Answer:
b) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature.
Explanation:
The solubility of NaCH₃CO₂ in water is ~1.23 g/mL. This means that at room temperature, we can dissolve 1.23 g of solute in 1 mL of water (solvent).
<em>What would be the best method for preparing a supersaturated NaCH₃CO₂ solution?</em>
<em>a) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at room temperature while stirring until all the solid dissolves.</em> NO. At room temperature, in 100 mL of H₂O can only be dissolved 123 g of solute. If we add 130 g of solute, 123 g will dissolve and the rest (7 g) will precipitate. The resulting solution will be saturated.
<em>b) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature. </em>YES. The solubility of NaCH₃CO₂ at 80 °C is ~1.50g/mL. If we add 130 g of solute at 80 °C and let it slowly cool (and without any perturbation), the resulting solution at room temperature will be supersaturated.
<em>c) add 1.23 g of NaCH₃CO₂ to 200 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature.</em> NO. If we add 1.23 g of solute to 200 mL of water, the resulting solution will have a concentration of 1.23 g/200 mL = 0.00615 g/mL, which represents an unsaturated solution.
Answer : The temperature will be, 392.462 K
Explanation :
According to the Arrhenius equation,
or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at 
=
= rate constant at 
= 
= activation energy for the reaction = 66.41 kJ/mole = 66410 J/mole
R = gas constant = 8.314 J/mole.K
= initial temperature = 293 K
= final temperature = ?
Now put all the given values in this formula, we get:
![\log (\frac{3K_1}{K_1})=\frac{66410J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{293K}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B3K_1%7D%7BK_1%7D%29%3D%5Cfrac%7B66410J%2Fmole%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B293K%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
Therefore, the temperature will be, 392.462 K
Answer:
1. D (24.0 moles CO2)
2. A (.239 moles H2)
Explanation:
1. First Balance the equation
1 C3H8 + 5 O2 ---> 3 CO2 + 4 H2O
Then set up a stoiciometric equation so that the moles of O2 cancel out
40mol O2 x 
= 24.0 moles CO2
2. Set up a stoichiometric equation
10 grams Fe x 
x 
= 0.239 moles H2
Answer:
4 mL
Explanation:
Use the density formula, d = m/v
Plug in the density and mass to solve for v:
4 = 16/v
4v = 16
v = 4
So, the volume is 4 mL
