Hello!
First you need to calculate q
<span>delta U is change in internal energy </span>
<span>delta U = q + w </span>
<span>q is heat and w work done </span>
<span>here work was done by the system means energy leaving the system so w is negative </span>
<span>delta U = q + w </span>
<span>q = delta U - w = 6865 J - (-346 J) = 7211 J = 7.211 KJ </span>
<span>q = m x c x delta T </span>
<span>7211 J = 80.0 g x c x (225-25) °C </span>
<span>c = 0.451 J /g °C
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Hope this Helps! Have A Wonderful Day! :)
<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>
Answer:
~1.5 g/cm3 and it does NOT float in water.
Explanation:
If you look at the graph, Object A weighs ~6 grams and is ~4 cm3 in volume
Density = Mass/Volume
So 6 grams/4 cm3 = 1.5 g/cm3
Water has a density of 1 g/cm3 and because Object A density is higher than that of water, it sinks.
:)
Explanation:
I am not understanding your question
