Answer:
12.44 g
Explanation:
2C4H10 + 13O2 = 8CO2 + 10H2O
n(C4H10) = m(C4H10)/M(C4H10) = 4.1 / 58g/mol = 0.0707 mol (excess).
n(O2) = m(O2)/M(O2) = 25.9 / 32g/mol = 0.809 mol (deficiency).
Since the ratio of O2 to octane is 13 : 2 we can divide 0.0707 by 2 to get 0.03535 and divide 0.809 by 13 to get 0.062.
mass of CO2 produced =
M = [0.0707 moles C4H10 x 8 moles CO2] / 2 moles C4H10 x 44 g CO2/mol
M = 0.5656/2 * 44
M = 0.2828 * 44
M = 12.44 of CO2
The chemicals are reactive with one another
Both figures are mixtures,
Figure II is a heterogenous mixture
Figure I is a homogenous mixture
<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 