Answer:
I'm assuming you mean how many atoms total and not mass or anything, so it seems there are 27 atoms altogether. Unless you meant to capitalize the o in (Po4) then it would be 28 atoms.
Answer:
Answer below
Explanation:
First to show combustion, you use O₂.
So C₆H₁₀O₅ + O₂ is the reaction. Assuming complete combustion, all you get as products are CO₂ and H₂O. Then you have to balance the full reaction.
C₆H₁₀O₅ + 6O₂ -----> 6CO₂ + 5H₂O
6 Carbons on each side
10 Hydrogens on each side
17 Oxygens on each side
Interesting problem. Thanks for posting.
C2H2 + (3/2)02 ====> H2O + 2CO2
CH4 + 2O2 =====> 2H2O + CO2
The molar mass of C2H2 = 2*12 + 2*1 = 26
The molar mass of CH4 = 1*12 + 4*1 = 16
The number of moles of C2H2 = x
The number of moles of CH4 = y
26x + 16y = 230.9 grams
For water we get (from the C2H2). Water has a molar mass of 2*1 + 16 = 18
x*18 See the balanced equation to see what it is the same number of moles as C2H2
From the methane we get
y*18
2*y* 18. Again see the balanced equation to see where that 2 came from.
18x + 36y is the total amount of water.
Now for the CO2. CO2 has a molar mass of 12 + 2*16 = 44
From C2H2 we get 2*44*x = 88x grams of CO2
From CH4 we get 1*y*44 grams of CO2
88x + 44y for CO2
Now we total to get the grand total of water and CO2
18x + 44y + 88x + 44y = 972.7 grams total.
106x + 88y = 972.7
Two equations, two unknowns, we should be able to solve this problem
26x + 16y = 230.9
106x + 88y = 972.7
I'm not going to go through the math unless you request me to do so.
x = 8.03 moles
y = 1.38 moles
The initial amount of C2H2 was 8.03 * 26 = 208.78
The initial amount of CH4 was 16*1.38 = 22.08
The total (as a check is 230.86 which is pretty close to the given amount.
So Methane's mass in the initial givens was 22.08 grams.
Answer:
See explanation.
Explanation:
Hello,
In this case, we could have two possible solutions:
A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:
That is the mass of copper (II) sulfate contained in 1 mol of substance.
B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:
So you can solve for the moles of the solute:
Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:
But this is just a supposition.
Regards.
Answer:
2HI + K2SO3=>2KI+H2SO3
Explanation:When aqueous hydroiodic acid and aqueous potassium sulfite are mixed the products obtained are potassium iodide and sulfurous acid.Both reactants are ionic compounds and they undergo double replacement reaction.In a double replacement reaction the parts of the ionic compounds are changed.The product is obtained by combinig cation of one compound with anion of other compound.so in above reaction sulfurous acid is obtained which is in gaseous form and potassium iodide is an ionic compound.
