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qwelly [4]
3 years ago
10

Sometimes the cell reaction of nickel-cadmium batteries is written with cd metal as the anode and solid nio2 as the cathode. ass

uming that the products of the electrode reactions are solid hydroxides of cd(ii) and ni(ii), respectively, write balanced chemical equations for the cathode and anode half-reactions and the overall cell reaction. include the phases of all species in the chemical equation.
Chemistry
1 answer:
Reptile [31]3 years ago
6 0
The cell reaction<span> of nickel-cadmium batteries are as follows:

At Anode   :     Cd(s)   +    2OH-(aq.)    </span>→    Cd(OH)2(s)    +     2e-
At cathode:   NiO2(s)   +   2H2O(l)   +   2e-   →   Ni(OH)2
_______________________________________________________________
Overall reaction:  Cd(s) + NiO2(s)   +   2H2O  →   Cd(OH)2(s)   +   Ni(OH)2 (s)
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It is difficult to measure the correct atomic radius. Why?
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Answer:

its kinda you just need to simplify

Explanation:

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The approximately 20 basic building blocks of protein are called
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Amino acids join end to end to form proteins since they are monomers.
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If I leave 750 mL of a 0.40 M sodium chloride solution uncovered on a windowsill
mel-nik [20]

Answer:

B. 0.5 molar

Explanation:

Given data:

Initial concentration = 0.40 M

Initial volume = 750 mL

Final volume =750  -  150 mL = 600 mL

Final concentration = ?

Solution:

Molarity is the number of moles of solutes in litter of solvent. In given problem it is stated that when the solution is uncovered solvent evaporate it means molarity is changed. we can calculate the new molarity with the following formula.

C₁V₁ = C₂V₂

C₁ = initial concentration

V₁ = initial volume

C₂ = final concentration

V₂ = final volume

Now we will put the values in formula.

C₁V₁ = C₂V₂

0.40 M × 750 mL = C₂ × 600 mL

300 M.mL / 600 mL = C₂

0.5 M = C₂

5 0
3 years ago
A 1.00 liter solution contains 0.43 M hydrofluoric acid and 0.56 M potassium fluoride. If 0.280 moles of potassium hydroxide are
kolbaska11 [484]

Answer:

Answers are in the explanation

Explanation:

Equlibrium of HF in H₂O is:

HF + H₂O ⇄ F⁻ + H₃O⁺

Now, the KOH reacts with HF, thus:

KOH + HF →  F⁻ + H₂O

<em>That means after reaction, concentration of HF decrease increasing F⁻ concentration.</em>

Now, seeing the equilibrium, as moles of HF decrease and F⁻ moles increase, the equilibrium will shift to the left decreasing H₃O⁺ concentration.

For the statements:

A. The number of moles of HF will increase. <em>FALSE</em>. HF react with KOH, thus, moles of HF decrease

B. The number of moles of F- will decrease. <em>FALSE</em>. The reaction produce  F⁻ increasing its moles.

C. The equilibrium concentration of H₃O⁺ will increase. <em>FALSE. </em>The equilibrium shift to the left decreasing concentration of H₃O⁺

D. The pH will decrease. <em>FALSE</em>. As the H₃O⁺ concentration decrease, pH will increase

E. The ratio of [HF] / [F-] will remain the same. <em>FALSE</em>. Because moles of HF are decreasing whereas F- moles are increasing changing, thus, ratio.

6 0
3 years ago
1)
MAXImum [283]

Answer:

Sulfuric acid + Zinc = reactant

Zinc sulfate + hydrogen = product

Lead oxide + carbon = reactant

lead + carbondioxide = product

Explanation:

Chemical equation:

Sulfuric acid + Zinc → Zinc sulfate + hydrogen

H₂SO₄ + Zn → ZnSO₄ + H₂

In this reaction sulfuric acid and zinc react to produce the zinc sulfate and hydrogen. That's why sulfuric acid and zinc are reactant. While zinc sulfate and hydrogen are products.

2nd reaction:

Lead oxide + carbon   →  lead + carbondioxide

2PbO + C →  2Pb + CO₂

In this reaction Lead oxide and carbon  react to produce the lead and carbondioxide. That's why Lead oxide and carbon are reactant. While lead and carbondioxide are products.

3 0
3 years ago
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