Answer:  kJ/mol
 kJ/mol
Explanation: <u>Enthalpy</u> <u>Change</u> is the amount of energy in a reaction - absorption or release - at a constant pressure. So, <u>Standard</u> <u>Enthalpy</u> <u>of</u> <u>Formation</u> is how much energy is necessary to form a substance.
The standard enthalpy of formation of HCl is calculated as:

 →
 → 
Standard Enthalpy of formation for the other compounds are:
Calcium Hydroxide:  -1002.82 kJ/mol
 -1002.82 kJ/mol
Calcium chloride:  -795.8 kJ/mol
 -795.8 kJ/mol
Water:  -285.83 kJ/mol
 -285.83 kJ/mol
Enthalpy is given per mol, which means we have to multiply by the mols in the balanced equation.
Calculating:
![-17.2=[-795.8+2(285.85)]-[-1002.82+2\Delta H]](https://tex.z-dn.net/?f=-17.2%3D%5B-795.8%2B2%28285.85%29%5D-%5B-1002.82%2B2%5CDelta%20H%5D)



So, the standard enthalpy of formation of HCl is -173.72 kJ/mol
 
        
             
        
        
        
Answer:
Explanation:
Not Many
1 mol of CO has a mass of
C = 12
O = 16
1 mol = 28 grams.
1 mol of molecules = 6.02 * 10^23
x mol of molecules = 3.14 * 10^15        Cross multiply
6.02*10^23 x = 1 * 3.14 * 10^15             Divide by 6.02*10^23
x = 3.14*10^15 / 6.02*10^23
x = 0.000000005 mols
x = 5*10^-9
1 mol of CO has a mass of 28
5*10^-9 mol of CO has a mass of x                        Cross Multiply
x = 5 * 10^-9 * 28
x = 1.46 * 10^-7 grams
Answer: there are 1.46 * 10-7 grams of CO if only 3.14 * 10^15 molecules are in the sample
 
        
             
        
        
        
<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 initial pressure, volume and temperature of the gas
 are the final 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 
 
        
             
        
        
        
Answer:
18.4 g 
Explanation:
M(He) = 4.0 g/mol
4.6 mol * 4.0 g/ 1 mol = 18.4 g