Answer:
The critical temperature of a substance is the temperature at and above which vapour of the substance cannot be liquefied, no matter how much pressure is applied.
Answer: -
IE 1 for X = 801
Here X is told to be in the third period.
So n = 3 for X.
For 1st ionization energy the expression is
IE1 = 13.6 x Z ^2 / n^2
Where Z =atomic number.
Thus Z =( n^2 x IE 1 / 13.6)^(1/2)
Z = ( 3^2 x 801 / 13.6 )^ (1/2)
= 23
Number of electrons = Z = 23
Nearest noble gas = Argon
Argon atomic number = 18
Number of extra electrons = 23 – 18 = 5
a) Electronic Configuration= [Ar] 3d34s2
We know that more the value of atomic radii, lower the force of attraction on the electrons by the nucleus and thus lower the first ionization energy.
So more the first ionization energy, less is the atomic radius.
X has more IE1 than Y.
b) So the atomic radius of X is lesser than that of Y.
c) After the first ionization, the atom is no longer electrically neutral. There is an extra proton in the atom.
Due to this the remaining electrons are more strongly pulled inside than before ionization. Hence after ionization, the radii of Y decreases.
When the specific heat capacity of the water is 4.18 J/g.°C so, we are going to use this formula to get the heat for cooling three phases changes from steam to liquid and from liquid to ice (solid) :
when Q = M*C*ΔT
Q is the heat in J
and M is the mass in gram = 1 mol H2O * 18 g/mol(molar mass) = 18 g
C is the specific heat J/g.°C
ΔT is the change in temperature
Q = Mw *[ ( Csteam * ΔTsteam)+(Cw*ΔTw) + (Cice * ΔT ice)]
= 18 g * [(2.01 * (155-100°C)) + (4.18 * (100-0°C)) + (2.09 * (0 - 55 °C))]
∴Q = 7444.8 J
and when we know that the heat of fusion for water = 334J/g
and heat of vaporization for water = 2260J/g
∴Q for the two phases changes = M * (2260+334)
= 18 * (2260+334)
= 46692 J
∴ Q total = 7444.8 + 46692 = 54136.8 J
Answer:
High levels of moisture are contained within an air mass.
Tell me if I'm correct, plz!
Explanation:
Answer: 0.151
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
![Rate=-\frac{1d[HBr]}{2dt}=+\frac{1d[H_2]}{2dt}=+\frac{1d[Br_2]}{dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7B1d%5BHBr%5D%7D%7B2dt%7D%3D%2B%5Cfrac%7B1d%5BH_2%5D%7D%7B2dt%7D%3D%2B%5Cfrac%7B1d%5BBr_2%5D%7D%7Bdt%7D)
Given: ![-\frac{d[HBr]}{dt}]=0.301](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BHBr%5D%7D%7Bdt%7D%5D%3D0.301)
Putting in the values we get:
![\frac{0.301}{2}=+\frac{1d[Br_2]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B0.301%7D%7B2%7D%3D%2B%5Cfrac%7B1d%5BBr_2%5D%7D%7Bdt%7D)
![+\frac{1d[Br_2]}{dt}=0.151](https://tex.z-dn.net/?f=%2B%5Cfrac%7B1d%5BBr_2%5D%7D%7Bdt%7D%3D0.151)
Thus the rate of appearance of 
is 0.151
