Answer:
0.062mol
Explanation:
Using ideal gas law as follows;
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821Latm/molK)
T = temperature (K)
Based on the information provided;
P = 152 Kpa = 152/101 = 1.50atm
V = 0.97L
n = ?
T = 12°C = 12 + 273 = 285K
Using PV = nRT
n = PV/RT
n = (1.5 × 0.97) ÷ (0.0821 × 285)
n = 1.455 ÷ 23.39
n = 0.062mol
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.
Hello!
We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100 g solution)
C: 83.7% = 83,7 g
H: 16.3% = 16.3 g
Let us use the above mentioned data (in g) and values will be converted to amount of substance (number of moles) by dividing by molecular mass (g / mol) each of the values, lets see:
We note that the values found above are not integers, so let's divide these values by the smallest of them, so that the proportion is not changed, let's see:
Note: So the ratio in the smallest whole numbers of carbon to hydrogen is 3:7, t<span>hus, the minimum or empirical formula found for the compound will be:
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I hope this helps. =)
Answer:
<em>Low tides would be lower and high tides would be higher and any low lying coastline would be flooded</em>
Explanation:
<em>Also...</em>
<em>If the moon got about 20 times closer it would make a gravitational force 400 times greater than what we are used to now</em>
Radio waves.
microwaves.
infrared waves.
light.
ultraviolet waves (rays)
x-rays.
gamma rays.
