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Kipish [7]
3 years ago
6

An electric wire produces a magnetic field. True or false? (GIVE PROOF)

Chemistry
1 answer:
kupik [55]3 years ago
3 0

Answer:

True

Explanation:

A magnetic field is made when an electric current flows through a wire.

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If the temperature of a liquid increases the rate of evaporation will do what?
konstantin123 [22]
The rate of evaporation will increase as well.
6 0
3 years ago
20)
anyanavicka [17]

Answer:

D

Explanation:

7 0
3 years ago
Explain why vanadium (radius=134 pm) and copper (radius=128 pm) have nearly identical atomic radii, even though the atomic numbe
Viktor [21]
<span>There are few main factors affecting the atomic radii, the outermost electrons and the protons in the nucleus and also the shielding of the internal electrons. I would speculate that the difference in radii is given by the electron clouds since the electrons difference in these two elements is in the d orbital and both has at least 1 electron in the 4s (this 4s electron is the outermost electron in all the transition metals of this period). The atomic radio will be mostly dependent of these 4s electrons than in the d electrons. Besides that, you can see that increasing the atomic number will increase the number of protons in the nucleus decreasing the ratio of the atoms along a period. The Cu is an exception and will accommodate one of the 4s electrons in the p orbital.

</span><span>Regarding the density you can find the density of Cu = 8.96g/cm3 and vanadium = 6.0g/cm3. This also correlates with the idea that if these two atoms have similar volume and one has more mass (more protons;  density is the relationship between m/V), then a bigger mass for a similar volume will result in a bigger density.</span>
7 0
3 years ago
The half-life of a radioactive isotope is the amount of time it takes for a quantity of that isotope to decay to one half of its
Sedaia [141]

Radio active decay reactions follow first order rate kinetics.

a) The half life and decay constant for radio active decay reactions are related by the equation:

t_{\frac{1}{2}} =\frac{ln 2}{k}

t_{\frac{1}{2}} = \frac{0.693}{k}

Where k is the decay constant

b) Finding out the decay constant for the decay of C-14 isotope:

Decay constant (k) = \frac{0.693}{t_{\frac{1}{2}}}

k = \frac{0.693}{5230 years}

k = 1.325 * 10^{-4} yr^{-1}

c) Finding the age of the sample :

35 % of the radiocarbon is present currently.

The first order rate equation is,

[A] = [A_{0}]e^{-kt}

\frac{[A]}{[A_{0}]} = e^{-kt}

\frac{35}{100} = e^{-(1.325 *10^{-4})t}

ln(0.35) = -(1.325 *10^{-4})(t)

t = 7923 years

Therefore, age of the sample is 7923 years.

3 0
3 years ago
a 160 milligram sample of a radioactive isotope decays to 10 kilograms in 12 years. what is the half life of this element
Shkiper50 [21]

Answer:

3 years

Explanation:

Given data:

Initial amount of sample = 160 Kg

Amount left after 12 years = 10 Kg

Half life = ?

Solution:

at time zero = 160 Kg

1st half life = 160/2 = 80 kg

2nd half life = 80/2 = 40 kg

3rd half life = 40 / 2 = 20 kg

4th half life = 20 / 2 = 10 kg

Half life:

HL = elapsed time / half life

12 years / 4 = 3 years

8 0
3 years ago
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