First you need to find out the Limiting reactant (LR). convert both reactants to the same thing. Check that the chemical equation is balanced. Now use stoichiometr and remember at moles, multiply: need moles, divide2 g / 42g/mol= 0.0477 mol propane mass propane/ Molar Mass propane = moles propane4 g / 32 g/mol= 0.125 mol oxygen X (1 mol/ 5 mol) = 0.025 mol propane oxygen is the LRmass O2 / MM O2 X (mol propane / mol O2)0.025 mol X (3 mol / 1 mol ) = .075 mol CO20.075 mol X (12 + 2*16) g /mol = 3.6 g CO2 In one step:2 g / 42g/mol X (3 mol / 1 mol ) X 48 g/mol = 6.86 g CO24 g / 32 g/mol X (3 mol / 5 mol ) X 48 g/mol = 3.6 g CO2mass/ MM X coefficient ratio X MM (new)
I am pretty sure the answer is B but correct me if I'm wrong hope this helps.
Answer : The value of
is 28.97 kJ/mol
Explanation :
To calculate
of the reaction, we use clausius claypron equation, which is:
![\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= vapor pressure at temperature
= 462.7 mmHg
= vapor pressure at temperature
= 140.5 mmHg
= Enthalpy of vaporization = ?
R = Gas constant = 8.314 J/mol K
= initial temperature = ![-21.0^oC=[-21.0+273]K=252K](https://tex.z-dn.net/?f=-21.0%5EoC%3D%5B-21.0%2B273%5DK%3D252K)
= final temperature = ![45^oC=[-41.0+273]K=232K](https://tex.z-dn.net/?f=45%5EoC%3D%5B-41.0%2B273%5DK%3D232K)
Putting values in above equation, we get:
![\ln(\frac{140.5mmHg}{462.7mmHg})=\frac{\Delta H_{vap}}{8.314J/mol.K}[\frac{1}{252}-\frac{1}{232}]\\\\\Delta H_{vap}=28966.6J/mol=28.97kJ/mol](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7B140.5mmHg%7D%7B462.7mmHg%7D%29%3D%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B252%7D-%5Cfrac%7B1%7D%7B232%7D%5D%5C%5C%5C%5C%5CDelta%20H_%7Bvap%7D%3D28966.6J%2Fmol%3D28.97kJ%2Fmol)
Therefore, the value of
is 28.97 kJ/mol
Answer: false 100
Explanation:
if this is the true or false test here
Answer:
35Cl = 75.9 %
37Cl = 24.1 %
Explanation:
Step 1: Data given
The relative atomic mass of Chlorine = 35.45 amu
Mass of the isotopes:
35Cl = 34.96885269 amu
37Cl = 36.96590258 amu
Step 2: Calculate percentage abundance
35.45 = x*34.96885269 + y*36.96590258
x+y = 1 x = 1-y
35.45 = (1-y)*34.96885269 + y*36.96590258
35.45 = 34.96885269 - 34.96885269y +36.96590258y
0.48114731 = 1,99704989y
y = 0.241 = 24.1 %
35Cl = 34.96885269 amu = 75.9 %
37Cl = 36.96590258 amu = 24.1 %