Osmosis is the diffusion of water from a high concentration to a lower concentration
Answer:
1. The electronic configuration of X is: 1s2 2s2 sp6 3s2
2. The configuration of the anion of Y (i.e Y^2-) is 1s2 2s2 2p6
3. The formula of the compound form by X and Y is given as: XY
Explanation:
For X to loss two electrons, it means X is a group 2 element. X can be any element in group 2. The electronic configuration of X is:
1s2 2s2 sp6 3s2
To get the electronic configuration of the anion of element Y, let us find the configuration of element Y. This is done as follows:
Y receives two electrons from X to complete its octet. Therefore Y is a group 6 element. The electronic configuration of Y is given below
1s2 2s2 2p4
The configuration of the anion of Y (i.e Y^2-) is 1s2 2s2 2p6
The formula of the compound form by X and Y is given below :
X^2+ + Y^2- —> XY
Their valency will cancel out thus forming XY
1) number of moles of N2 = n/2
2) Number of moles of CH4 = n/2
3) Total number of moles of the mixture = n/2 + n/2 = n
4) Kg of N2
mass in grams = number of moles * molar mass
molar mass of N2 = 2 * 14.0 g/mol = 28 g/mol
=> mass of N2 in grams = (n/2) * 28 = 14n
mass of N2 in Kg = mass of N2 in grams * [1 kg / 1000g] = 14n/1000 kg = 0.014n kg
Answer: mass of N2 in kg = 0.014n kg
Answer:
Option B. 2096.1 K
Explanation:
Data obtained from the question include the following:
Enthalpy (H) = +1287 kJmol¯¹ = +1287000 Jmol¯¹
Entropy (S) = +614 JK¯¹mol¯¹
Temperature (T) =.?
Entropy is related to enthalphy and temperature by the following equation:
Change in entropy (ΔS) = change in enthalphy (ΔH) / Temperature (T)
ΔS = ΔH / T
With the above formula, we can obtain the temperature at which the reaction will be feasible as follow:
ΔS = ΔH / T
614 = 1287000/ T
Cross multiply
614 x T = 1287000
Divide both side by 614
T = 1287000/614
T = 2096.1 K
Therefore, the temperature at which the reaction will be feasible is 2096.1 K
Answer:
n = 2
l = 1
m = 1
s = +1/2
Explanation:
₇N tiene la configuración electrónica;
1s2 2s2 2p3
Esto implica que este último electrón tiene los siguientes números cuánticos;
n = 2
l = 1
m = 1
s = +1/2
Este último electrón estará en un orbital de 2pz como lo muestran los números cuánticos enumerados anteriormente.
