Answer: The volume of 0.640 grams of 
gas at Standard Temperature and Pressure (STP) is 0.449 L.
Explanation:
Given: Mass of gas = 0.640 g
Pressure = 1.0 atm
Temperature = 273 K
As number of moles is the mass of substance divided by its molar mass.
So, moles of (molar mass = 32.0 g/mol) is as follows.
Now, ideal gas equation is used to calculate the volume as follows.
PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.
Thus, we can conclude that the volume of 0.640 grams of gas at Standard Temperature and Pressure (STP) is 0.449 L.
Answer:
Explanation:
Here, we want to calculate the number of formula units in the given molecule
We start by getting the number of moles
To get the number of moles, we have to divide the mass given by the molar mass
The molar mass is the mass per mole
The molar mass of calcium bromide is 200 g/mol
Thus, we have the number of moles as follows:
The number of formula units in a mole is:
The number of formula units in 0.2075 mole will be:
Answer:
<h2>Density = 0.46 g/mL</h2>
Explanation:
Density of a substance can be found by using the formula
<h3>
</h3>
From the question
mass = 5.52 g
volume = 12 mL
Substitute the values into the above formula and solve for the Density
That's
<h3>
</h3>
We have the final answer as
<h3>Density = 0.46 g/mL</h3>
Hope this helps you
A
Is the correct answer
I’m 95% sure
Answer is: the approximate freezing point of a 0.10 m NaCl solution is -2x°C.
V<span>an't
Hoff factor (i) for NaCl solution is approximately 2.
</span>Van't Hoff factor (i) for glucose solution is 1.<span>
Change in freezing point from pure solvent to
solution: ΔT = i · Kf · m.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
m - molality, moles of solute per
kilogram of solvent.
</span>Kf and molality for this two solutions are the same, but Van't Hoff factor for sodium chloride is twice bigger, so freezing point is twice bigger.
