Answer:
New pressure is 5.40 atm
Explanation:
When T° and moles remain constant, we should use these relation to determine pressure, or also volume
P₁ . V₁ = P₂ . V₂
1.80 atm . 3L = P₂ . 1L
(1.80 atm . 3L ) / 1L = P₂
5.40 atm = P₂
Answer:
Bottle is sitting on the shelf for 15 days.
Explanation:
Given data:
Co-60 half life = 5 days
Total amount = 30.0 g
Amount left = 3.75 g
Time taken = ?
Solution:
AT time zero = 30 g
AT first half life= 30g /2 = 15 g
At 2nd half life = 15 g/ 2 = 7.5 g
At 3rd half life = 7.5 g/2 = 3.75 g
Now we will calculate the sitting time of bottle.
Half life = Time taken / number of half lives
3× 5 days = time taken
Time taken = 15 days
Bottle is sitting on the shelf for 15 days.
I don't have a calculator with me right now, but that mass would be 1200 grams. Divide the given amount of grams by the molar mass of NH3, which is 17.031g/mol. (Nitrogen + 3(hydrogen)). Again, sorry I didn't have a calculator. But all you would need to do is divide 1200 by 17.031. If you need to use sig figs, your answer should have 2 because the 1.2 x 10^3 limits your amount of sig figs.
<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>