Answer:
The answer is "670.176 km".
Explanation:
Volume of the occupies of one M & M= 0.5 \ cm^3\\\\
M&M 1 mole
Calculating volume of M& M mole
Calculating the cube mole

Therefore 18 tractor trailers wouldn't be sufficient.
To solve this problem, we should recall that
the change in enthalpy is calculated by subtracting the total enthalpy of the reactants
from the total enthalpy of the products:
ΔH = Total H of products – Total H of reactants
You did not insert the table in this problem, therefore I
will find other sources to find for the enthalpies of each compound.
ΔHf CO2 (g) = -393.5 kJ/mol
ΔHf CO (g) = -110.5 kJ/mol
ΔHf Fe2O3 (s) = -822.1 kJ/mol
ΔHf Fe(s) = 0.0 kJ/mol
Since the given enthalpies are still in kJ/mol, we have to
multiply that with the number of moles in the formula. Therefore solving for ΔH:
ΔH = [<span>3 mol </span><span>( − </span><span>393.5 </span>kJ/mol<span>) + 1 mol (</span>0.0
kJ/mol)<span>] − [</span><span>3 mol </span><span>( − </span><span>110.5 </span>kJ/mol<span>) + </span><span>2 mol </span><span>( − </span><span>822.1 </span>kJ/mol<span>)]</span>
ΔH = <span>795.2
kJ</span>
Answer:
The standard moisture content specification for hard gelatin capsules is between 13 % w/w and 16 % w/w. This value can vary depending upon the conditions to which they are exposed: at low humidity's they will lose moisture and become brittle, and at high humidity's they will gain moisture and soften.
Explanation:
Hope this helps!
Answer:
T2 = 29.79°C
Explanation:
Equliibrium signifies that heat loss = heat gained
Heat gained by Ice;
H = ML
Mass, M = Number of moles * Molar mass = 1 * 18 = 18g
l = 6.01 k J m o l = 334 J/g
C = 4.186 J/g
H = 18(334)
H = 6012
Heat lost by water
H = MCΔT
H = 18 * 4.186 * (50 - T2)
H = 3767.4 - 75.348T2
Since H = H, we have;
6012 = 3767.4 - 75.348T2
- 75.348T2 = 3767 - 6012
T2 = 2245 / 75.348
T2 = 29.79°C
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