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3 years ago

Question 10 OT 20

Chemistry

1 answer:

Airida [17]3 years ago

8 0

Answer:

B) is reduced.

Explanation:

Oxidation:

Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.

Reduction:

Reduction involve the gain of electron and oxidation number is decreased.

Consider the following reactions.

4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl

the oxidation state of copper is changed from +2 to +1 so copper get reduced and it is oxidizing agent.

CO + H₂O → CO₂ + H₂

the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized and it is reducing gent.

Oxidizing agents:

Oxidizing agents oxidize the other elements and itself gets reduced.

Reducing agents:

Reducing agents reduced the other element are it self gets oxidized.

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What is the theoretical mass of CO2 + H2O lost per gram of NaHCO3?

Mama L [17]

Answer:

The theoretical mass of CO2 = 2.09 grams

The theoretical mass of H2O = 0.429 grams

The theoretical mass of CO2+ H2O = 2.519 grams

Explanation:

Step 1: Data given

Mass of NaHCO3 = 1 gram

Molar mass NaHCO3 = 84.00 g/mol

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation

4NaHCO3 + O2 → 4CO2 + 2H2O + 2Na2O2

Step 3: Calculate moles of NaHCO3

Moles NaHCO3 = mass NaHCO3 / molar mass NaHCO3

Moles NaHCO3 = 1.0 gram / 84.00 g/mol

Moles NaHCO3 = 0.0119 moles

Step 4: Calculate moles CO2 and H2O

For 4 moles NaHCO3 we need 1 mol O2 to produce 4 moles CO2 and 2 moles H2O and 2 moles Na2O2

For 0.0119 moles NaHCO3 we'll have 4*0.0119= 0.0476 moles CO2

and 2* 0.0119 = 0.0238 moles H2O

Step 5: Calculate mass CO2 and H2O

Mass = moles * molar mass

Mass CO2 = 0.0476 moles * 44.01 g/mol

Mass CO2 = 2.09 grams

Mass H2O = 0.0238 moles ¨18.02 g/mol

Mass H2O = 0.429 grams

The theoretical mass of CO2 = 2.09 grams

The theoretical mass of H2O = 0.429 grams

The theoretical mass of CO2+ H2O = 2.519 grams

6 0

4 years ago

PLEASE HELP !! BRAINLIEST AND 15 POINTS

WARRIOR [948]

Answer:

is your answer there u go

Explanation:

7 0

3 years ago

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

Plant A gets no fertilizer, plnt B-5mg/day, and plnt C-10mg/day. Which is the control group?

prisoha [69]

Answer:

Plant A

Explanation:

This is because Plant A is what you will compare to the other plants. Because it has no fertilizer is is the independent variable.

8 0

3 years ago

What did Ernest Rutherford’s gold foil experiment demonstrate about atoms? Their positive charge is located in a small region th

sergey [27]

Their positive charge is located in a small region that is called the nucleus.

Explanation:

Ernest Rutherford in his gold foil experiment was able to demonstrate that the nucleus is made up of positive charges which occupies a small and tiny nucleus.

Rutherford bombarded a thin gold foil with alpha particles from a radioactive source.

He observed that all the particles passed through but a small portion was deflected back.
This led to his proposition of the nuclear model the atom.

learn more:

Ernest Rutherford brainly.com/question/1859083

#learnwithBrainly

5 0

3 years ago