Answer: 14.3%Explanation: In order to find the mass percent of hydrogen in this compound, you must determine how many grams of hydrogen you'd get in 100 g of compound.
In your case, you know that an unknown mass of hydrogen reacts with 0.771 g of carbon to form 0.90 g of hydrocarbon, which is a compound that contains only carbon and hydrogen.
Use the total mass of the hydrocarbon to determine how many grams of hydrogen reacted with the carbon.
Now, if 0.90 g of this compound contain 0.129 g of hydrogen, it follows that 100 g of this compound will contain.
So, if 100 g of this compound contain 14.33 g of hydrogen, it follows that the mass percent of hydrogen is 14.3%
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Answer:
work = 3.45 Joules
Explanation:
Work from the 1st Law of Thermodynamics is PΔV in the equation of internal energy => ΔE = ΔH + PΔV.
Therefore, work = PΔV = 2.07atm·(2.334L - 0.666L) = 3.45 L·atm = 3.45 Joules
Answer:
It uses sodium nitrate, so that the nitrate group reacts with silver and precipitates as a salt, thus forming silver nitrate
And as for the precipitation of barium, I would use sodium sulfate, since we would use the nitrate group, barium nitrate would form, which is very dangerous and is considered high risk.
Explanation:
All the compounds that we name are reactions where as a product they give salts that are highly complex, where they are solid at room temperature, precipitate and are whitish in color.
These salts so that they precipitate it is important that they are in supersaturation since otherwise they will be in constant dynamism dissolving with the water, since they are soluble in them and the decantation will not be perceived at the naked eye of the human eye.
Answer:
moles Pb(NO₃)₂ ≅ 0.0213 mole (three sig. figs.)
Explanation:
converting...
grams to moles => divide by formula weight
moles to grams => multiply by formula weight
for this problem ...
moles Pb(NO₃)₂ = 7.04g / 331.2g/mol = 0.021256 (calculator answer)
≅ 0.0213 mole (three sig. figs.)
The energy released by 1 atom of Uranium is Q=185.*10^6-25.
The energy needed for U.S. that I will name Q2, in eV is: 1*10^19J* (1eV/(1.602*10^-19))=6.2422*10^37eV
Q=mc^2
Q2=mc^2
m is the mass of Uranium in kg that we calculated at 3.90*10^-25. We need to fund m2 because that is the unknown mass in the equation for Q2 to give us the energy for the U.S. and remember, Q is the energy that 1 atom of Uranium gives out which is why we took a ratio.
m2=Q2/Q*m=1.3*10^5 kg (final answer)
