Ok to answer this question we firsst need to fin the number of mol of Urea (CH4N2O). to do this we simply :
1 mol of urea =15/60.055 = 0.25mol
therefore 200g of water contain 0.25mol
the next step is to determine the malality of our solution in 200g of water, to do this we say:
200 g = 1Kg/1000g = 0.2kg
therefor 0.25mol/0.2Kg = 1.25mol/kg
and from the equation:
we know that i = 1
we are given Kf
b is the molality that we just calculated
therefore;
the solutions freezing point is -2.325°C
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
Answer:
ρ = 1.08 g/cm³
Explanation:
Step 1: Given data
Mass of the substance (m): 21.112 g
Volume of the substance (V): 19.5 cm³
Step 2: Calculate the density of the substance
The density (ρ) of a substance is equal to its mass divided by its volume.
ρ = m / V
ρ = 21.112 g / 19.5 cm³
ρ = 1.08 g/cm³
The density of the substance is 1.08 g/cm³.
Its an ore of uraninte i think.
According to Balance chemical equation,
N₂ + 3 H₂ → 2 NH₃
1 mole of Nitrogen reacts with 3 moles of Hydrogen to produce 2 mole of Ammonia.
It is known that i mole of any gas at standard temperature and pressure occupies 22.4 L of Volume. So, we can also say,
22.4 L (1 × 22.4) of Nitrogen gas (in question it is taken in excess) reacts with 67.2 L (22.4 × 3) of Hydrogen gas to produce 44.8 L (22.4 × 2) of Ammonia.
Result:
44.8 L is the correct answer.