Answer:
Molality = 8.57 m
Explanation:
Given data:
Molarity of solution = 5.73 M
density = 0.9327 g/mL
Molality of solution = ?
Solution:
Molality = moles of solute / kg of solvent.
Kg of solvent:
Mass of 1 L solution = density× volume
Mass of 1 L solution = 0.9327 g/mL × 1000 mL
Mass of 1 L solution = 932.7 g
Mass of solute:
Mass of 1 L = number of moles × molar mass
Mass = 5.73 mol × 46.068 g/mol
Mass = 263.97 g
Mass of solvent:
Mass of solvent = mass of solution - mass of solute
Mass of solvent = 932.7 g - 263.97 g
Mass of solvent = 668.73 g
In Kg = 668.73 /1000 = 0.6687 Kg
Molality:
Molality = number of moles of solute / mass of solvent in Kg
Molality = 5.73 mol / 0.6687 Kg
Molality = 8.57 m
As we know that there are avogadro no. of atoms in 9 g of beryllium.
1 mole of beryllium = 6.02 * 10^23 atoms
so 2.5 mole= 6.02*10^23*2.5 i.e = <span>15.055 × 10^23 atoms </span>
Answer:
A) chlorine
Explanation:
To solve this question we can use:
PV = nRT
In order to solve the moles of the gas. With the moles and the mass we can find the molar mass of the gas to have an idea of its identy:
PV = nRT
PV / RT = n
<em>Where P is pressure: 603mmHg * (1atm / 760mmHg) = 0.7934atm</em>
<em>V = 100mL = 0.100L</em>
<em>R is gas constant = 0.082atmL/molK</em>
<em>T is absolute temperature = 14°C + 273.15 = 287.15K</em>
0.7934atm*0.100L / 0.082atmL/molK*287.15K = n
3.37x10⁻³ moles of the gas
In 0.239g. The molar mass is:
0.239g / 3.37x10⁻³ moles = 70.9g/mol
The gas with this molar mass is Chlorine, Cl₂:
<h3>A) chlorine
</h3><h3 />
Answer:
(a) Cu²⁺ +2e⁻ ⇌ Cu
(c) 0.07 V
Explanation:
(a) Cu half-reaction
Cu²⁺ + 2e⁻ ⇌ Cu
(c) Cell voltage
The standard reduction potentials for the half-reactions are+
<u> E°/V
</u>
Cu²⁺ + 2e⁻ ⇌ Cu; 0.34
Hg₂Cl₂ + 2e⁻ ⇌ 2Hg + 2Cl⁻; 0.241
The equation for the cell reaction is
E°/V
Cu²⁺(0.1 mol·L⁻¹) + 2e⁻ ⇌ Cu; 0.34
<u>2Hg + 2Cl⁻ ⇌ Hg₂Cl₂ + 2e⁻; </u> <u>-0.241
</u>
Cu²⁺(0.1 mol·L⁻¹) + 2Hg + 2Cl⁻ ⇌ Cu + Hg₂Cl₂; 0.10
The concentration is not 1 mol·L⁻¹, so we must use the Nernst equation
(ii) Calculations:
T = 25 + 273.15 = 298.15 K
![Q = \dfrac{\text{[Cl}^{-}]^{2}}{ \text{[Cu}^{2+}]} = \dfrac{1}{0.1} = 10\\\\E = 0.10 - \left (\dfrac{8.314 \times 298.15 }{2 \times 96485}\right ) \ln(10)\\\\=0.010 -0.01285 \times 2.3 = 0.10 - 0.03 = \textbf{0.07 V}\\\text{The cell potential is }\large\boxed{\textbf{0.07 V}}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BCl%7D%5E%7B-%7D%5D%5E%7B2%7D%7D%7B%20%5Ctext%7B%5BCu%7D%5E%7B2%2B%7D%5D%7D%20%3D%20%5Cdfrac%7B1%7D%7B0.1%7D%20%3D%2010%5C%5C%5C%5CE%20%3D%200.10%20-%20%5Cleft%20%28%5Cdfrac%7B8.314%20%5Ctimes%20298.15%20%7D%7B2%20%5Ctimes%2096485%7D%5Cright%20%29%20%5Cln%2810%29%5C%5C%5C%5C%3D0.010%20-0.01285%20%5Ctimes%202.3%20%3D%200.10%20-%200.03%20%3D%20%5Ctextbf%7B0.07%20V%7D%5C%5C%5Ctext%7BThe%20cell%20potential%20is%20%7D%5Clarge%5Cboxed%7B%5Ctextbf%7B0.07%20V%7D%7D)
