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777dan777 [17]
3 years ago
14

Balance the following equation in acidic conditions. phases are optional.

Chemistry
1 answer:
Elanso [62]3 years ago
4 0
This is an oxidation reduction reaction
oxidation happens when the species gives out electrons.Oxidation state of the element Cu increases from 0 to +2
Cu  ---> Cu²⁺ +2e --1)
Reduction happens when the species gains electrons.
Oxidation state of N reduces from +5 to +2
4H⁺ + NO₃⁻ + 3e ---> NO + 2H₂O --2)
To balance the reactions, number of electrons need to be balanced. 
Multiply 1st reaction by 3
Multiply 2nd reaction by 2
3Cu  ---> 3Cu²⁺ + 6e
8H⁺ + 2NO₃⁻ + 6e ---> 2NO + 4H₂O
add the 2 equations 
3Cu + 8H⁺ + 2NO₃⁻ --> 3Cu²⁺ + 2NO + 4H₂O
add 6NO₃⁻ ions to each side 
3Cu + 8HNO₃ --> 3Cu(NO₃)₂ + 2NO + 4H₂O
the balanced redox reaction equation is as follows;

3Cu(s) + 8HNO₃(aq) --> 3Cu(NO₃)₂ (aq)+ 2NO(g) + 4H₂O(l)
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How is the rate of appearance of h2o related to the rate of disappearance of o2?
erastovalidia [21]
If we consider a combustion reaction of Methane:

The balanced equation is:

CH4 + 2O2 ---> 2H2O + CO2

The rate of appearance of H2O is rH2O, rate of disappearance of O2 is -rO2

(rH2O)^2 = (-rO2)^2
rH2O = -rO2
6 0
3 years ago
Tarnished silver can be cleaned with toothpaste. Give reason
vfiekz [6]
Toothpaste is an abrasive that contains many things, but mainly phosphates.
6 0
3 years ago
What is the mass of 3.77mol of K3N?
AleksAgata [21]
<h3>Answer:</h3>

495 g K₃N

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.77 mol K₃N

<u>Step 2: Identify Conversions</u>

Molar Mass of K - 39.10 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of K₃N - 3(39.10) + 14.01 = 131.31 g/mol

<u>Step 3: Convert</u>

  1. Set up:                       \displaystyle 3.77 \ mol \ K_3N(\frac{131.31 \ g \ K_3N}{1 \ mol \ K_3N})
  2. Multiply/Divide:         \displaystyle 495.039 \ g \ K_3N

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

495.039 g K₃N ≈ 495 g K₃N

5 0
3 years ago
1. a.) Calculate the wavelength of light which has a frequency of 5.25 x 10 14 Hz.
Scilla [17]
<h3>Answer:</h3>

5.71 × 10² nm

<h3>Explanation:</h3>

The product of wavelength and frequency of a wave gives the speed of the wave.

Therefore;

Velocity of wave = Wavelength × Frequency

c = f ×λ

In our case;

Frequency = 5.25 × 10^14 Hz

Speed of light = 2.998 × 10^8m/s

But;

λ = c ÷ f

  = 2.998 × 10^8m/s ÷  5.25 × 10^14 Hz

  = 5.71 × 10^-7 m

But; 1 M = 10^9 nm

Therefore;

wavelength = 5.71 × 10^-7 × 10^9

                  = 5.71 × 10² nm

The wavelength of light wave 5.71 × 10² nm

3 0
3 years ago
Write the balanced electrochemical reaction for when zinc reacts with a copper solution. Label what is being oxidized and what i
Fed [463]

The balanced equation would be Zn + Cu^{2+} ---- > Cu + Zn^{2+}

<h3>Electrochemical equations</h3>

Zn reacts with Cu solution according to the following equation:

Zn + Cu^{2+} ---- > Cu + Zn^{2+}

In the reaction, Cu^{2+} is reduced according to the following: Cu^{2+} + 2 e^- -- > Cu

While Zn is oxidized according to the following: Zn - 2e^- --- > Zn^{2+}

Thus, giving the overall equation of; Zn + Cu^{2+} ---- > Cu + Zn^{2+}

More oxidation-reduction equations can be found here: brainly.com/question/13699873

#SPJ1

5 0
2 years ago
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