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)
A. A piece of Iron being heated until it melts
Because it goes from solid to liquid.
A physical change is changing from solid to a liquid, a liquid to a gas, a gas to a liquid, a liquid to a solid, a solid to a gas, or a gas to a solid.
<u>Answer:</u> The mass of water produced in the reaction is 97.2 grams
<u>Explanation:</u>
We are given:
Moles of calcium hydroxide = 2.70 moles
The chemical equation for the reaction of calcium hydroxide and HCl follows:
By Stoichiometry of the reaction:
1 mole of calcium hydroxide produces 2 moles of water
So, 2.70 moles of calcium hydroxide will produce = 
of HCl
To calculate mass for given number of moles, we use the equation:
Molar mass of water = 18 g/mol
Moles of water = 5.40 moles
Putting values in above equation, we get:
Hence, the mass of water produced in the reaction is 97.2 grams
Answer:
1. <u>No, you cannot calculate the solubility of X in water at 26ºC.</u>
Explanation:
You cannot calculate the solubility of X in <em>water at 26 degrees Celsius </em>because you do not know whether the solution formed by dissolving the crystals in 3.00 liters of water is saturaed or not.
The only way to determine the solubility of the compound X is by dissolving the crystals in certain (measured) amount of water and making sure that some crystals remain undissolved, as a solid on the bottom of the beaker.
Next, you should filter the solution to remove the undissolved crystals. Then, weigh the solution, evaporate, wash, dry, and weigh the crystals.
Then you have the mass of the crystals dissolved and the mass of the solution which will let you calculate the mass of pure water, and then the solubility.
Answer:
C
Explanation:
This experiment by Rutherford involved the firing of alpha particles at gold foils. It is also. called the gold foil experiment.
He fired these alpha particles at different points. He noticed that at some points, there were deflections, while at some other points, there were no deflections. It is necessary to state that these alpha particles are positively charged. For there to be a deflection, there must have been a kind of repulsion between the gold foil and the alpha particles.
From the basic physics of like repels like, he knew for sure that there must be dense positive core in the atom that is causing the deflection of the alpha particles. This enabled him to come up with the theory that the atom contained a small dense positive core called the nucleus
