All are the same. It equals to the same thing.
Information would you need to know about the H₂O₂ solution is through stoichiometry experiment
The ideal gas constant R can be found experimentally by determining the number of moles of gas that occupies a particular measured volume at a known pressure and temprature and the H₂O₂ is a chemical compound used un various chemical reactions and is slightly viscous than water and the experiment by decomposition of hydrogen peroxide and using the ideal gas law rearrangement equation we can calculate the value of r and we will need the information such as concentration, volume and moles of H₂O₂ to determine its stoichiometry
Know more about ideal gas constant
brainly.com/question/20372603
#SPJ4
Answer:
0.57 moles (NH4)3PO4 (2 sig. figs.)
Explanation:
To quote, J.R.
"Note: liquid ammonia (NH3) is actually aqueous ammonium hydroxide (NH4OH) because NH3 + H2O -> NH4OH.
H3PO4(aq) + 3NH4OH(aq) ==> (NH4)3PO4 + 3H2O
Assuming that H3PO4 is not limiting, i.e. it is present in excess
1.7 mol NH4OH x 1 mole (NH4)3PO4/3 moles NH4OH = 0.567 moles = 0.57 moles (NH4)3PO4 (2 sig. figs.)"
Answer:
13.44dm^3
Explanation:
To calculate this we first need to know the number of mole produced. We will first need to balance the equation to know the theoretical mole ratio.
C2H6 (g) + 3.5O2 (g) → 2CO2 (g) + 3H2O (g)
From the balanced equation, we can deduce that one mole of ethane yielded 2 moles of carbon iv oxide. We use this information to calculate the actual number of moles yielded.
24g were reacted. Now to know the number of moles reacted, we simply divide the mass by the molar mass. The molar mass of ethane is 2(12) +6(1)= 40g/mol
The number of moles is thus 24/40 = 0.6 moles
Like we said earlier, one mole yielded 2 moles of carbon iv oxide, hence, 0.6 moles will yield 0.6 * 2 = 0.12 moles of carbon iv oxide.
Now, at stp, one mole of a gas occupies a volume of 22.4dm^3 thus, 0.6 mole will occupy 0.6 * 22.4 = 13.44dm^3
