Answer:
<em> ionic equation : </em>3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)
<em> net ionic equation: </em>3Fe(2+)(aq) + 2PO4 (3-)(aq) → Fe3(PO4)2(s)
Explanation:
The balanced equation is
3FeSO4(aq)+ 2Na3PO4(aq) → Fe3(PO4)2(s)+ 3Na2SO4(aq)
<em>Ionic equations: </em>Start with a balanced molecular equation. Break all soluble strong electrolytes (compounds with (aq) beside them) into their ions
. Indicate the correct formula and charge of each ion. Indicate the correct number of each ion
. Write (aq) after each ion
.Bring down all compounds with (s), (l), or (g) unchanged. The coefficents are given by the number of moles in the original equation
3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)
<em>Net ionic equations: </em>Write the balanced molecular equation. Write the balanced complete ionic equation. Cross out the spectator ions, it means the repeated ions that are present. Write the "leftovers" as the net ionic equation.
3Fe(2+)(aq) + 2PO4 (3-)(aq) → Fe3(PO4)2(s)
When baking powder is added to a cake, the acid base reaction it undergoes with the acidic butter produces the carbon dioxide that makes the cake rise.
The answer is: A molecule with a difference in electrical charge between two ends.
Electronegativity (χ) is a property that describes the tendency of an atom to attract a shared pair of electrons.
Atoms with higher electronegativity attracts more electrons towards it, electrons are closer to that atom.
For example fluorine has electronegativity approximately χ = 4 and oxygen χ = 3,5, fluorine attracts electron and he has negative charge and oxygen has positive charge.
Answer:
K = 10
Explanation:
Using Hess's law, it is possible to obtain the equilibrium constant, K, of a reaction using K of similar reactions. For example:
<em> If A ⇄ B K = X</em>
B ⇄ A K = 1/X
2A ⇄ 2B K = X².
Thus, if A(g) ⇄ 2B(g) K = 0.010
2B(g) ⇄ A(g) K = 1 / 0.010; K = 100
B(g) ⇄ A(g) K = √100 = 10
<h3>K = 10</h3>
Answer:
Explanation:
Hello!
In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:
We plug in the mass of water, temperature change and specific heat to obtain:
Now, this enthalpy of reaction corresponds to the combustion of propyne:
Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:
However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:
Now, we solve for the enthalpy of formation of C3H4 as shown below:
So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):
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