If excess carbon disulfide reacts with 450 mL of oxygen, 150 mL of carbon dioxide and 300 mL of sulfur dioxide gases would be produced respectively.
<h3>Stoichiometric calculation</h3>
The reaction between liquid carbon disulfide and oxygen is represented by the equations below:
The mole ratio of oxygen to carbon dioxide and sulfur dioxide produced is 3:1:2.
Thus, for 450 mL oxygen, 1/3 x 450 = 150 mL of carbon dioxide will be required.
Also for 450 mL of oxygen, 2/3 x 450 = 300 mL of sulfur dioxide will be required.
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Answer: Potassium to fluorine, fluorine to nitrogen, bromine to hydrogen, carbon to hydrogen, lithium to chlorine, sodium to chlorine.
Explanation:
Ionic bond is greater when the electronegativity difference existing between the two atoms are large causing the bonding to be more polar.
Answer:
My Neighbor Totoro
Princess Mononoke
Pom Poko
Only Yesterday
Kiki's Delivery Service
Nausicaä of the Valley of the Wind
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).
<u>Answer:</u> The amount of sample left after 20 years is 288.522 g and after 50 years is 144.26 g
<u>Explanation:</u>
We are given a function that calculates the amount of sample remaining after 't' years, which is:
Putting values in above equation:
Hence, the amount of sample left after 20 years is 288.522 g
Putting values in above equation:
Hence, the amount of sample left after 50 years is 144.26 g
