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klemol [59]
3 years ago
11

At what temperature will aluminum have a resistivity that is three times the resistivity of tungsten at room temperature? (Assum

e room temperature is 20° C.)
Chemistry
1 answer:
umka2103 [35]3 years ago
3 0

Explanation:

At room temperature, resistivity of tungsten is 5.6 \times 10^{-8} and the resistivity of aluminium is 2.8 \times 10^{-8}.

Temperature coefficient of aluminium (\alpha) = 3.9 \times 10^{-3}

R_{a} = 3R_{c} = 5.1 \times 10^{-8} \ohm m

and,      R_{a} = R_{o}(1 + \alpha \times \Delta T)

                       = 2.82 \times 10^{-8} (1 + 3.9 \times 10^{-8} \times \Delta T)

                  \Delta T = 207.3^{o}C

                             T = (207.3 + 20)^{o}C

                                 = 227.3^{o}C

So, at 227.3^{o}C aluminium has resistivity which is three times the resistivity of tungsten at room temperature.

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when the volume of a gas is changed from 3.75 L to 6.52 L the temperature will change from 100k to _K
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The temperature will change from 100K to 173.87 K

calculation

by use of    law  that is V1/T1=V2/T2

V1=3.75 L

T1=100k

V2=6.53 L

T2=?

make T2 the subject of the formula

T2=(V2 xT1)V1

=6.52 x100/3.75=173.87K


4 0
3 years ago
Read 2 more answers
What might happen when a gas is exposed to a small flame?
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The fire may grow bigger, depending on the gas.

Explanation:

If you expose more air to a small flame then it could grow larger because air keeps fire alive.

3 0
3 years ago
Which description best matches Democritus idea of the atom?
puteri [66]
Democritus said that the atom was just a sphere.
4 0
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Do CO2 and h2o have the same geometry
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7 0
3 years ago
What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom
Anastaziya [24]

Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

As, 207.2 g of lead contains 6.022\times 10^{23} atoms

So, 5.5 g of lead contains \frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22} atoms

and,

As, 118.71 g of lead contains 6.022\times 10^{23} atoms

So, 94.5 g of lead contains \frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23} atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100

\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%

and,

\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100

\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

6 0
3 years ago
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