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maw [93]
3 years ago
15

Balance the following skeleton reaction in acidic solution (if the coefficient is 1, put 1; if the coefficient is 0, put 0) and

identify the oxidizing and reducing agents. For the oxidizing and reducing agent, enter numbers and +/- directly for superscripts and subscripts. (eg. enter HCO3- for HCO3-, Cu2+ for Cu2+) CrO42- (aq) + N2O (g) Cr3+ (aq) + NO (g) 2 CrO42–(aq) + 3 N2O(g) + 10 H+(aq) + 0 H2O(l) 2 Cr3+(aq) + 6 NO(g) + 0 H+(aq) + 5 H2O(l) (acidic) Oxidizing agent: CrO42- Reducing agent: N2O
Chemistry
1 answer:
Mashcka [7]3 years ago
7 0

Answer:

2 CrO42- + 3N2O + 10 H+ -----> 2Cr3+ + 6NO + 5H2O

Explanation:

2 CrO42- + 3N2O + 10 H+ -----> 2Cr3+ + 6NO + 5H2O

Oxidizing agent: -----------------------------> CrO42-

Reducing agent: ----------------------------> N2O

explanation:

in CrO4-2 oxdiation state of Cr = +6

in Cr+3 oxidation state = +3

+6 oxidation state changed from +3 it is reduction .

so CrO4-2 is oxidizing agent

atomatically

N2O should be reducing agent

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B.liquids thanks have a great day
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If you only have 36.5 grams of Potassium Carbonate (K2CO3 molar mass = 138.2 g/mole). How many liters of a 0.52 M solution can y
kirza4 [7]

Answer:

0.508 L of solution.

Explanation:

Always a safe bet to convert to moles:

n= {m \over MM}\\

Where n is moles, m is mass, and MM is molar mass.

Now remember the equation for concentration (molarity):

C={n \over V}

Where C is the concentration, n is moles, and V is volume.

To make this easy, combine the two equations (note n appears in both, so you can do a substitution) and solve for V as the question asks:

C={m \over MM \times V}\\0.52={36.5 \over 138.2 \times V}\\V={36.5 \over 138.2 \times 0.52}\\V=0.508

Therefore we can make 0.508 L of solution.

8 0
3 years ago
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podryga [215]
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8 0
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Read 2 more answers
A 27.86 ml sample of 0.1744 m hno3 is tirades with 29.4ml of a job solution. What is the molarity of the koh
Arte-miy333 [17]

Answer:

0.165 mol·L⁻¹

Explanation:

1. Write the <em>chemical equation</em> for the reaction.

HNO₃ + KOH ⟶ KNO₃ + H₂O

===============

2. Calculate the <em>moles of HNO₃</em>

c = n/V                                               Multiply each side by V and transpose

n = Vc

V = 0.027 86 L

c = 0.1744 mol·L⁻¹                             Calculate the moles of HNO₃

Moles of HNO₃ = 0.027 86 × 0.1744

Moles of HNO₃ = 4.859 × 10⁻³ mol HNO₃

===============

3. Calculate the <em>moles of KOH </em>

1 mol KOH ≡ 1 mol HNO₃                 Calculate the moles of KOH

Moles of KOH = 4.859 × 10⁻³× 1/1

Moles of KOH = 4.859 × 10⁻³ mol KOH

===============

4. Calculate the <em>molar concentration</em> of the KOH

V = 29.4 mL = 0.0294 L                   Calculate the concentration

c = 4.859 × 10⁻³/0.0294

c = 0.165 mol·L⁻¹

4 0
4 years ago
For the reaction 2NH3(g) + 2O2(g)N2O(g) + 3H2O(l) H° = -683.1 kJ and S° = -365.6 J/K The standard free energy change for the rea
BlackZzzverrR [31]

Answer:

\Delta G^{0} = -457.9 kJ and reaction is product favored.

Explanation:

The given reaction is associated with 2 moles of NH_{3}

Standard free energy change of the reaction (\Delta G^{0}) is given as:

           \Delta G^{0}=\Delta H^{0}-T\Delta S^{0}   , where T represents temperature in kelvin scale

So, \Delta G^{0}=(-683.1\times 10^{3})J-(273K\times -365.6J/K)=-583291.2J

So, for the reaction of 1.57 moles of NH_{3}, \Delta G^{0}=(\frac{1.57}{2})\times -583291.2J=-457883.592J=-457.9kJ

As, \Delta G^{0} is negative therefore reaction is product favored under standard condition.

6 0
3 years ago
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