To disprove something
you have to have a....
1)hypothesis/observed
2)facts
3)experiment
that is all part of the scientific theory <span />
Answer:
0.925 atm
Explanation:
By Dalton's Law, the total pressure of a gas mixture is the sum of the partial pressure of its components. The vapor pressure of the water is the pressure that some molecules that evaporated do under the liquid surface. The gas and the liquid are at equilibrium. So, the gas mixture is water vapor and hydrogen gas.
Ptotal = Pwater + PH₂
745 = 42 + PH₂
PH₂ = 703 torr
Transforming to atm:
1 atm ------------------760 torr
x ------------------ 703 torr
By a simple direct three rule
760x = 703
x = 0.925 atm
Explanation:
Ionic bonds involve a cation and an anion. The bond is formed when an atom, typically a metal, loses an electron or electrons, and becomes a positive ion, or cation. Another atom, typically a non-metal, is able to acquire the electron(s) to become a negative ion, or anion.
Moles are used conveniently in chemistry especially in stoichiometric calculations involving reactions. The unit of mole is a collective term that holds 6.022×10^23 particles. These particles is a general term for any small units of matter including molecules, atoms and sub-particles. This ratio of 6.022×10^23 particles to 1 mole is known to be the Avogadro's number. Its exact number is actually <span>6.0221409</span>×10^23. We use this constant in our stoichiometric calculation as follows:
15 moles oxygen * (6.022×10^23 molecules/ 1 mole oxygen) = 9.033×10^24 molecules of oxygen
Answer:
0.35 V
Explanation:
(a) Standard reduction potentials
<u> E°/V</u>
Fe²⁺ + 2e- ⇌ Fe; -0.41
Cr³⁺ + 3e⁻ ⇌ Cr; -0.74
(b) Standard cell potential
<u> E°/V</u>
2Cr³⁺ + 6e⁻ ⇌ 2Cr; +0.74
<u>3Fe ⇌ 3Fe²⁺ + 6e-; </u> <u>-0.41
</u>
2Cr³⁺ + 3Fe ⇌ 2Cr + 3Fe²⁺; +0.33
3. Cell potential
2Cr³⁺(0.75 mol·L⁻¹) + 6e⁻ ⇌ 2Cr
<u>3Fe ⇌ 3Fe²⁺(0.25 mol·L⁻¹) + 6e-
</u>
2Cr³⁺(0.75 mol·L⁻¹) + 3Fe ⇌ 2Cr + 3Fe²⁺(0.25 mol·L⁻¹)
The concentrations are not 1 mol·L⁻¹, so we must use the Nernst equation
(a) Data
E° = 0.33 V
R = 8.314 J·K⁻¹mol⁻¹
T = 298 K
z = 6
F = 96 485 C/mol
(b) Calculations:
![Q = \dfrac{\text{[Fe}^{2+}]^{3}}{ \text{[Cr}^{3+}]^{2}} = \dfrac{0.25^{3}}{ 0.75^{2}} =\dfrac{0.0156}{0.562} = 0.0278\\\\E = 0.33 - \left (\dfrac{8.314 \times 298}{6 \times 96485}\right ) \ln(0.0278)\\\\=0.33 -0.00428 \times (-3.58) = 0.33 + 0.0153 = \textbf{0.35 V}\\\text{The cell potential is }\large\boxed{\textbf{0.35 V}}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BFe%7D%5E%7B2%2B%7D%5D%5E%7B3%7D%7D%7B%20%5Ctext%7B%5BCr%7D%5E%7B3%2B%7D%5D%5E%7B2%7D%7D%20%3D%20%5Cdfrac%7B0.25%5E%7B3%7D%7D%7B%200.75%5E%7B2%7D%7D%20%3D%5Cdfrac%7B0.0156%7D%7B0.562%7D%20%3D%200.0278%5C%5C%5C%5CE%20%3D%200.33%20-%20%5Cleft%20%28%5Cdfrac%7B8.314%20%5Ctimes%20298%7D%7B6%20%5Ctimes%2096485%7D%5Cright%20%29%20%5Cln%280.0278%29%5C%5C%5C%5C%3D0.33%20-0.00428%20%5Ctimes%20%28-3.58%29%20%3D%200.33%20%2B%200.0153%20%3D%20%5Ctextbf%7B0.35%20V%7D%5C%5C%5Ctext%7BThe%20cell%20potential%20is%20%7D%5Clarge%5Cboxed%7B%5Ctextbf%7B0.35%20V%7D%7D)
