Answer:- 3. 
and 
Explanations:- An empirical formula is the simplest whole number ratio of atoms of each element present in the molecule/compound.
For example, the molecular formula of benzene is 
. The ratio of C to H in it is 6:6 that could be simplified to 1:1. So, an empirical formula of benzene is CH.
In the first pair, the ratio of C to H in first molecule is 2:4 that could be simplified to 1:2 and the empirical formula is 
. In second molecule the ratio of C to H is 6:6 and it could be simplified to 1:1. and the empirical formula is CH. Empirical formulas are different for both the molecules of first pair and so it is not the right choice.
In second pair, C to H ratio in first molecule is 1:2, so the empirical formula is . The C to H ratio for second molecule is 1:4, so the empirical formula is 
. Here also, the empirical formulas are not same and hence it is also not the right choice.
In third pair, C to H ratio in first molecule is 1:3, so the empirical formula is . In second molecule the C to H ratio is 2:6 and it is simplified to 1:3. So, the empirical formula for this one is also . Hence. this is the correct choice.
In fourth pair, first molecule empirical formula is CH. Second molecule has 2:4 that is 1:2 mole ratio of C to H and so its empirical formula is . As the empirical formulas are different, it is not the right choice.
So, the only and only correct pair is the third one. 3. and
Answer:
moles of carbon dioxide produced are 410.9 mol.
Explanation:
Given data:
Mass of C₆H₁₄O₂ = 16.5 g
Moles of O₂ = 499 mol
Moles of CO₂ = ?
First of all we will write the balance chemical equation.
2C₆H₁₄O₂ + 17O₂ → 14CO₂ + 12H₂O
moles of C₆H₁₄O₂ = mass × molar mass
moles of C₆H₁₄O₂ = 16.5 g × 118 g/mol
moles of C₆H₁₄O₂ = 1947 mol
Now we compare the moles of CO₂ with moles of O₂ and C₆H₁₄O₂ from balance chemical equation.
O₂ : CO₂
17 : 14
499 : 14/17× 499 = 410.9 moles
C₆H₁₄O₂ : CO₂
2 : 14
1947 : 14/2× 1947 = 13629 moles
Oxygen will be limiting reactant so moles of carbon dioxide produced are 410.9 mol.
1 kg = 1000g
2.43 kg *1000g/1kg = 2430 g
Answer:
Y = 62.5%
Explanation:
Hello there!
In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:
Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:
Best regards!
For the first one, you have to find one that has both a metal and a nonmetal in it, plus potassium. Bonds between nonmetals are called covalent because they share electrons and a bond between a metal and a nonmetal is an ionic bond because they exchange electrons.
Potassium chloride is KCl. Only an ionic bond.
Potassium hydride is KH. Only an ionic bond.
Potassium nitrate is KNO3. There we go, that has a covalent bond between the nitrogen and oxygen, and an ionic bond between potassium and the nitrogen and oxygen.
For the second one, potassium chloride is the answer because the other three also have covalent bonds. Chloride is the only one that isn't a compound.
