<span>Important information to solve the exercise :
Substance ΔHf (kJ/mol):
HCl(g)= −92.0 </span><span>kJ/mol
Al(OH)3(s)= −1277.0 </span><span><span>kJ/mol
</span> H2O(l)= −285.8 </span><span>kJ/mol
AlCl3(s) =−705.6 </span><span>kJ/mol
</span><span>Al(OH)3(s)+3HCl(g)→AlCl3(s)+3H2O(l)
reactants products
products- reactants:</span><span>
(−705.6) + (3 x −285.8) - ( −1277.0 ) - (3 x −92.0 ) = - 10.0 </span>kJ per mole at 25°C
<span>
</span>
Answer:
Work = 90.65 j
Explanation:
Given data:
Mass = 0.500 Kg
Distance = 18.5 m
Work done = ?
Solution:
Work = force . distance
Force = mg
Work = mg.distance
Work = mgh
Work = 0.500 Kg × 9.8 m/s²× 18.5 m
Work = 90.65 Kg .m²/s²
Kg .m²/s² = j
Work = 90.65 j
Recall that density is Mass/Volume. We are given the mL of liquid which is volume so all we need is mass now. We are given the mass of the granulated cylinder both with and without the liquid, so if we subtract them, we can get the mass of the liquid by itself. So, 136.08-105.56= 30.52g. This is the mass of the liquid. We now have all we need to find the density. So, let’s plug these into the density formula. 30.52g/45.4mL= 0.672 g/mL. This is our final answer since the problem requests the answer in g/mL, but be careful, because some problems in the future may ask for g/L requiring unit conversions. Also note that 30.52 was 4 sigfigs and 45.4 was 3 sigfigs, and so dividing them required an answer that was 3 sigfigs as well, hence why the answer is in the thousandths place
A) 120 mm
B) 127 mm
C) 914.4 mm
D) 1000 mm
E) 3048 mm