Answer:
the empirical (lowest raios) is
C2H4Cl
Explanation:
A compound is known to consist solely of carbon, hydrogen, and chlorine. Through elemental analysis, it was determined that the compound is composed of 24.27% carbon.
What is the empirical formula of this compound?
the compound has ONLY C, H, and Cl
the % Cl = 100% - 24.27% -4.03% = 71.7%
in 100 gm, there are 71.7 gm Cl, 24.27 gm C, and 4.03 gm H
the number of moles are Cl=71.7/70.91 =1.01= ~ 1
C = 24.27/12.0 = 2.02 =~ 2
H = 403/1.01 = 3.97 =~ 4
so the empirical (lowest raios) is
C2H4Cl
<span>Answer:
Correct answers are- Electron affinity decreases; Cl has 7 valence electrons but Na has only 1. So Na is going to lose its e, Cl is going to gain an e.</span>
Answer:
Explanation:
Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.
Metal X can form 2 oxides (A and B).
A + B = 3g
The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.
The mass of metal X in the two oxides will be the same because it's the same metal.
Thus, we represent the mass of the metal in the two oxides as 2X.
2X + 0.72 + 1.16 = 3
2X + 1.88 = 3
2X = 3 - 1.88
2X = 1.12
X = 0.56
<u>Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.</u>
Thus, mass of metal (X) in 1g of oxygen in A is
0.56g ⇒ 0.72g
X ⇒ 1
X = 1 × 0.56/0.72
X = 0.78 g
Hence, 0.78g of the metal will combine with 1g of oxygen for A
Also, mass of metal (X) in 1g of oxygen in B is
0.56g ⇒ 1.16g
X ⇒ 1g
X = 1×0.56/1.16
X = 0.48 g
Thus, 0.48g of the metal will combine with 1g of oxygen for B
Answer:If each side of the equation has the same number of atoms of a given element, that element is balanced. If all elements are balanced, the equation is balanced. - online resource
Explanation: if not im sorry
Answer:
27%
Explanation:
Hello,
The following information is missing, but I found it: "1.92 g of sodium sulfate is produced from the reaction of 4.9 g of sulfuric acid and 7.8 g of sodium hydroxide" so the undergoing chemical reaction is:

Now, to compute the percent yield, we must first establish the limiting reagent to subsequently determine the theoretical yield of sodium sulfate because the real (1.92g) is already given, thus, we consider the following procedure:

- The moles of sodium hydroxide that completely react with 0.05 moles of sulfuric acid are:

As this number is higher than the previously computed 0.05 moles of available sulfuric acid, one states that the sulfuric acid is the limiting reagent. Now, the theoretical grams of sodium sulfate are found via:

Finally, the percent yield turns out into:

%
Best regards.