The CaCO3 produced if 47.5 moles of NH3 produced is calculated as follows
CaCN2 +3H2O = CaCO3 + 2NH3
by use of mole ratio between CaCO3 to NH3 which is 1:2 the moles of CaCO3 is therefore = 47.5 /2= 23.75 moles
mass of CaCO3 is therefore = moles x molar mass
= 23.75 moles x 100g/mol= 2375 grams which is approximate 2380 grams(answer 6)
Answer:
6.2moles of Gold
Explanation:
To solve this problem, we are going to use the mole concept approach.
Given that;
Number of atoms of gold is 3.73 x 10²⁴ atoms
Now;
In 1 mole of any substance, we have 6.02 x 10²³ atoms;
So;
If there 6.02 x 10²³ atoms in 1 mole of any substance;
3.73 x 10²⁴ atoms will contain 
= 6.2moles of Gold
Answer:
Explanation:
The possible energy states for the particles are 0, 10 and 20 J.
The constraint in the system is that the total energy of the particles must be 20 J.
One given configuration where the total energy is 20 J is if both the particles occupy the 10 J state.
Hence, (10;10) is the given configuration.
Another possibility is if one of the particle is in 0 J state and another is in 20 J state. Hence, the system has a total energy of 0+20 = 20 J.
Hence, the possible configuration can be written as (0;20) or (20;0) which are energetically equivalent to the given configuration. Note that if the circles are indistinguishable, then the configuration (0,20) and (20,0) is the same thing.
