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4 years ago

For a concentration cell, the standard cell potential is always:________.

Chemistry

1 answer:

oksano4ka [1.4K]4 years ago

4 0

Answer:

C. Zero

Explanation:

A concentration cell is defined as a type of galvanic cell which has two equivalent half-cells of the same constituent in the electrodes but different concentration in the electrolytes.

For a concentration cell, the standard cell potential is always Zero and the cell potential is usually measured with the use of the Nernst equation.

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Solve and show work. Li2S + 2 HNO3 --> 2 LiNO3 + H2S (a) Calculate the mass of lithium sulfide that will react with 250 mL of

Elenna [48]

Li2S + 2 HNO3 --> 2 LiNO3 + H2S

Li2 S + H2 N2 O2 --> Li2 N2 O5 + H2 S

Li S + H2 N2 O5 -> Li N2 O5 + H2 S

Li2 S2 + H4 N4 O10 --> Li2 N4 O10 + H4 S2

Li^2 S^2 + H^4 N^4 O^10 --> Li^2 N^4 O^10 + H^4 S^2

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3 years ago

Consider the reaction of A(g) + B(g) + C(g) => D(g) for whichthe following data were obtained:

AveGali [126]

Answer:

rate= k[A]²[B]²[C]

Explanation:

When concentration of A is increased two times ,keeping other's concentration constant , rate of reaction becomes 4 times .

So rate is proportional to [A]²

When concentration of B is increased two times , keeping other's concentration constant,rate of reaction becomes 4 times.

So rate is proportional to [B]²

When concentration of C is increased two times , keeping other's concentration constant, rate of reaction becomes 2 times.

So rate is proportional to [C]

So rate= k[A]²[B]²[C]

5 0

4 years ago

The combustion of 40.10 g of a compound which contains only C, H, Cl and O yields 58.57 g of CO2 and 14.98 g of H2O. Another sam

elena-14-01-66 [18.8K]

Answer:

The empirical formula is C4H5ClO2

Explanation:

Step 1: Data given

Mass of the sample = 40.10 grams

Mass of CO2 produced = 58.57 grams

Mass of H2O produced = 14.98 grams

Molar mass CO2 = 44.01 g/mol

Molar mass H2O = 18.02 g/mol

Atomic mass C= 12.01 g/mol

Atomic mass O = 16.0 g/mol

Atomic mass H = 1.01 g/mol

In experiment 2, mass = 75.00 grams and 22.06 grams is Cl

Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 58.57 grams / 44.01 g/mol

Moles CO2 = 1.33 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol CO2

For 1.33 moles CO2 we have 1.33 moles C

Step 4: Calculate mass C

Mass C = 1.33 grams * 12.01 g/mol

Mass C = 15.97 grams

Step 5: Calculate moles H2O

Moles H2O= 14.98 grams /18.02 g/mol

Moles H2O = 0.831 moles

Step 6: Calculate moles H

For 1mol H2O we have 2 moles H

For 0.831 moles H2O we have 2*0.831 = 1.662 moles H

Step7: Calculate mass H

Mass H = 1.662 moles * 1.01 g/mol

Mass H = 1.68 grams

Step 8: Calculate mass %

%C = (15.97 grams / 40.10) * 100 %

%C = 39.8 %

%H = (1.68 / 40.10 ) *100%

%H = 4.2 %

%Cl = (22.06 / 75.00 ) * 100%

%Cl = 29.4 %

%O = 100 % - 39.8% - 4.2 % - 29.4 %

%O = 26.6 %

Step 9: Calculate moles in compound

We assume the compound has a mass of 100 grams

Mass C = 39.8 grams

MAss H = 4.2 grams

MAss Cl = 29.4 grams

Mass O = 26.6 grams

Moles C = 39.8 grams / 12.01 g/mol

Moles C = 3.314 moles

Moles H = 4.2 moles / 1.01 g/mol

Moles H = 4.158 moles

Moles Cl =29.4 grams / 35.45 g/mol

Moles Cl = 0.829 moles

Moles O = 26.6 grams / 16.0 g/mol

Moles O = 1.663 moles

Step 10: calculate the mol ratio

We divide by the smallest amount of moles

C: 3.314 moles / 0.829 moles = 4

H: 4.158 moles / 0.829 moles = 5

Cl: 0.829 moles /0.829 moles = 1

O: 1.663 moles / 0.829 moles = 2

This means For each Cl atom we have 4 C atoms, 5 H atoms and 2 O atoms

The empirical formula is C4H5ClO2

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3 years ago

How did Rutherford’s atomic model fix the shortcomings of Thomson’s atomic model?

dedylja [7]

<u>Plum Pudding Model(Thomson's atomic model)</u>

Thomson's atomic model states that an atom has a positive sphere charge with electrons embedded inside it. He compared the atom with a plum pudding,as the electrons according to him seemed like the dry fruits embedded in the spherical pudding.

<u>Rutherford's Model</u>

However Rutherford bombarded high energy streams of α-particles on a thin gold foil of 100 nm thickness. The deflection produced by the trajectory of these high energy α-particles after interaction with the thin sheet of gold was studied by placing a screen made up of zinc sulfide around the gold foil.
The major observations made by Rutherford were that a very huge fraction of α-particles passed through the gold sheet without getting deflected. Thus he concluded that the major part of an atom must be empty.
Very few α-particles got deflected minutely or at very small angles by the gold sheet when they were bombarded against it. Also very few particles got deflected at large angles. This made him conclude that the positive charge is concentrated in a very small region and is not distributed uniformly.

From the above observations he gave the following postulates:

An atom is made up of positively charged particles. The mass of an atom was concentrated in small region which is named as the nucleus of an atom.
The nucleus is surrounded by electrons which are negatively charged particles which revolve around the nucleus in a fixed circular path called as “orbits.”
An atom is electrically neutral because electrons are negatively charged and the nucleus is positively charged. The electrons are held by the nucleus due to a strong electrostatic force.
Compared to the total size of an atom the size of the nucleus is very small.

7 0

3 years ago

What role does heat energy from the sun play influencing ocean currents

alexandr1967 [171]

The sun heats up the atmosphere at the equator. The dense water sinks under and forms deep water masses under the ocean current. Then it gets wind due to the air current and this brings currents at the surface if the ocean.

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3 years ago