

We have to remember the nomenclature for redox reactions:

<u>Oxidation</u>:

M~->~M^++e^-

<u>Reduction</u>:

M^+ + e^-~ ->M

So, all the answer options are written as <u>reductions</u>. Therefore the voltage values, in this case, would be the <u>reduction potentials</u>. If we have a higher (<u>positive</u>) potential the compound would be <u>more inclined to a reduction</u> and vice-versa if we have a smaller potential (<u>negative</u>) the compound would be <u>more inclined to oxidation</u>.

In this case, the <u>most negative value</u> is for Na+, so this atom would be more easily <u>oxidized</u>.

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Cytokinesis happens differently for plant and animal cells. Both separate cytoplasm between two new daughter cells. However, whi
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Answer:

Animal Cell Cytokinesis

Explanation:

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3 years ago
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Water has a high specific heat because: Select one: a. it is a poor insulator b. hydrogen bonds must be broken to raise its temp
Crank
<h2>The required "option is b) hydrogen bonds must be broken to raise its temperature.</h2>

Explanation:

  • Water has high specific heat due to hydrogen bonds present in it.
  • The Ionisation of water does not affect the specific heat of the water.
  • On decreasing the temperature, there is the formation of bonds hence option (d) is wrong.
  • On increasing the temperature, there is the breaking of bonds hence option (b) is correct.
3 0
3 years ago
A 13.30 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.00 grams of CO2 and 2
a_sh-v [17]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is CHO_2 and C_2H_2O_4

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=13.00g

Mass of H_2O=2.662g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 13.00 g of carbon dioxide, \frac{12}{44}\times 13.00=3.54g of carbon will be contained.

<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.662 g of water, \frac{2}{18}\times 2.662=0.296g of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.295 moles.

For Carbon = \frac{0.295}{0.295}=1

For Hydrogen = \frac{0.296}{0.295}=1

For Oxygen = \frac{0.603}{0.295}=2.044\approx 2

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is CHO_2

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 90.04 g/mol

Mass of empirical formula = 45 g/mol

Putting values in above equation, we get:

n=\frac{90.04g/mol}{45g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4

Hence, the empirical and molecular formula for the given organic compound is CHO_2 and C_2H_2O_4

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3 years ago
What is20 grams of HNO^3
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26. Balance the following equations:<br> Ca(s) + H3PO4(aq)Ca3(PO4)2(s) + H2(g)
yKpoI14uk [10]

Hey there!

Ca + H₃PO₄ → Ca₃(PO₄)₂ + H₂

Balance PO₄.

1 on the left, 2 on the right. Add a coefficient of 2 in front of H₃PO₄.

Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + H₂

Balance H.

6 on the left, 2 on the right. Add a coefficient of 3 in front of H₂.

Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂

Balance Ca.

1 on the right, 3 on the right. Add a coefficient of 3 in front of Ca.

3Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂

Our final balanced equation:

3Ca + 2H₃PO₄ → Ca₃(PO₄)₂ + 3H₂

Hope this helps!

3 0
3 years ago
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