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:
did you mean to add or attach a paper to this? We need more info to help
Explanation:
1 mole ------------ 6.02x10²³ atoms
5.88 moles ---- ?
atoms = 5.88 * 6.02x10²³ / 1
= 3.539x10²⁴ atoms
hope this helps!
1. P = F/A; weight is a force (the force of gravity on an object), so divide the weight by the area given. P = 768 pounds/75.0 in² = 10.2 pounds/in².
2. Using the same equation from question 1, rearrange it to solve for A: A = F/P. We're given the force (the weight) and the pressure, so A = 125 pounds/3.25 pounds/in² = 38.5 in².
3. Again, using the same equation from question 1, rearrange it this time to solve for F: F = PA = (4.33 pounds/in²)(35.6 in²) = 154 pounds.
4. We can set up a proportion given that 14.7 PSI = 101 KPa. This ratio should hold for 23.6 PSI. In other words, 14.7/101 = 23.6/x; to solve for x, which would be your answer, we compute 23.6 PSI × 101 kPa ÷ 14.7 PSI = 162 kPa.
5. We are told that 1.00 atm = 760. mmHg, and we want to know how many atm are equal to 854 mmHg. As we did with question 4, we set up a proportion: 1/760. = x/854, and solve for x. 854 mmHg × 1.00 atm ÷ 760. mmHg = 1.12 atm.
6. The total pressure of the three gases in this container is just the sum of the partial pressures of each individual gas. Since our answer must be given in PSI, we should convert all our partial pressures that are not given in PSI into PSI for the sake of convenience. Fortunately, we only need to do that for one of the gases: oxygen, whose partial pressure is given as 324 mmHg. Given that 14.7 PSI = 760. mmHg, we can set up a proportion to find the partial pressure of oxygen gas in PSI: 14.7/760. = x/324; solving for x gives us 6.27 PSI oxygen. Now, we add up the partial pressures of all the gases: 11.2 PSI nitrogen + 6.27 PSI oxygen + 4.27 PSI carbon dioxide = 21.7 PSI, which is our total pressure.
