Answer:
35.68g CO2
Explanation:
we use the combustion equation with CH4:
CH4+ O2= CO2 + H2O
And then balance it:
CH4+ 2O2= CO2 + 2H2O
Using this equation we can use sociometry:
We know that 16.032 is how many grams there are in one mole of CH4 by adding the weights of the atoms (12 +1.008+1.008+1.008+1.008). These weights can be found on the periodic table. The same goes for the amount of grams per CO2.
The important thing about sociometry is to make sure your units cancel out until you are only left with the unit you want. If grams of CH4 is in the numerator, the next fraction you multiply by should have grams of Ch4 in the denominator. If moles of CO2 are in the numerator, the next fraction should have moles of CO2 in the denominator.
Answer:
Molar mass→ 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
Explanation:
Let's apply the formula for freezing point depression:
ΔT = Kf . m
ΔT = 74.2°C - 73.4°C → 0.8°C
Difference between the freezing T° of pure solvent and freezing T° of solution
Kf = Cryoscopic constant → 5.5°C/m
So, if we replace in the formula
ΔT = Kf . m → ΔT / Kf = m
0.8°C / 5.5 m/°C = m → 0.0516 mol/kg
These are the moles in 1 kg of solvent so let's find out the moles in our mass of solvent which is 0.125 kg
0.0516 mol/kg . 0.125 kg = 6.45×10⁻³ moles. Now we can determine the molar mass:
Molar mass (mol/kg) → 0.930 g / 6.45×10⁻³ mol = 144.15 g/mol
Answer:
8 moles
Explanation:
When we are asked to convert from grams of a substance into moles, we have to use the substance's molar mass.
Meaning that for this problem, we'll <em>use the molar mass of hydrogen peroxide</em> (H₂O₂), as follows:
There are 8 moles in 272 grams of hydrogen peroxide.
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Answer: 100 times
Explanation: Since logaritms are about exponent of base ten.The concentration will be 10^2 or 100 times greater concentration.
