Answer:
Faster hydrogen burning means the star will grow larger and more luminous to allow the extra energy created to escape
Answer:
Tollen's test or silver mirror test is use to distinguish between ketone and aldehyde.
Explanation:
Tollen's test is use to distinguish aldehyde from ketone because Tollen reagent consist of silver nitrate and ammonia and it oxidized aldehydes to carboxylic acid and the silver Ag+ is reduced to solid silver Ag. It form a inner film wall on the tube that look like mirror. Ketone do not react with Tollen reagent and therefore does not form silver mirror on the tube wall.
Answer:
4.45 atm
Explanation:
Applying,
PV = P'V'............ Equation 1
Where P = Initial pressure of the container, V = Initial volume of the container, P' = Final pressure of the container, V' = Final volume of the container.
make P the subject of the equation
P = P'V'/V........... Equation 2
From the question,
Given: V = 55.2 L, P' = 8.53 atm, V' = 28.8 L
Substitute these values into equation 2
P = (8.53×28.8)/55.2
P = 4.45 atm
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.
Third quarter (or last quarter)
