Answer: 530 hours
Explanation:
The reduction of Nickel ions to nickel is shown as:
of electricity deposits 1 mole of Nickel
1 mole of Nickel weighs = 58.7 g
Given quantity = 18.0 kg = 18000 g (1kg=1000g)
58.7 g of Nickel is deposited by 193000 C of electricity
18000 g of Nickel is deposited by = 
of electricity
where Q= quantity of electricity in coloumbs = 59182282.8C
I = current in amperes = 31.0 A
t= time in seconds = ?
(1h=3600 sec)
Thus 530 hours are required to plate 18.0 kg of nickel onto the cathode if the current passed through the cell is held constant at 31.0 A
Answer:
An alkali metal present in period 2 have larger first ionization energy.
Explanation:
Ionization energy:
The amount of energy required to remove the electron from the atom is called ionization energy.
Trend along period:
As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required.
Trend along group:
As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.
As the size of atom increases the ionization energy from top to bottom also decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus. Thus alkali metal present in period 2 have larger ionization energy because of more nuclear attraction as compared to the alkali metal present in period 4.
Answer:
18.2 g.
Explanation:
You need to first figure out how many moles of nitrogen gas and hydrogen (gas) you have. To do this, use the molar masses of nitrogen gas and hydrogen (gas) on the periodic table. You get the following:
0.535 g. N2 and 1.984 g. H2
Then find out which reactant is the limiting one. In this case, it's N2. The amount of ammonia, then, that would be produced is 2 times the amount of moles of N2. This gives you 1.07 mol, approximately. Then multiply this by the molar mass of ammonia to find your answer of 18.2 g.
V1 = 2.0 L
T1 = 25.0 oC = 298 K V2 = V1T2 = (2.0 L)(244 K) = 1.6 L
V2 = ? t1(298 K)
T2 = –28.9 oC = 244 K
