Answer: Iron
Explanation: There are little minnerals that make it magnetic.
The answer is: Dividing the number of molecules in the sample by Avogadro's number.
The Avogadro’s number is the number of atoms in 12 grams of the isotope carbon-12 (¹²C).
Na is Avogadro number or Avogadro constant (the number of particles, in this example carbon, that are contained in the amount of substance given by one mole).
The Avogadro number has value 6.022·10²³ 1/mol in the International System of Units; Na = 6.022·10²³ 1/mol.
For example:
N(Ba) = 2.62·10²³; number of atoms of barium.
n(Ba) = N(Ba) ÷ Na.
n(Ba) = 1.3·10²⁴ ÷ 6.022·10²³ 1/mol.
n(Ba) = 2.158 mol; amount of substance of barium.
Answer:
About 1.301 atm
Explanation:
The formula that you should is PV=nRT, where P stands for pressure, V stands for volume, n stands for the number of moles, R stands for the universal gas constant, and T stands for temperature in Kelvin. Since the volume, number of moles, and universal gas constant don't change, you don't need to worry about them.
1.07V=393nR
PV=498nR
P=1.301 atm. Hope this helps!
Answer:
The cell potential is 0.609 V. Given E > 0 the electrochemical cell is spontaneous as written.
Explanation:
Let's consider the oxidation and reduction half-reactions and the global reaction.
Anode (oxidation): Sn²⁺(0.0023 M) ⇒ Sn⁴⁺(0.13 M) + 2 e⁻
Cathode (reduction): 2 Fe³⁺(0.11 M) + 2 e⁻ ⇒ 2 Fe²⁺(0.0037 M)
Global reaction: Sn²⁺(0.0023 M) + 2 Fe³⁺(0.11 M) ⇒ Sn⁴⁺(0.13 M) + 2 Fe²⁺(0.0037 M)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red,cat - E°red,an
E° = 0.771 V - 0.154 V = 0.617 V
The Nernst equation allows us to calculate the cell potential (E) under the given conditions.
![E=E\° -\frac{0.05916}{n} logQ\\E=E\° -\frac{0.05916}{n} log\frac{[Sn^{+4}].[Fe^{2+}]}{[Sn^{2+}].[Fe^{3+} ]} \\E=0.617V-\frac{0.05916}{2} log\frac{(0.13).(0.0037)}{(0.0023).(0.11)} \\E=0.609V](https://tex.z-dn.net/?f=E%3DE%5C%C2%B0%20-%5Cfrac%7B0.05916%7D%7Bn%7D%20logQ%5C%5CE%3DE%5C%C2%B0%20-%5Cfrac%7B0.05916%7D%7Bn%7D%20log%5Cfrac%7B%5BSn%5E%7B%2B4%7D%5D.%5BFe%5E%7B2%2B%7D%5D%7D%7B%5BSn%5E%7B2%2B%7D%5D.%5BFe%5E%7B3%2B%7D%20%5D%7D%20%5C%5CE%3D0.617V-%5Cfrac%7B0.05916%7D%7B2%7D%20log%5Cfrac%7B%280.13%29.%280.0037%29%7D%7B%280.0023%29.%280.11%29%7D%20%5C%5CE%3D0.609V)
The cell potential is 0.609 V. Given E > 0 the electrochemical cell is spontaneous as written.
