Answer:
Number of moles = 0.12 mol
Explanation:
Given data:
Mass of carbonate = 6.5 g
Moles of carbonate = ?
Solution:
Number of moles = mass / molar mass
Molar mass of carbonate = 60 g/mol
Now we will put the values in formula:
Number of moles = 6.5 g/ 60 g/mol
Number of moles = 0.12 mol
B is correct....................
Writing the Configuration
Starting at the top of the periodic table and moving from left to right across the rows, write the row number, the block letter and how many squares are in each block's section until you get to the desired element. To calculate the electron configuration for phosphorus (P), which is in the third row, p-block, third element in that block, write: 1s2 2s2 2p6 3s2 3p3. Check your work by adding the electron numbers to see if they equal the atomic number of the element; for this example, you would write: 2+2+6+2+3=15, which is the atomic number of phosphorus.
Special Instructions
Because an atom's orbitals sometimes overlap in real life, you must account for that in your configurations. For the d-block, change the row number to one number less than it actually is. For example, the electron configuration for germanium (Ge) is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p2. Notice even though you are on row four, it is called "3d" to account for overlap.
Hope i helped you :) Brainliest?
Answer:
See the answer and the resolution below
Explanation:
The equation of the decomposition of magnesium carbonate is as follows:
Mg C03 (white solid) ---> C02 (gas) + Mg= (solid)
To calculate the mass of carbon dioxide produced, the weight of 1 mol of this and 1 mol of MgC03 is calculated. Then a simple rule of 3 is made.
Weight 1 mol of MgC03= Weight Mg + Weight C+ 3x(Weight 0)= 24,305g+ 12,017+3x (15,999)= 84, 318g/mol
Weight 1 mol of C02= Weight C + 2x(Weight 0)=12,017+2x (15,999)= 44, 008g/mol
84, 319 g MgC03------44,008g C02
36,0 g MgC03------X= (36,0 g MgC03x44,008g C02)/84, 319 g MgC03= 18,8 g C02
Answer:
b. Add NO gas
Explanation:
According with Le Chatelier´s principle, when a reactant is added the equilibrium is displaced to the product(s), to achieve the equilibrium again. Therefore, if NO is added in the reaction, it will be produced more of NO2(g).
