Answer:
This is a precipitation reaction: AgCl is the formed precipitate.
<u>Answer:</u> The fraction of atom's mass contributed by nucleus is 0.99
<u>Explanation:</u>
Nucleons are defined as the sub-atomic particles which are present in the nucleus of an atom. Nucleons are protons and neutrons.
The isotopic symbol of Helium-4 atom is 
Number of electrons = 2
Number of protons = 2
Number of neutrons = 4 - 2 = 2
We are given:
Mass of He-4 atom = 
Mass of 1 electron = 
Calculating the mass contributed by the nucleus = 
Mass of the nucleus of He-4 atom = 
To calculate the fraction of atom's mass contributed by the nucleus, we use the equation:
Putting values in above equation, we get:
Hence, the fraction of atom's mass contributed by nucleus is 0.99
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
Answer:
Explanation:
When you divide exponentials, you subtract the powers. For the numbers infront, just use a basic calculator for.
7.95/6.02 = 1.32
10^22/10^23 = 10^-1
1.32 x 10^-1 is your answer
