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Answer:
Remain unchanged.
Explanation:
The total number of moles of liquid remain unchanged as the some moles of species B are added to the system because specie B that is added in the liquid phase is again restored after addition. If the specie B did not restored after addition to the liquid phase so the total number of moles increases in the liquid phase so that's why we can say that the liquid phase remain unchanged.
Questions:
The questions or computes to do are:
<span>a- a massa, em
kg, de cada placa de alumínio;
b- a quantidade mínima de viagens
necessárias para que apenas um veículo de transporte entregue o material
solicitado ao cliente.
Dado: densidade do alumínio = 2,7 g/cm3
Answer:
a) mass in kg of every aluminum plate
Dimensions of every aluminum plate: </span>
<span>2 M X 50 Cm X 2cm
Volume: 200 cm * 50 cm * 2 cm = 20,000 cm^3
Mass:
density = mass / volume => mass = density * volume = 2.7 g/cm^3 * 20,000 cm^3 = 54,000 g = 54 kg.
Answer: the mass of everyplate of aluminum is 54 kg.
b) number of travels required for one truck deliver all the material:
number of travels = amount requested / amount that a truck can deliver in one travel.
amount requested: 100 plates
mass of 100 plates = 100 plates * 54 kg / plate = 5,400 kg
limit of transport per travel: 3 tons = 3,000 kg
number of travels = 5,400 kg / 3,000 kg/travel = 1.8 travels => 2 travels.
Answer: at least 2 travels.
</span>
This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:
Substituting in our values, we get
![\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].](https://tex.z-dn.net/?f=%5C%5BM_2%3D%5Cfrac%7B%5Cleft%20%28%2050%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%5Cleft%20%28%200.235%20%5Ctext%7B%20M%7D%20%5Cright%20%29%7D%7B%5Cleft%20%28%20200.0%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%7D%3D%200.05875%20%5Ctext%7B%20M%7D%5C%5D.)
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.
