Answer:
Explanation:
We are given the amounts of two reactants, so this is a limiting reactant problem.
1. Assemble all the data in one place, with molar masses above the formulas and other information below them.
Mᵣ: 58.44
NaCl + AgNO₃ ⟶ NaNO₃ + AgCl
m/g: 0.245
V/mL: 50.
c/mmol·mL⁻¹: 0.0180
2. Calculate the moles of each reactant
3. Identify the limiting reactant
Calculate the moles of AgCl we can obtain from each reactant.
From NaCl:
The molar ratio of NaCl to AgCl is 1:1.
From AgNO₃:
The molar ratio of AgNO₃ to AgCl is 1:1.
AgNO₃ is the limiting reactant because it gives the smaller amount of AgCl.
4. Calculate the moles of excess reactant
Ag⁺(aq) + Cl⁻(aq) ⟶ AgCl(s)
I/mmol: 0.900 4.192 0
C/mmol: -0.900 -0.900 +0.900
E/mmol: 0 3.292 0.900
So, we end up with 50. mL of a solution containing 3.292 mmol of Cl⁻.
5. Calculate the concentration of Cl⁻
![\text{[Cl$^{-}$] } = \dfrac{\text{3.292 mmol}}{\text{50. mL}} = \textbf{0.066 mol/L}\\\text{The concentration of chloride ion is $\large \boxed{\textbf{0.066 mol/L}}$}](https://tex.z-dn.net/?f=%5Ctext%7B%5BCl%24%5E%7B-%7D%24%5D%20%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B3.292%20mmol%7D%7D%7B%5Ctext%7B50.%20mL%7D%7D%20%3D%20%5Ctextbf%7B0.066%20mol%2FL%7D%5C%5C%5Ctext%7BThe%20concentration%20of%20chloride%20ion%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.066%20mol%2FL%7D%7D%24%7D)
Answer:
3.55atm
Explanation:
We will apply Boyle's law formula in solving this problem.
P1V1 = P2V2
And with values given in the question
P1=initial pressure of gas = 1.75atm
V1=initial volume of gas =7.5L
P2=final pressure of gas inside new piston in atm
V2=final volume of gas = 3.7L
We need to find the final pressure
From the equation, P1V1 = P2V2,
We make P2 subject
P2 = (P1V1) / V2
P2 = (1.75×7.5)/3.7
P2=3.55atm
Therefore, the new pressure inside the piston is 3.55atm
<span>a) The strong nuclear force
</span>
Answer:
Ionic
Explanation:
Lithium is an alkali metal and form an ionic bond by donating an electron.
Fluorine is a halogen and forms ionic bonds by accepting an electron.
The electronegativity difference between lithium (0.98) and fluorine (3.98) is 3 Pauling units. Usually, when elements have a difference of 1.7 (or 2.0) Pauling units the bond is classified as ionic.
Hope this helps! If you have any more questions or need more explanation please leave a comment down below and I will reply ASAP. Good luck!
