The question is incomplete, here is the complete question.
A chemist prepares a solution of copper(II) fluoride by measuring out 0.0498 g of copper(II) fluoride into a 100.0mL volumetric flask and filling the flask to the mark with water.
Calculate the concentration in mol/L of the chemist's copper(II) fluoride solution. Round your answer to 3 significant digits.
<u>Answer:</u> The concentration of copper fluoride in the solution is 
<u>Explanation:</u>
To calculate the molarity of solute, we use the equation:

We are given:
Given mass of copper (II) fluoride = 0.0498 g
Molar mass of copper (II) fluoride = 101.54 g/mol
Volume of solution = 100.0 mL
Putting values in above equation, we get:

Hence, the concentration of copper fluoride in the solution is 
Answer: (1). There are 0.0165 moles of gaseous arsine (AsH3) occupy 0.372 L at STP.
(2). The density of gaseous arsine is 3.45 g/L.
Explanation:
1). At STP the pressure is 1 atm and temperature is 273.15 K. So, using the ideal gas equation number of moles are calculated as follows.
PV = nRT
where,
P = pressure
V = volume
n = number of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.

2). As number of moles are also equal to mass of a substance divided by its molar mass.
So, number of moles of Arsine
(molar mass = 77.95 g/mol) is as follows.

Density is the mass of substance divided by its volume. Hence, density of arsine is calculated as follows.

Thus, we can conclude that 0.0165 moles of gaseous arsine (AsH3) occupy 0.372 L at STP and the density of gaseous arsine is 3.45 g/L.
Answer:
Explanation:
Ionic bond
Ionic bond, also called electrovalent bond, type of linkage formed from the electrostatic attraction between oppositely charged ions in a chemical compound.
Answer:
5.81L
Explanation:
N1 = 1.70 moles
V1 = 3.80L
V2 = ?
N2 = 2.60 moles
Mole - volume relationship,
N1 / V1 = N2 / V2
V2 = (N2 × V1) / N1
V2 = (2.60 × 3.80) / 1.70
V2 = 9.88 / 1.70
V2 = 5.81 L
The volume of the gas is 5.81L
It would begin osmosis on it. If you put it in vinegar for example water would begin to seep through the membrane.