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There are 4 lone pairs of electrons present in the carbon dioxide molecule
Answer:
Concentration of sodium carbonate in the solution before the addition of HCl is 0.004881 mol/L.
Explanation:

Molarity of HCl solution = 0.1174 M
Volume of HCl solution = 83.15 mL = 0.08315 L
Moles of HCl = n



According to reaction , 2 moles of HCl reacts with 1 mole of sodium carbonate.
Then 0.009762 mol of HCl will recat with:

Moles of Sodium carbonate = 0.004881 mol
Volume of the sodium carbonate containing solution taken = 1L
Concentration of sodium carbonate in the solution before the addition of HCl:
![[Na_2CO_3]=\frac{0.004881 mol}{1 L}=0.004881 mol/L](https://tex.z-dn.net/?f=%5BNa_2CO_3%5D%3D%5Cfrac%7B0.004881%20mol%7D%7B1%20L%7D%3D0.004881%20mol%2FL)
Answer:The volume of the remaining gas that is ammonia is 23.85 L.
Explanation:

Moles of 
Moles of HCl of gas = 
According to reaction 1 mole of HCl reacts with 1 mol of
then 2.06 moles of HCl will react with = 2.06 moles of
Moles left of ammonia left = 4.43 - 2.06 = 2.36 moles
Volume of the gas will be given by Ideal gas equation: PV=nRT
Pressure = 752 mmHg = 752 × 0.0031 atm = 2.33 atm
R = 0.08026 L atm/K mol
V = ? , n = number of moles of ammonia
Temperature = 14 °C = 14 + 273 K = 287 K(0°C = 273K)

The volume of the remaining gas that is ammonia is 23.85 L.
A=Mass number=24
N=neutrons=13
Z=atomic number.
A=Z+N
24=Z+13
Z=24-13
Z=11
The atomic number is 11, and this atom is sodium.