Answer : Option A) HF.
Explanation : Assuming the complete question as per the attachment.
H-F bonds are formed because of hydrogen bonding between H atom and F atom and is considered to be strongest amongst the rest other options given.
As given in the table it is clear that the molar mass of HF molecule is the lowest. It shows that, when dispersion increases in the molecule molecular weight also increases.
c. Diluted, and
d. soluble
the answer will be b)
Answer:
Ld₂Mz
General Formulas and Concepts:
Chemistry - Compounds
Explanation:
Step 1: Define
"Lindenium" Ld w/ oxidation charge of +1
"Mendezine" Mz w/ oxidation charge of -2
Step 2: Write Compound
To balance out the charges to be neutral (0), we must counter balance the amount of Ld's with Mz's. We need 2 Ld's with 1 Mz to create a compound with a neutral charge:
2Ld¹⁺ + Mz²⁻
2(+1) + (-2) = (+2) + (-2) = 0 (neutral charge)
Therefore, our compound must be Ld₂Mz.
The quantity of nitrogen the cylinder contains is 4477.8 g
Pressure is a force exerted in a perpendicular direction in any item.
By ideal gas law
PV = nRT
w = mass
Volume is 20.0 l
Pressure is
The molar mass of nitrogen is 28 g/mol
R is gas constant = 0.0821
Temperature is 28 converted into kelvin that is 301 k
Putting the values
Thus, the mass of nitrogen is 4477.8 g.
Learn more about pressure, here:
Answer : The mass of gas is, 4477.8 g
Solution :
using ideal gas equation,
where,
n = number of moles of gas
w = mass of gas
P = pressure of the gas =
conversion :
T = temperature of the gas =
M = molar mass of gas = 28 g/mole
R = gas constant = 0.0821 Latm/moleK
V = volume of gas = 20 L
Now put all the given values in the above equation, we get the mass of gas.
Therefore, the mass of gas is, 4477.8 g
The balanced chemical reaction is:
2Al2O3 = 4Al +3 O2
We are given the amount of the reactant. This will be the starting point of our calculations.
26.5 mol Al2O3 ( 3 mol O2 / 2 mol Al2O3 ) = 39.75 mol O2 needed
Thus, 0.014 mol of I2 is needed to form the given amount of NI3.
the balanced chemical equation for the decomposition of Al₂O₃ is as follows
2Al₂O₃ --> 4Al + 3O₂
stoichiometry of Al₂O₃ to O₂ is 2:3
when 2 mol of Al₂O₃ decomposes - 3 mol of O₂ are formed
therefore when 26.5 mol of Al₂O₃ decomposes - 3/2 x 26.5 mol = 39.75 mol
the number of moles of O₂ formed are 39.75 mol
Answer:
Till
Explanation:
Till is unsorted glacial sediment showing no stratfaction.
Answer: Glacier Deposition
Explanation: