Answer:
0.107 mole of SO2.
Explanation:
1 mole of a gas occupy 22.4 L at standard temperature and pressure (STP).
With the above information, we can simply calculate the number of mole of SO2 that will occupy 2.4 L at STP.
This can be obtained as follow:
22.4 L contains 1 mole of SO2.
Therefore, 2.4 L will contain = 2.4/22.4 = 0.107 mole of SO2.
Therefore, 0.107 mole of SO2 is present in 2.4 L at STP.
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.
A 10.0g
ok hope this helps
Answer:
See answer below
Explanation:
The picture below will show you the final product and mechanism.
In the first step, the NaNH₂ is a strong base, so, this base will substract the hydrogen from carbon 2, to generate a negative charge there, and then, carbon 2 becomes a nucleophyle.
As a nucleophyle it will attack to the CH₃I in the next step, and it will attach to the CH₃.
The second step is just a regular step to reduce the triple bond of the alkyne to alkane or alkene, this will depend on the quantity of the reactant. In this case, an alkene.
Hope this helps,,,,,,k
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