NH3 +HCl ----> NH4Cl
moles of HCl used = (0.8 x 17.4) /1000= 0.0139 moles
by use of reacting ratio between HCl to NH4Cl which is 1:1 therefore the moles of NH4Cl is also = 0.0139 moles
molar concentration = moles /volume in liters
molar concentration is therefore= (0.0139/5) x1000 = 2.7 M
Answer:
ΔH°rxn = - 433.1 KJ/mol
Explanation:
- CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)
⇒ ΔH°rxn = 4ΔH°HCl(g) + ΔH°CCl4(g) - 4ΔH°Cl2(g) - ΔH°CH4(g)
∴ ΔH°Cl2(g) = 0 KJ/mol.....pure element in its reference state
∴ ΔH°CCl4(g) = - 138.7 KJ/mol
∴ ΔH°HCl(g) = - 92.3 KJ/mol
∴ ΔH°CH4(g) = - 74.8 KJ/mol
⇒ ΔH°rxn = 4(- 92.3 KJ/mol) + (- 138.7 KJ/mol) - 4(0 KJ/mol) - (- 74.8 KJ/mol)
⇒ ΔH°rxn = - 369.2 KJ/mol - 138.7 KJ/mol - 0 KJ/mol + 74.8 KJ/mol
⇒ ΔH°rxn = - 433.1 KJ/mol
Answer:
Explanation:
Strontium chlorate appears as a moist solid or semi-solid slurry of white crystals. May explode under exposure to heat or fire. Used in pyrotechnics
strontium chlorate | Sr(ClO3)2 - PubChem.
Description: Strontium chlorate appears as a ...
Synonyms: STRONTIUM CHLORATE7791-10-...
Molecular Formula: Sr(ClO3)2 or Cl2O6Sr
Answer:
3.8 M
Explanation:
Volume of acid used VA= 57.0 - 37.5 = 19.5 ml
Volume of base used VB= 67.8 - 45.0 = 22.8 ml
Equation of the reaction
2HNO3(aq) + Ca(OH)2(aq) --------> Ca(NO3)2(aq) + 2H2O(l)
Number of moles of acid NA= 2
Number of moles of base NB= 1
Concentration of acid CA= ???
Concentration of base CB= 1.63 M
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
CA= CBVBNA/VANB
CA= 1.63 × 22.8 × 2/ 19.5 × 1
CA= 3.8 M
HENCE THE MOLARITY OF THE ACID IS 3.8 M.
