Answer:
The molarity of the solution is 0,31 M
Explanation:
We calculate the weight of 1 mol of NaCl from the atomic weights of each element of the periodic table. Then, we calculate the molarity, which is a concentration measure that indicates the moles of solute (in this case NaCl) in 1000ml of solution (1 liter)
Weight 1 mol NaCl= Weight Na + Weight Cl= 23 g + 35, 5 g= 58, 5 g
58, 5 g-----1 mol NaCl
13,1 g ---------x= (13,1 g x 1 mol NaCl)/58, 5 g= 0, 224 mol NaCl
727 ml solution------ 0, 224 mol NaCl
1000ml solution------x= (1000ml solutionx0, 224 mol NaCl)/727 ml solution
x=0,308 mol NaCl---> <em>The solution is 0,31 molar (0,31 M)</em>
First, we need to get moles of NaOH:
when moles NaOH = volume * molarity
= 0.02573L * 0.11 M
= 0.0028 moles
from the reaction equation:
H3PO4(aq) + 3NaOH → 3 H2O(l) + Na3PO4(aq)
we can see that when 1 mol H3PO4 reacts with→ 3 mol NaOH
∴ X mol H3PO4 reacts with → 0.0028 moles NaOH
∴ moles H3PO4 = 0.0028 mol / 3 = 9.4 x 10^-4 mol
now we can get the concentration of H3PO4:
∴[H3PO4] = moles H2PO4 / volume
= 9.4 x 10^-4 / 0.034 L
= 0.028 M
The answer is D, reactant.
A stable arrangement of eight valence electrons : ³⁵Cl⁻¹
<h3>Further explanation</h3>
Chlorine is a halogen gas, located in group 17, p block
Chlorine has an atomic number of 17 and an atomic mass of 35
Electron configuration: [Ne] 3s²3p⁵
If we look at the electron configuration, then Cl will bind 1 more electron so that the configuration is stable like Argon (atomic number 18)
So by binding this one electron, chlorine forms negative ions (anions)
³⁵Cl⁻¹
B. Cl⁻² binds 2 electrons, exceeding the octet rule
C. Cl⁺¹, releases 1 electron, remains unstable
D. Cl, the neutral form of Cl, is still unstable with a 7-electron valence configuration
