Answer:
a) 381.2 g
b) 39916 g
c) 0.0013 lb mol
d) 29.6 g mol
Explanation:
The molecular weight (mw) of a compound is the mass of it per mole, so it's the ratio of the mass (m) per mole (n).
a) The molecular weight of one mol is found at the periodic table. So, for Mg, mw = 24.3 g/mol, for Cl = 35.5 g/mol, so for MgCl2, mw = 24.3 + 2*35.5 = 95.3 g/mol. The g mol is the mass divided by the molecular weight:
g mol = m/mw
4 = m/95.3
m = 381.2 g
b) The pound (lb) is a unity of mass, and the lb mol is a unity of the mass divided by the molecular weight. So, by the periodic table, the molecular weight of C3H8 is 3*12 (of C) + 8*1 (of H) = 44 lb/mol.
lb mol = m/mw
2 = m/44
m = 88 lb
1 lb = 453.592 g
So, m = 88*453.592 = 39916 g
c) The molecular weight of N2 is 2*14 (of N) = 28 lb/mol.
m = 16/453.592 = 0.0353 lb
lb mol = m/mw
lb mol = 0.0353/28
lb mol = 0.0013 lb mol
d) The molecular weight is 2*12 (of C) + 6*1(of H) + 1*16(of O) = 46 g/mol
3 lb = 1360.78 g
g mol = m/mw
g mol = 1360.78/46
g mol = 29.6 g mol
Maybe
A. 400 ml of 5.0% glucose solution
Answer:
The time required to melt the frost is 3.25 hours.
Explanation:
The time required to melt the frost dependes on the latent heat of the frost and the amount of heat it is transfered by convection to the air .
The heat transferred per unit area can be expressed as:
being hc the convective heat transfer coefficient (2 Wm^-2K^-1) and ΔT the difference of temperature (20-0=20 °C or K).
If we take 1 m^2 of ice, with 2 mm of thickness, we have this volume
The mass of the frost can be estimated as
Then, the amount of heat needed to melt this surface (1 m²) of frost is
The time needed to melt the frost can be calculated as
Answer:
Maintaining a high starting-material concentration can render this reaction favorable.
Explanation:
A reaction is <em>favorable</em> when <em>ΔG < 0</em> (<em>exergonic</em>). ΔG depends on the temperature and on the reaction of reactants and products as established in the following expression:
ΔG = ΔG° + R.T.lnQ
where,
ΔG° is the standard Gibbs free energy
R is the ideal gas constant
T is the absolute temperature
Q is the reaction quotient
To make ΔG < 0 when ΔG° > 0 we need to make the term R.T.lnQ < 0. Since T is always positive we need lnQ to be negative, what happens when Q < 1. Q < 1 implies the concentration of reactants being greater than the concentration of products, that is, maintaining a high starting-material concentration will make Q < 1.
