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Arada [10]
2 years ago
14

What are three limitations of current cloaking technology.

Chemistry
1 answer:
kirill [66]2 years ago
4 0

Explanation:

1.undetectable to electromagnetic waves

2.hiding an object from an illumination containing diffre t wave lengths become difficult as the object sizes grow.

3. reduce the scattering by two orders.

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What is the halflife of a radioisotope if a 20-g sample becomes 10g after 16 hours
ohaa [14]

Answer:

T½ = 16hours

Explanation:

Final mass (N) = 10g

Initial mass (No) = 20g

Time (t) = 16hours

T½ = ?

T½ = In2 / λ

But λ = ?

In(N/No) = -λt

In(10/20) = -(λ * 16)

In(0.5) = -16λ

-0.693 = -16λ

λ = 0.693 / 16

λ = 0.0433

Note : λ is known as the disintegration constant

T½ = In2 / λ

T½ = 0.693 / 0.0433

T½ = 16hours

The half-life of the sample is 16hours

5 0
2 years ago
Based on the crystal-field strengths cl- < f- < h2o < nh3 < h2nc2h4nh2, which octahedral ti (iii) complex below has
kati45 [8]
<span>Based on the crystal field strength, Cl ligand would give the longest d-d transition when complexed with Ti(III). as this is the weak field ligand and would cause minimum splitting of d orbitals.</span>
8 0
2 years ago
Calculate the cell potential E at 25°C for the reaction 2 Al(s) + 3 Fe2+(aq) → 2 Al3+(aq) + 3 Fe(s) given that [Fe 2+] = 0.020 M
Elodia [21]

Answer:

1.18 V

Explanation:

The given cell is:

Al(s)/Al^{3+}(0.10M)||Fe^{2+}(0.020M)/Fe(s)

Half reactions for the given cell follows:

Oxidation half reaction: Al(s)\rightarrow Al^{3+}(0.10M)+2e^-;E^o_{Al^{3+}/Al}=-1.66V

Reduction half reaction: Fe^{2+}(0.020M)+2e^-\rightarrow Fe(s);E^o_{Fe^{2+}/Fe}=-0.45V

Multiply Oxidation half reaction by 2 and Reduction half reaction by 3

Net reaction: 2Al(s)+3Fe^{2+}(0.020M)\rightarrow 2Al^{3+}(0.10M)+3Fe(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

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

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

Putting values in above equation, we get:

E^o_{cell}=-0.45-(-1.66)=1.21V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}

where,

E_{cell} = electrode potential of the cell = ?V

E^o_{cell} = standard electrode potential of the cell = +1.21 V

n = number of electrons exchanged = 6

Putting values in above equation, we get:

E_{cell}=1.21-\frac{0.059}{6}\times \log(\frac{0.10^2}{0.020^3})\\\\E_{cell}=1.18V

5 0
2 years ago
balance the following reaction using LCM method by showing each steps Pb (N3)2 + Cr(MnO4)2  Cr2O3 + MnO2 + Pb3O4+ NO​
blsea [12.9K]

here's the answer to your question

5 0
1 year ago
why does every human measurement have uncertainty associated with it? Why must we manage the uncertainty during calculations?
dimulka [17.4K]
Because there are many numbers of the human measurement . we must manage the uncertainly doing calculations because we can know what we are calculating.
3 0
2 years ago
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