Two electrons is your answer glad to help!
Answer:The formulas of ionic compounds are:
a)
b)
c)
d)
Explanation:
Formulas for the an ionic compounds is determine by:
Criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.
(a) Copper bromide :Given that it contains
ion.

(b) Manganese oxide : Given that it contains
ion.

(c)Mercury iodide :Given that it contains 

(d) Magnesium phosphate :Given that it contains 

Acetaminophen as a chemical formula of C8H9NO2. The molar
masses are:
C8H9NO2 = 151.163 g/mol
C = 12 g/mol
H = 1 g/mol
N = 14 g/mol
O = 16 g/mol
<span>TO get the mass percent, simply multiply the molar mass of
each elements with the number of the
element divide by the molar mass of acetaminophen, that is:</span>
%C = [(12 * 8) / 151.163] * 100% = 63.50%
%H = [(1 * 9) / 151.163] * 100% = 5.954%
%N = [(14 * 1) / 151.163] * 100% = 9.262%
<span>%O = [(16 * 2) / 151.163] * 100% = 21.17% </span>
Answer:
From the following enthalpy of reaction data and data in Appendix C, calculate ΔH∘f for CaC2(s): CaC2(s)+2H2O(l)→Ca(OH)2(s)+C2H2(g)ΔH∘=−127.2kJ
ΔHf°(C2H2) = 227.4 kJ/mol
ΔHf°(H2O) = -285.8 kJ/mol and
ΔHf°(Ca(OH)2) = -985.2 kJ/mol
(Ans)
ΔHf° of CaC2 = -59.0 kJ/mol
Explanation:
CaC2(s) + 2 H2O(l) → Ca(OH)2(s) + C2H2 (g) = −127.2kJ
ΔHrxn = −127.2kJ
ΔHrxn = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - ΔHf°(CaC2)- 2ΔHf°(H2O);
ΔHf°(CaC2) = ΔHf°(C2H2) + ΔHf°(Ca(OH)2) - 2ΔHf°(H2O) – ΔHrxn
Where
ΔHf°(C2H2) = 227.4 kJ/mol
ΔHf°(H2O) = -285.8 kJ/mol and
ΔHf°(Ca(OH)2) = -985.2 kJ/mol
ΔHf°(CaC2) =227.4 - 985.2 + 2x285.8 + 127.2 = -59.0 kJ/mol
ΔHf°(CaC2) = -59.0 kJ/mol
<span>Answer:
It depends on what came after "0.5440 M H...".
If it was a monoprotic acid, like HCl, the calculation would go like this:
(55.25 mL) x (0.5440 M acid) x (1 mol KOH / 1 mol acid) / (0.2450 M KOH) =
122.7 mL KOH
If it was a diprotic acid, like H2SO4, like this:
(55.25 mL) x (0.5440 M acid) x (2 mol KOH / 1 mol acid) / (0.2450 M KOH) =
245.4 mL KOH
If it was a triprotic acid, like H3PO4, like this:
(55.25 mL) x (0.5440 M acid) x (3 mol KOH / 1 mol acid) / (0.2450 M KOH) =
368.0 mL KOH</span>