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SIZIF [17.4K]
3 years ago
15

How does energy move

Chemistry
1 answer:
Gekata [30.6K]3 years ago
6 0

energy is transferred through vibrations of electric and magnetic fields  

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Knowing that the reaction of potassium in water is extremely exothermic, what would be the problem with adding a large piece of
Alenkasestr [34]

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e  a s p o r t s

Explanation:

Its in the game

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3 years ago
How many moles of CO are produced when 1.8 moles C reacts?
olasank [31]

Answer:

1.44mole of CO

Explanation:

The reaction equation is given as:

         5C  + 2SO₂ →  CS₂ + 4CO

We check to see if the expression is balanced and it is so;

 Now;

 Given;

           1.8mole of C reacted; how many moles of CO are produced;

  From the balanced reaction equation:

             5 mole of C is expected to produce 4 mole of CO

             1.8 mole of C will then produce \frac{4  x 1.8}{5} = 1.44mole of CO

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3 years ago
An acidified solution was electrolyzed using copper electrodes. A constant current of 1.18 A caused the anode to lose 0.584 g af
Alexxx [7]

Answer:

\boxed{\text{(a) 209 mL; (b) } 6.09 \times 10^{23}}

Explanation:

(a) Gas produced at cathode.

(i). Identity

The only species known to be present are Cu, H⁺, and H₂O.

Only the H⁺ and H₂O can be reduced.

The corresponding reduction half reactions are:

(1) 2H₂O + 2e⁻ ⇌ H₂ + 2OH⁻;     E° = -0.8277 V

(2) 2H⁺ +2e⁻ ⇌ H₂;                     E° =  0.0000 V

Two important points to remember when using a table of standard reduction potentials:

  • The higher up a species is on the right-hand side, the more readily it will lose electrons (be oxidized).
  • The lower down a species is on the left-hand side, the more readily it will accept electrons (be reduced}.

H⁺ is below H₂O, so H⁺ is reduced to H₂.

The cathode reaction is 2H⁺ +2e⁻ ⇌ H₂, and the gas produced at the cathode is hydrogen.

(ii) Volume

a. Anode reaction

The only species that can be oxidized are Cu and H₂O.

The corresponding half reactions  are:

(3) Cu²⁺ + 2e⁻ ⇌ Cu;                E° =  0.3419 V

(4) O₂ + 4H⁺ + 4e⁻ ⇌ 2H₂O     E° =   1.229   V

Cu is above H₂O, so Cu is more easily oxidized.

The anode reaction is Cu ⇌ Cu²⁺ + 2e⁻.

b. Overall reaction:

Cu           ⇌ Cu²⁺ + 2e⁻

<u>2H⁺ +2e⁻ ⇌ H₂            </u>        

Cu + 2H⁺ ⇌ Cu²⁺ + H₂

c. Moles of Cu lost

n_{\text{Cu}} = \text{0.584 g } \times \dfrac{\text{1 mol}}{\text{63.55 g}} = 9.190 \times 10^{-3}\text{ mol Cu}

d. Moles of H₂ formed

n_{\text{H}_{2}}} = 9.190 \times 10^{-3}\text{ mol Cu} \times \dfrac{\text{1 mol H}_{2}}{\text{1 mol Cu}} =9.190 \times 10^{-3}\text{ mol H}_{2}

e. Volume of H₂ formed

Volume of 1 mol at STP (0 °C and  1 bar) = 22.71 mL

V = 9.190 \times 10^{-3}\text{ mol}\times \dfrac{\text{22.71 L}}{\text{1 mol}}  = \text{0.209 L} = \boxed{\textbf{209 mL}}

(b) Avogadro's number

(i) Moles of electrons transferred

\text{Moles of electrons} = 9.190 \times 10^{-3}\text{ mol Cu}\times \dfrac{\text{2 mol electrons}}{\text{1 mol Cu}}\\\\\\= \text{0.018 38 mol electrons}

(ii) Number of coulombs

Q  = It  

Q = \text{1.18 C/s} \times 1.52 \times 10^{3} \text{ s} = 1794 C

(iii). Number of electrons

n = \text{ 1794 C} \times \dfrac{\text{1 electron}}{1.6022 \times 10^{-19} \text{ C}} = 1.119 \times 10^{22} \text{ electrons}

(iv) Avogadro's number

N_{\text{A}} = \dfrac{1.119 \times 10^{22} \text{ electrons}}{\text{0.018 38 mol}} = \boxed{6.09 \times 10^{23} \textbf{ electrons/mol}}

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Explanation:

What would a biologist see if she examined a small piece of a leaf from a plant and a small piece of a root from the same plant through a microscope? Both the leaf and the root would be made up of cells. The leaf would be made up of cells, but the root would not be made up of cells.

5 0
3 years ago
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