Answer:
true
because with the both states we increase the surface of reaction
To determine the moles in 40 grams of magnesium, we need the atomic weight. This can easily be found on a periodic table. For this problem, let's use 24.305 grams/mole.
We are going to set up an equation to determine this problem. In this equation, we want all our units to cancel out except for 'moles.'
In this, we can see that the unit 'grams' will cancel out to leave us with moles.
In solving the equation, we determine that there are approximately 1.65 moles of Magnesium.
Answer:
-800 kJ/mol
Explanation:
To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).
First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:
Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol
moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄
Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:
ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol
Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).
First, we use avogadro's number to convert atoms into moles. Then, relate the number of moles from elemental to the compound. Lastly, we use conditions at STP to calculate the volume. We do as follows:
<span>9.86 x 10²⁸ O-atoms ( 1 mol / 6.022x10^23 atoms O) ( 1 mol N2O2 / 2 mol O ) ( 22.4 L / 1 mol ) = 1833809.37 L needed</span>