Stoichiomety:
1 moles of C + 1 mol of O2 = 1 mol of CO2
multiply each # of moles times the atomic molar mass of the compund to find the relation is weights
Atomic or molar weights:
C: 12 g/mol
O2: 2 * 16 g/mol = 32 g/mol
CO2 = 12 g/mol + 2* 16 g/mol = 44 g/mol
Stoichiometry:
12 g of C react with 32 g of O2 to produce 44 g of CO2
Then 18 g of C will react with: 18 * 32/ 12 g of Oxygen = 48 g of Oxygen
And the result will be 12 g of C + 48 g of O2 = 60 g of CO2.
You cannot obtain 72 g of CO2 from 18 g of C.
May be they just pretended that you use the law of consrvation of mass and say that you need 72 g - 18g = 54 g. But it violates the proportion of C and O2 in the CO2 and is not possible.
Answer:
C₃H₄O₄
Explanation:
In order to get the empirical formula of a compound, we have to follow a series of steps.
Step 1: Divide the percent by mass of each element by its atomic mass.
C: 34.6/12.01 = 2.88
H: 3.9/1.01 = 3.86
O: 61.5/16.00 = 3.84
Step 2: Divide all the numbers by the smallest one, i.e., 2.88
C: 2.88/2.88 = 1
H: 3.86/2.88 ≈ 1.34
O: 3.84/2.88 ≈ 1.33
Step 3: Multiply all the numbers by a number that makes all of them integer
C: 1 × 3 = 3
H: 1.34 × 3 = 4
O: 1.33 × 3 = 4
The empirical formula is C₃H₄O₄.
Answer:
I think its A. It accepted radiation in a chemical reaction, hope this helped.
Au, N, O ( give me brainliest please)
