Answer:
18.06 × 10²³ molecules
Explanation:
Add the two amounts of molecules together.
6.02 × 10²³ + 12.04 × 10²³ = 18.06 × 10²³
You will have 18.06 × 10²³ molecules in the vessel when the reaction is complete. This is because of the Law of Conservation of Mass. Mass is neither created nor destroyed in chemical reactions. You will have the exact number of molecules in the reaction vessel as you did in the beginning. The types of molecules may change, but the number will stay the same.
Answer:
4
Explanation:
1 is correct
Liquids have no definite shape as they take up the shape of the container. Thus, we can say a liquid has no shape of its own but rather has the shape of the container in which it is filled.
2 is correct
When the atmospheric temperature is increased, it also will increase the boiling point of the liquid
3 is correct
This is an extension of the statement 2. While we decrease the atmospheric pressure, we are also decreasing the boiling point
4 is incorrect
A liquid have a definite volume. When we say a volume is definite, it means the volume is fixed and does not change. The volume of liquids is definite for a particular mass of the liquid and does not change
Answer:
Q1: 9
Q2:10
Q3: 50
Q4:True
Q5: False
Q6:False
Q7:True
Explanation: Just took the quiz
Answer:
I guess 4 weeks is a little late.
Explanation:
Answer:
Final concentration of C at the end of the interval of 3s if its initial concentration was 3.0 M, is 3.06 M and if the initial concentration was 3.960 M, the concentration at the end of the interval is 4.02 M
Explanation:
4A + 3B ------> C + 2D
In the 3s interval, the rate of change of the reactant A is given as -0.08 M/s
The amount of A that has reacted at the end of 3 seconds will be
0.08 × 3 = 0.24 M
Assuming the volume of reacting vessel is constant, we can use number of moles and concentration in mol/L interchangeably in the stoichiometric balance.
From the chemical reaction,
4 moles of A gives 1 mole of C
0.24 M of reacted A will form (0.24 × 1)/4 M of C
Amount of C formed at the end of the 3s interval = 0.06 M
If the initial concentration of C was 3 M, the new concentration of C would be (3 + 0.06) = 3.06 M.
If the initial concentration of C was 3.96 M, the new concentration of C would be (3.96 + 0.06) = 4.02 M