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Tcecarenko [31]

2 years ago

5

Calculate the density of the aluminum cylinder with a diameter 0f 1.3 cm weighing 18 grams. Height of the cylinder is 5.2 Cm. Fi

nd percent error (Actual density of Aluminum is 2.7 g/cm3). (Formula for volume of cylinder is πr2h)

1 answer:

Masja [62]2 years ago

6 0

Answer:

Percent error = 3.7%

Explanation:

Given data:

Density of Al cylinder = ?

Weight of cylinder = 18 g

Diameter = 1.3 cm

Height = 5.2 cm

Actual density of Al = 2.7 g/cm³

Percent error = ?

Solution:

First of all we will calculate the volume of cylinder through given formula.

V = πr²h

r = diameter /2

V = 22/7 × (0.65 cm)²× 5.2 cm

V = 22/7 × 0.4225cm²× 5.2 cm

V = 6.89 cm³

Now we will calculate the density.

d = m/v

d = 18 g/ 6.89 cm³

d = 2.6 g/cm³

Percent error:

Percent error = measured value - actual value /actual value × 100

Percent error = 2.6g/cm³ - 2.7g/cm³ /2.7g/cm³ × 100

Percent error = 3.7%

Negative sign shows that measured or experimental value is less than actual value.

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2 years ago

What volume of gold would be equal in mass to a piece of copper with a volume of 141 ml? the density of gold is 19.3 g/ml; the d

sergejj [24]

We are tasked to solve for the volume of Gold given that the mass of Gold is equivalent to that of a copper.

Details:

Density of Copper= 8.96 g/ml
Volume of Copper=141 ml
mass of Gold = mass of Copper
Density of Gold=19.3 g/ml

In order to solve for the mass of copper, we need to use the density formula
Density= mass/volume

Since we are to solve for the mass of copper
mass of copper= Density of Copper * Volume of Copper
mass of copper= 8.96 g/ml* 141 ml= 1263.36 g
Thus,
mass of gold=mass of copper=1263.36 g
Hence,
Using still the density formula to solve for the volume of gold,
Volume of gold=mass of gold/ Density of gold
Volume of gold=1263.36 g/ 19.3 g/ml = 65.46 mL

Therefore, the volume of the gold must be 65.46 mL in order to have the same mass of a copper

8 0

2 years ago

PLSSSSS HEEEELP!!!!

Naily [24]

Following is the balanced <span>radioactive decay series:
</span><span>
Particle/radiations generated during the reaction are labeled in bold at end of reaction.

Care must be taken that, atomic number and atomic mass number should be balanced in each of these reactions.

1) 92 238U </span>→ <span> 90 234Th + 2 4He(</span>α particle<span>)

A = </span>90 234Th because alpha particle is emitted along with it. So atomic number of daughter element has to be 92 - 2 = 90. This corresponds to Th. <span>

2) 90 234Th </span>→<span> 91 234Pa + -1 0e (electron)

B = -1 0e i.e electron because after radioactive disintegration atomic number of daughter element (Pa) is +1 as compared to parent element (Th)

3) 91 234Pa </span>→<span> 92 234U + –1 0e (electron)

</span>C = 92 234U because electron is emitted along with it. So atomic number of daughter element has to be 91 - (-1) = 92. This corresponds to U. <span>

4) 92 234U </span>→ 90 230Th + 2 4He (α particle<span>)

</span><span>In this case, 92 234U undergoes nuclear disintegration to generate 90 230Th and alpha particle

5) 90 230Th </span>→<span> 88 226Ra + 2 4He </span>(α particle)

D = 88 226Ra because alpha particle is emitted along with it. So atomic number of daughter element has to be 90 - 2 = 88. This corresponds to Ra.

<span>6) 88 266Ra </span>→ 86 222Rn + 2 4He (α particle)

E = alpha particle because during nuclear disintegration, 88 266Ra is converted into 86 222Rn. Hence, for mass balance we have 88 - 86 = 2. It corresponds to alpha particles.
<span>
7) 86 222Rn </span>→<span> 84 218Po + 2 4He </span>(α particle)

Again, F = alpha particle because during nuclear disintegration, 86 222Rn is converted into 84 218Rn. Hence, for mass balance we have 86 - 84 = 2. It corresponds to alpha particles.
<span>
8) 84 218Po </span>→<span> 82 214Pb + 2 4He </span>(α particle)

G = 82 214Pb because alpha particle is emitted along with it. So atomic number of daughter element has to be 84 - 2 = 82. This corresponds to Pb.
<span>
9) 82 214Pb </span>→<span> 83 214Bi + -1 0e (electron)

H = </span>-1 0e because after radioactive disintegration atomic number of daughter element (Bi) is +1 as compared to parent element (Pb)<span>

10) 83 214Bi </span>→<span> 84 214Po + –1 0e (electron)

I = </span>84 214Po because electron is emitted along with it. So atomic number of daughter element has to be 83 - (-1) = 84. This corresponds to Po.<span>

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8 0

2 years ago

An electrochemical cell at 25°C is composed of pure copper and pure lead solutions immersed in their respective ionis. For a 0.6

ExtremeBDS [4]

Answer :

(a) The concentration of $Pb^{2+}$ is, 0.0337 M

(b) The concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

Solution :

<u>(a) As per question, lead is oxidized and copper is reduced.</u>

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Pb\rightarrow Pb^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

The balanced cell reaction will be,

$Pb(s)+Cu^{2+}(aq)\rightarrow Pb^{2+}(aq)+Cu(s)$

Here lead (Pb) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Cu^{2+}/Cu]}-E^o_{[Pb^{2+}/Pb]}$

$E^o=0.34V-(-0.13V)=0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Pb^{2+}]}{[Cu^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = 0.507 V

Now put all the given values in the above equation, we get:

$0.507=0.47-\frac{0.0592}{2}\log \frac{[Pb^{2+}]}{(0.6)}$

$[Pb^{2+}]=0.0337M$

Therefore, the concentration of $Pb^{2+}$ is, 0.0337 M

<u>(b) As per question, lead is reduced and copper is oxidized.</u>

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction: $Pb^{2+}+2e^-\rightarrow Pb$

The balanced cell reaction will be,

$Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)$

Here Copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Pb^{2+}/Pb]}-E^o_{[Cu^{2+}/Cu]}$

$E^o=-0.13V-(0.34V)=-0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}]}{[Pb^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = 0.507 V

Now put all the given values in the above equation, we get:

$0.507=-0.47-\frac{0.0592}{2}\log \frac{(0.6)}{[Pb^{2+}]}$

$[Pb^{2+}]=6.093\times 10^{32}M$

Therefore, the concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

6 0

2 years ago

List all the physical properties of the iceberg.

Semmy [17]

Answer: My answer is in the explanation

Explanation:

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3 0

2 years ago