The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is
C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is
6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)
Using the equation for the standard enthalpy change of formation
ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}
C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
= -1273.3 - (0 + 0 + 0)
= -1273.3
The type of the bond is present Na₃PO₄ is the ionic bond. the Na₃PO₄ is the ionic compound. yes the Na₃PO₄ is the polyatomic ion.
The Na₃PO₄ is Na⁺ and PO₄³⁻. the phosphorus is the non metal and the oxygen atom is the non metal. the non meta and non meta form the covalent or molecular bond. the bond between the PO₄³⁻ bond is the covalent bond but the overall present in the Na₃PO₄ is the ionic bond . the bons in between the Na⁺ and PO₄³⁻ is the the ionic bond. the PO₄³⁻ id the polyatomic ion .
The bond between the positively charged ion and the negatively charged ion are called as the ionic bond and the compound form is the ionic compound.
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Answer:
-Aluminum Oxide: The cation is Al3+ and the anion is O2-. The sum of the charges for aluminum oxide is 2(3+) + 3(2-) = 0. Thus, the formula is Al2O3. An ionic compound is named using the name of the cation followed by the name of the anion, eliminating the word ion from each.
Explanation:
Warmer air is less dense than cold air.As air warm it rises while the cold air sink. Warmer air masses forces the cooler air to move which causes wind. These is illustrated when you open a hot oven the hotter air inside the oven rises into cooler air outside the oven.
Answer:
26.0 g/mol is the molar mass of the gas
Explanation:
We have to combine density data with the Ideal Gases Law equation to solve this:
P . V = n . R .T
Let's convert the pressure mmHg to atm by a rule of three:
760 mmHg ____ 1 atm
752 mmHg ____ (752 . 1)/760 = 0.989 atm
In density we know that 1 L, occupies 1.053 grams of gas, but we don't know the moles.
Moles = Mass / molar mass.
We can replace density data as this in the equation:
0.989 atm . 1L = (1.053 g / x ) . 0.082 L.atm/mol.K . 298K
(0.989 atm . 1L) / (0.082 L.atm/mol.K . 298K) = 1.053 g / x
0.0405 mol = 1.053 g / x
x = 1.053 g / 0.0405 mol = 26 g/mol
