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3 years ago

A discharging lead-acid battery is best described as

Chemistry

2 answers:

dalvyx [7]3 years ago

6 0

Answer:

chemical cells that produce an electrical current

Explanation:

lead acid battery is a type of chemical cell, that discharges to produce electrical current.

lead acid battery uses lead peroxide for the conversion of the chemical energy into electrical power when it is discharging. It is also rechargeable, that is, it can convert electrical energy back to chemical energy during charging.

Thus, a discharging lead acid battery is best described as a chemical cells that produce an electrical current

yKpoI14uk [10]3 years ago

6 0

Answer:

The correct option is;

2.) Chemical cells that produce an electric current

Explanation:

A discharging lead-acid battery is an electrochemical cell that produces electrical energy through the chemical reaction taking place at the battery electrodes which have lead as the anode and lead oxide as the cathode which result in the flow of an electric current, thereby converting chemical energy of the lead, lead oxide and aqueous sulfuric acid system into electrical energy, therefore, it is a chemical cell that produces an electric current.

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3 years ago

The combustion of glucose (c6h12o6) with oxygen gas produces carbon dioxide and water. this process releases 2803 kj per mole of

V125BC [204]

Answer:- 335 kcal of heat energy is produced.

Solution:- The balanced equation for the combustion of glucose in presence of oxygen to give carbon dioxide and water is:

$C_6H_1_2O_6+6O_2\rightarrow 6CO_2+6H_2O$

From given info, 2803 kJ of heat is released bu the combustion of 1 mol of glucose. We need to calculate the energy produced when 3.00 moles of oxygen react with excess of glucose.

We could solve this using dimensional analysis as:

$3.00mol O_2(\frac{1mol glucose}{6mol O_2})(\frac{2803 kJ}{1mol glucose})$

= 1401.5 kJ

Now, let's convert kJ to kcal.

We know that, 1kcal = 4.184kJ

So, $1401.5kJ(\frac{1kcal}{4.184kJ})$

= 335 kcal

Hence, 335 kcal of heat energy is produced by the use of 3.00 moles of oxygen gas.

8 0

3 years ago

A student performs an experiment to find the percentage of water in a hydrate. He determines that the hydrate contains 22% water

Stels [109]

Answer:

A feasible error could have been the removal of the sample before all water evaporated.

Explanation:

In order to determine the percentage of water in an hydrate, an experiment that could be performed is the heating of the sample until the mass does not change. If the student heated the sample an insufficient amount of time, water will be present in the sample, thus reducing the percentage reported.

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3 years ago

Calculate ΔrG∘ at 298 K for the following reactions.CO(g)+H2O(g)→H2(g)+CO2(g)2-Predict the effect on ΔrG∘ of lowering the temper

KonstantinChe [14]

Answer:

1) ΔG°r(298 K) = - 28.619 KJ/mol

2) ΔG°r will decrease with decreasing temperature

Explanation:

CO(g) + H2O(g) → H2(g) + CO2(g)

1) ΔG°r = ∑νiΔG°f,i

⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)

from literature, T = 298 K:

∴ ΔG°CO2(g) = - 394.359 KJ/mol

∴ ΔG°CO(g) = - 137.152 KJ/mol

∴ ΔG°H2(g) = 0 KJ/mol........pure substance

∴ ΔG°H2O(g) = - 228.588 KJ/mol

⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )

⇒ ΔG°r(298 K) = - 28.619 KJ/mol

2) K = e∧(-ΔG°/RT)

∴ R = 8.314 E-3 KJ/K.mol

∴ T = 298 K

⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6

⇒ ΔG°r = - RTLnK

If T (↓) ⇒ ΔG°r (↓)

assuming T = 200 K

⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)

⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol

6 0

3 years ago