Answer:
A, B, and C
Explanation:
Atoms do in fact have mass! They are also the basic unit of a chemical element. Matter is in fact made of atoms, but atoms do not take up space.
Atoms are filled with empty space, excluding it out from the rest of the answers.
So yes, A, B, and C should be the correct answers
Answer: 3 moles solute x 1 dm^3/0.60 moles solute = 5 dm^3
Explanation:
Answer:
24e⁻ are transferred by the reaction of respiration.
Explanation:
C₆H₁₂O₆ + 6O₂ → 6 H₂O + 6CO₂
This is the reaction for the respiration process.
In this redox, oxygen acts with 0 in the oxidation state on the reactant side, and -2 in the product side - REDUCTION
Carbon acts with 0 in the glucose (cause it is neutral), on the reactant side and it has +4, on the product side - OXIDATION
6C → 6C⁴⁺ + 24e⁻
In reactant side we have a neutral carbon, so as in the product side we have a carbon with +4, it had to lose 4e⁻ to get oxidized, but we have 6 carbons, so finally carbon has lost 24 e⁻
6O⁻² + 6O₂ + 24e⁻ → 6O₂²⁻ + 6O⁻²
In reactant side, we have 6 oxygen from the glucose (oxidation state of -2) and the diatomic molecule, with no charge (ground state), so in the product side, we have the oxygen from the dioxide with -2 and the oxygen from the water, also with -2 at the oxidation state. Finally the global charge for the product side is -36, and in reactant side is -12, so it has to win 24 e⁻ (those that were released by the C) to be reduced.
Answer:- Volume of the gas in the flask after the reaction is 156.0 L.
Solution:- The balanced equation for the combustion of ethane is:
From the balanced equation, ethane and oxygen react in 2:7 mol ratio or 2:7 volume ratio as we are assuming ideal behavior.
Let's see if any one of them is limiting by calculating the required volume of one for the other. Let's say we calculate required volume of oxygen for given 36.0 L of ethane as:
= 126 L 
126 L of oxygen are required to react completely with 36.0 L of ethane but only 105.0 L of oxygen are available, It means oxygen is limiting reactant.
let's calculate the volumes of each product gas formed for 105.0 L of oxygen as:
= 60.0 L 
Similarly, let's calculate the volume of water vapors formed:
= 90.0 L 
Since ethane is present in excess, the remaining volume of it would also be present in the flask.
Let's first calculate how many liters of it were used to react with 105.0 L of oxygen and then subtract them from given volume of ethane to know it's remaining volume:
= 30.0 L 
Excess volume of ethane = 36.0 L - 30.0 L = 6.0 L
Total volume of gas in the flask after reaction = 6.0 L + 60.0 L + 90.0 L = 156.0 L
Hence. the answer is 156.0 L.
<span>Avogadro's number
represents the number of units in one mole of any substance. This has the value
of 6.022 x 10^23 units / mole. This number can be used to convert the number of
atoms or molecules into number of moles. We calculate as follows:
</span>1.40x10^23 molecules of N2 ( 1 mol / 6.022 x 10^23 molecules ) ( 28.02 g / mol ) = 6.51 g N2
