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3 years ago

Why are atoms always bonded

1 answer:

6 0

Atoms are always bonded so that nearby atoms have to form into a compound and can be classified into different forms of compounds or molecules. It is also integral in our everyday life.

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How do human beings prepare their food

Rudiy27

Well they buy food then they cook it then they eat it

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3 years ago

Read 2 more answers

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⁻

2H⁺ +2e⁻ ⇌ H₂

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

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

6 0

3 years ago

Chlorine has two natural isotopes: Cl-37 and Cl-35. Hydrogen reacts with chlorine to form the compound HCl. Would a given amount

jolli1 [7]

Answer: Yes, a given amount of Hydrogen would react with different masses of the two isotopes of chlorine, and no, this does not conflict with the Law of Definite proportions

Explanation:

About 76% percent of Cl is found in the Cl-35 isotope, and about 24% in the Cl.37 isotope. that means that about 24% of Cl nuclei have 2 more neutrons than the average Cl nucleus.

So, if $H_2$ reacts with $Cl_2$ , 76% of the Hydrogen that reacted will react with Cl-35, and the rest will react with Cl-37. Why does this not conflict with the law of definite proportions? Because each Hydrogen atom ends up paired to a single Chlorine atom! Moreover, the proportion of Cl-35 to Cl-37 remains constant in all samples of Chlorine that are naturally found, thus we will always find the same proportion of Chlorine to Hydrogen in any HCl sample we come across. Thus the weight of a mol of Cl will always be $35*0.76+37*0.24=35.48$ or 35.45 if we had done this calculation with more significant digits.