To convert from grams to moles, you divide the number of grams by the molar mass To convert from moles to grams, you multiply by the molar mass. You must first calculate the molar mass of the substance
Ionic bonds usually occur between metal and nonmetal ions. For example, sodium (Na), a metal, and chloride (Cl), a nonmetal, form an ionic bond to make NaCl. In a covalent bond, the atoms bond by sharing electrons. Covalent bonds usually occur between nonmetals.
1) Chemical reaction:
Fe2O3 + 2Al ---> Al2O3 + 2Fe
2) molar ratios
1 mol Fe2O3 : 2Al : 1 mol Al2O3 : 2 mol Fe
3) Convert 15.0 g of iron into moles
atomic mass Fe = 55.8 g/mol
moles = mass in grams / atomic mass = 15.0 g / 55.8 g/mol = 0.269 mol
4) Use proportions to determine the moles of Fe2O3, Al, and Al2O3
a) 1mol Fe2O3 / 2 mol Fe = x / 0.269 mol Fe
x =
=> x = 0.269 mol Fe * 1 mol Fe2O3 / 2 mol Fe = 0.134 mol Fe2O3
b) 2 mol Al / 2 mol Fe = x / 0.269 mol Fe
=> x = 0.269 mol Al
c) 2 mol Fe / 1 mol Al2O3 = 0.269 mol Fe / x
=> x = 0.269 mol Fe * 1 mol Al2O3 / 2 mol Fe
x = 0.134 mol Al2O3
5) Convert moles to grams
a) Fe2O3
molar mass Fe2O3 = 2* 55.8 g/mol + 3*16g/mol = 159.6 g/mol
mass = molar mass * number of moles
mass = 159.6 g/mol * 0.134 mol = 21.4 g
b) Al
atomic mass = 27.0 g/mol
mass = number of moles * atomic mass = 0.269 mol * 27.0 g/mol = 7.26 g
c) Al2O3
molar mass = 2 * 27.0 g/mol + 3*16.0 g/mol = 102.0g/mol
mass Al2O3 = numer of moles * molar mass = 0.134 mol * 102.0 g/mol = 13.7 g
Answers:
21.4 g Fe2O3
7.26 g Al
13.7 g Al2O3
Answer:
they have delocolised electrons which can carry electrical charges through the metals.
Explanation:
Answer:
W = - 10.5943 KJ, work is negative because it is carried out by the system towards the surroundings.
Explanation:
heat engine:
∴ Th = 485°C
∴ Tc = 42°C
∴ Qh = 9.75 KJ......heat from hot source
first law:
∴ ΔU = Q + W = 0 .........cyclic process
⇒ Q = - W
∴ Q = Qh + Qc = Qh - (- Qc)
∴ Qc: heat from the cold source ( - )
⇒ Qh - ( - Qc) = - W..............(1)
⇒ Qc/Qh = - Tc/Th...........(2)
from (2):
⇒ Qc = - (Tc/Th)(Qh) = - (42°C/485°C)(9.75 KJ)
⇒ Qc = - 0.8443 KJ
replacing in (1):
⇒ - W = 9.75 KJ - ( - 0.8443 KJ)
⇒ - W = 10.5943 KJ
