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3 years ago

16. A baseball has a mass of 1500 grams and displaces 500 ml. of water. What is it's density?

Chemistry

1 answer:

Debora [2.8K]3 years ago

7 0

Answer:

3.00 g/mL

Explanation:

Step 1: Given data

Mass of the baseball (m): 1500 g

Volume of the baseball = Volume of water displaced (V): 500 mL

Density of the baseball (ρ): ?

Step 2: Calculate the density of the baseball

The density is equal to the mass divided by the volume.

ρ = m/V

ρ = 1500 g/500 mL

ρ = 3.00 g/mL

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Many substances can go through physical and chemical changes. Which of the following is an example of a physical change? Questio

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Mixing baking soda with vinegar to form CO2

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3 years ago

The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:

german

Answer:

For 1: The largest positive cell potential is of cell having 1st and 4th half reactions.

For 2: The standard electrode potential of the cell is 1.539 V

For 3: The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

Explanation:

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

We are given:

$Ag^++(aq.)+e^-\rightarrow Ag(s);E^o_{Ag^+/Ag}=0.799V\\\\Cu^{2+}+(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.337V\\\\Ni^{2+}(aq.)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.28V\\\\Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o_{Cr^{3+}/Cr}=-0.74V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Cell having 1st and 2nd half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Copper will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-0.337=0.462V$

Cell having 1st and 3rd half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.28)=1.079V$

Cell having 1st and 4th half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.74)=1.539V$

Cell having 2nd and 3rd half reactions:

Copper has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.337-(-0.28)=0.617V$

Cell having 3rd and 4th half reactions:

Nickel has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=-0.28-(-0.74)=0.46V$

Hence,

For 1: The largest positive cell potential is of cell having 1st and 4th half reactions.

For 2: The standard electrode potential of the cell is 1.539 V

For 3: The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

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3 years ago

Starting with lead (ii) carbonate describe how a solid sample of lead (ii) sulphate can be prepared

tatuchka [14]

The preparation of lead (ii) sulphate from lead (ii) carbonate occurs in two steps:

insoluble lead carbonate is converted to soluble lead (ii) nitrate
soluble lead (ii) nitrate is reacted with sulphuric acid to produce lead (ii) sulphate.

<h3>How can a solid sample of lead (ii) sulphate be prepared from lead (ii) carbonate?</h3>

Lead (ii) carbonate and lead (ii) sulphate are both insoluble salts of lead.

In order to prepare lead (ii) sulphate, a two step process is performed.

In the first step, Lead (ii) carbonate is reacted with dilute trioxonitrate (v) acid to produce lead (ii) nitrate.

PbCO₃ + 2HNO₃ → Pb(NO₃)₂ + CO₂ + H₂O

In the second step, dilute sulfuric acid is reacted with the lead (ii) nitrate to produce insoluble lead (ii) sulphate which is filtered and dried.

Pb(NO₃)₂ + H₂SO₄ → PbSO₄ + 2HNO₃

In conclusion, lead (ii) sulphate is prepared in two steps.

Learn more about lead (ii) sulphate at: brainly.com/question/188055

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