Answer:
im pretty sure the correct option here id c
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Answer:
15 moles of ammonium sulfate would be formed from 30 moles of ammonia.
Explanation:
Given data:
Number of moles of ammonium sulfate formed = ?
Number of moles of ammonia = 30.0 mol
Solution:
Chemical equation:
2NH₃ + H₂SO₄ → (NH₄)₂SO₄
Now we will compare the moles of ammonium sulfate with ammonia.
NH₃ : (NH₄)₂SO₄
2 : 1
30.0 : 1/2×30.0 = 15.0 mol
So 15 moles of ammonium sulfate would be formed from 30 moles of ammonia.
Answer: 94.13 L
Explanation: In STP in an ideal gas there is a standard value for both temperature and pressure. At STP,pressure is equal to 1atm and the temperature at 0°C is equal to 273.15K. This problem is an ideal gas so we use PV=nRT where R is a constant R= 0.08205 L.atm/mol.K.
To find volume, derive the equation, it becomes V=nRT/P. Substitute the values. V= 4.20 mol( 0.08205L.atm/mol.K)(273.15K) / 1 atm = 94.13 L. The mole units, atm and K will be cancelled out and L will be the remaining unit which is for volume.
<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>
