Oxygen has to be involved when methanol is ignited
<span>The student should
follow following steps to make 1 L of </span>2.0 M CaCl₂.<span>
<span>
1. First he should
calculate the number of moles of 2.0 M CaCl</span></span>₂ in 1 L solution.<span>
</span>Molarity of the solution = 2.0 M<span>
Volume of solution which should be prepared = 1 L
Molarity =
number of moles / volume of the solution
Hence, number of moles in 1 L = 2 mol
2. Find
out the mass of dry CaCl</span>₂ in 2 moles.<span>
moles =
mass / molar mass
Moles of CaCl₂ =
2 mol</span><span>
Molar mass of CaCl₂ = </span><span>110.98 g/mol
Hence, mass of CaCl</span>₂ = 2 mol x <span>110.98 g/mol
= 221.96
g
3. Weigh the mass
accurately
4. Then take a cleaned and dry1 L volumetric flask and place a funnel top of it. Then carefully add the salt into the volumetric flask and
finally wash the funnel and watch glass
with de-ionized water. That water also should be added into the volumetric
flask.
5. Then add some
de-ionized water into
the volumetric flask and swirl well until all salt are
dissolved.
<span>6. Then top up to
mark of the volumetric flask carefully.
</span></span>
7. As the final step prepared solution should be labelled.
you forgot the well answers
<u>Answer:</u> The volume when the pressure and temperature has changed is 
<u>Explanation:</u>
To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.
The equation follows:

where,
are the initial pressure, volume and temperature of the gas
are the final pressure, volume and temperature of the gas
Let us assume:
![P_1=1.20atm\\V_1=795mL\\T_1=116^oC=[116+273]K=389K\\P_2=0.55atm\\V_2=?mL\\T_2=75^oC=[75+273]K=348K](https://tex.z-dn.net/?f=P_1%3D1.20atm%5C%5CV_1%3D795mL%5C%5CT_1%3D116%5EoC%3D%5B116%2B273%5DK%3D389K%5C%5CP_2%3D0.55atm%5C%5CV_2%3D%3FmL%5C%5CT_2%3D75%5EoC%3D%5B75%2B273%5DK%3D348K)
Putting values in above equation, we get:

Hence, the volume when the pressure and temperature has changed is 