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3 years ago

What role does the wire play in voltaic cell?

Chemistry

1 answer:

Triss [41]3 years ago

7 0

<h3>Answer;</h3>

a) It allows electrons to flow from the anode to the cathode.

<h3>Explanation;</h3>

Voltaic cell is an electrochemical cell in which a spontaneous chemical reaction produces the flow of electrons.
Electrons are produced by the oxidation reaction occurring at the anode. Electrons flow through the conducting wire from the anode to the cathode. At the cathode these electrons are used to reduce copper(II) ions to copper atoms.
A conducting wire or a wire play connects the two electrodes allowing electrons to flow from the anode to the cathode.

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A large glycogen molecule is converted into may smaller glucose molecules.what type of chemical reaction is this

likoan [24]

Answer is: catabolism.

Missing question:
Synthesis.
Catabolism.
Rearrangement.
Anabolism.
Catabolism (the set of metabolic pathways) breaks down large molecules (in this example glycogen, a polysaccharide) into smaller units (in this example glucose, a monosaccharide).
Glycogen is the main storage form of glucose in the body.

5 0

3 years ago

Looking at the periodic table how can you tell how many valence electrons an atom will have?

jeka57 [31]

The Group number of a non-transition metal can be used to find the number of valence electrons in an atom of that element. The ones place of the group number is the number of valence electrons in an atom of these elements

7 0

4 years ago

Hydroxylapatite, Ca10(PO4)6(OH)2, has a solubility constant of Ksp = 2.34 x 10^-59. Solid hydroxylapatite is dissolved in water

nadezda [96]

Answer:

The concentration of [Ca²⁺] is 8.47 x 10⁻³ M

Explanation:

We consider the solubility of hydroxyapatite,

Ca₁₀(PO₄)₆(OH)₂ ⇔ 10Ca²⁺ + 6PO₄³⁻ + 2 OH⁻

Assumed that there is a mol of hydoxyapatite disolved in water, yielding 10a mol Ca²⁺ of and 6a mol of PO₄³⁻

We also have Ksp equation,

Ksp = [Ca²⁺]¹⁰ x [PO₄³⁻]⁶ x [OH⁻]² = 2.34 x 10⁻⁵⁹

⇔ 10a¹⁰ x 6a⁶ x (5.30 x 10⁻⁶)² = 2.24 x 10⁻⁵⁹

⇔ 60a¹⁶ = 2.24 x 10⁻⁵⁹ / 5.30 x 10⁻¹²

⇔ a¹⁶ = 0.007 x 10⁻⁴⁷ = 7 x 10⁻⁵⁰

⇔ a = $\sqrt[16]{7 . 10^{-50} }$ = 8.47 x 10⁻⁴

Hence,

[Ca²⁺] = 10a = 8.47 x 10⁻³ M

3 0

4 years ago

Which of the following elements is most likely to form an ion that will then form an ionic bond with an ion of a Group 1A elemen

solniwko [45]

Group 1A (the alkali metals) almost always form cations (positive ions). They'd need anions (negative ions) to ionic bond with. Beryllium (Be) is group 1A already and forms Be+ cation. Bromine is a halogen, and forms Br-, an anion. Platinum is a metal, and usually won't ionic bond with anything. Francium is rare and highly radioactive, plus it so happens to be group 1A as well. Only bromine can form the anion that the group 1A cations need.

3 0

3 years ago

Fish breathe the dissolved air in water through their gills. Assuming the partial pressures of oxygen and nitrogen in air to be

amid [387]

Answer:

X(O₂) = 0.323

X(N₂) = 0.677

Explanation:

We have the partial pressures of oxygen (O₂) and nitrogen (N₂):

P(O₂) = 0.20 atm

P(N₂) = 0.80 atm

In order to solve the problem, you need the solubilities of each gas in water at 298 K. We can consider 1.3 x 10⁻³ mol/(L atm) for oxygen (O₂) and 6.8 x 10⁻⁴mol/(L atm) for nitrogen (N₂) from the bibliography.

s(O₂) = 1.3 x 10⁻³ mol/(L atm)

s(N₂) = 6.8 x 10⁻⁴mol/(L atm)

So, we calculate the concentration (C) of each gas as the product of its partial pressure (P) and the solubility (s):

C(O₂) = P(O₂) x s(O₂) = 0.20 atm x 1.3 x 10⁻³ mol/(L atm) = 2.6 x 10⁻⁴mol/L

C(N₂) = P(N₂) x s(N₂) = 0.80 atm x 6.8 x 10⁻⁴mol/(L atm) = 5.44 x 10⁻⁴ mol/L

In 1 liter of water, we have the following number of moles (n):

n(O₂) = 2.6 x 10⁻⁴ mol

n(N₂) = 5.44 x 10⁻⁴ mol

Thus, the total number of moles (nt) is calculated as the sum of the number of moles of the gases in the mixture:

nt = n(O₂) + n(N₂) = 2.6 x 10⁻⁴ mol + 5.44 x 10⁻⁴ mol = 8.04 x 10⁻⁴ mol

Finally, the mole fraction of each gas is calculated as the ratio between the number of moles of each gas and the total number of moles:

X(O₂) = n(O₂)/nt = 2.6 x 10⁻⁴ mol/(8.04 x 10⁻⁴ mol) = 0.323

X(N₂) = n(N₂)/nt = 5.44 x 10⁻⁴ mol/(8.04 x 10⁻⁴ mol) = 0.677

5 0

3 years ago