The answer is
<span>2PbS(s) + 3O2(g) = 2PbO(s) + 2SO2(g)
Your answer is not yet balanced because you have 3 oxygen atoms. it should be balanced by multiplying both side by 2 such as the balanced equation I made. To check it, I will explain why your answer is not yet balanced.
check: (from your equation)
</span> 1-Pb-1
1-S-1
2 -O-3
the difference between the reactant and the product of Oxygen will prove that it is not yet balanced.
If you use 2PbS(s) + 3O2(g) = 2PbO(s) + 2SO2(g), to check it:
2-Pb-2
2-S-2
6 -O-6
then this is now balance
Answer:
-177.9 kJ.
Explanation:
Use Hess's law. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2Ca(s) + O2(g) → 2CaO(s) ΔH = -1269.8 kJ We need to get rid of the Ca and O2 in the equations, so we need to change the equations so that they're on both sides so they "cancel" out, similar to a system of equations. I changed the second equation. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ The sign changes in the second equation above since the reaction changed direction. Next, we need to multiply the first equation by two in order to get the coefficients of the Ca and O2 to match those in the second equation. We also multiply the enthalpy of the first equation by 2. 2Ca(s) + 2CO2(g) + O2(g) → 2CaCO3(s) ΔH = -1625.6 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ Now we add the two equations. The O2 and 2Ca "cancel" since they're on opposite sides of the arrow. Think of it more mathematically. We add the two enthalpies and get 2CaO(s) + 2CO2(g) → 2CaCO3(s) and ΔH = -355.8 kJ. Finally divide by two to get the given equation: CaO(s) + CO2(g) → CaCO3(s) and ΔH = -177.9 kJ.
Answer:
Calcium for 2+ charge and Fluorine forms 1- charge
Explanation:
C. 4 decorated cupcakes.
Because 12 divided by 3 = 4, so you can have 4 cupcakes each with 3 cherries
<span>The answer is Cs, no. 2.
Cesium (Cs) is an element which has an outermost electron with most energy in
the ground state.</span>
Cesium (Cs), Lithium (Li), Potassium
(K), and Sodium (Na) are elements which belong to the group 1 family, the
alkali metals. Each has a valance of 1 and wants to release/ lose this e-
(electron) to be isoelectronic with the nearest noble gas. Based on the
periodic trends, an atom’s radii will raise going down a group.