Answer:
1) 950 mL
2) 625 mmHg
3) 426 mL
Explanation:
1) This is the relationship between pressure and volume. This relationship looks like this:
P1*V1 = P2*V2
This means the first pressure times the initial volume is equal to the second pressure times the second volume. We are solving for the second volume. First, convert the mmHg to atm and the mL to L.
1 L * 1 atm = 1.053 atm * X
X = 0.95 L or 950 mL
2) This is the same concept as the last one. :) We don't have to convert the mmHg to atm since the answer wants it in mmHg.
750 mmHg * 0.25 L = 0.3 L * X
X = 625 mmHg
3) The relationship between volume and temperature is similar to the one between pressure and temperature (like the problem in your last question). Remember to convert degrees C to Kelvin and mL to L.
V1 / T1 = V2 / T2
0.4 L / 303 K = X / 323 K
X = 0.426 L pr 426 mL
These problems become much easier once you learn the relationships between the different variables (temp, pressure, volume, etc.) When you have a problem like this, I like to first determine what relationship I am dealing with and then write out what I have and what I am solving for. This helps with organizing the problem. Then just solve it like a normal algebra problem. Always remember to convert temp to Kelvin, mL to L, and pressure to atm (unless it wants it in a different unit, then just make sure all the units match).
Good luck with you studies! :)
Answer:
C
Explanation:
b/c when copper and zinc metal are addedto solution,then the solution will be consider under redox reaction
Explanation: cell in the body is enclosed by a cell (Plasma) membrane. The cell membrane separates the material outside the cell, extracellular, from the material inside the cell, intracellular. ... All materials within a cell must have access to the cell membrane (the cell's boundary) for the needed exchange
**Answer**: The answer would be Yes I believe
Answer is: 127 grams <span>rams of metallic copper can be obtained.
</span>Balanced chemical reaction: 2Al + 3CuSO₄ → Al₂(SO₄)₃ + 3Cu.
m(Al) = 54.0 g.
n(Al) = m(Al) ÷ M(Al).
n(Al) = 54 g ÷ 27 g/mol.
n(Al) = 2 mol.
m(CuSO₄) = 319 g.
n(CuSO₄) = 319 g ÷ 159.6 g/mol.
n(CuSO₄) = 2 mol; limiting reactant.
From chemical reaction: n(CuSO₄) : n(Cu) = 3 : 3 (1 : 1).
n(Cu) = 2 mol.
n(Cu) = 2 mol · 63.55 g/mol.
n(Cu) = 127.1 g.
