Answer:
acidic because of electrical issues and the body of electrical equipment
In the laboratory you dissolve 18.7 g of copper(II) bromide in a volumetric flask and add water to a total volume of 375mL.
Required:
a. What is the molarity of the solution?
b. What is the concentration of the copper(II) cation?
c. What is the concentration of the acetate anion?
Answer:
a) - 0.2 M
b) - 0.2 M
c)- 0
Explanation:
The chemical formula of copper (II) bromide is CuBr₂. Its molar mass (MM) is calculated as follows:
MM(CuBr₂)= MM(Cu) + (2 x MM(Br) = 63.5 g/mol + (2 x 80 g/mol)= 223.5 g/mol
a). Molarity = moles CuBr₂/1 L solution
moles CuBr₂ = mass/MM = 18.7 g x 1 mol/223.5 g = 0.084 mol
Volume in L = 375 mL x 1 L/1000 mL = 0.375 L
M = 0.084 mol/(0.375 L) = 0.223 M ≅ 0.2 M
b). When is added to water, CuBr₂ dissociates into ions as follows:
CuBr₂ ⇒ Cu²⁺ + 2 Br⁻
We have 1 mol Cu²⁺ (copper (II) cation) per mol of CuBr₂. Thus, the concentration of copper (II) cation is:
0.2 mol CuBr₂ x 1 mol Cu²⁺/mol CuBr₂ = 0.2 M
c). The concentration of acetate anion is 0. There is no acetate anion in the solution (the anion from CuBr₂ is bromide Br⁻).
Solid diarsenic trioxide reacts with fluorine gas (F2) to produce liquid arsenic pentafluoride and oxygen gas (O2). Write the Qc for this reaction.
Answer:
QC= [O2]^3/[F2]^10
Explanation:
What is the most highly populated rotational level of Cl2 (i) 25deg C and (ii) 100 deg C? Take B=0.244cm-1.This question should not be resubmitted, it is a textbook question from the Atkins physical chemistry txtbook. 10 e.
Answer:
i
[tex]J_{m} = 20 [/tex]
ii
[tex]J_{m} = 22.5 [/tex]
Explanation:
From the question we are told that
The first temperatures is [tex]T_1 = 25^oC = 25 +273 =298 \ K[/tex]
The second temperature is [tex]T_2 = 100^oC = 100 +273 = 373 \ K[/tex]
Generally the equation for the most highly populated rotational energy level is mathematically represented as
[tex]J_{m} = [ \frac{RT}{2B}] ^{\frac{1}{2} } - \frac{1}{2}[/tex]
Here R is the gas constant with value [tex]R =8.314 \ J\cdot K^{-1} \cdot mol^{-1}[/tex]
Also
B is given as [tex]B=\ 0.244 \ cm^{-1}[/tex]
Generally the energy require per mole to move 1 cm is 12 J /mole
So [tex]0.244 \ cm^{-1}[/tex] will require x J/mole
[tex]x = 0.244 * 12[/tex]
=> [tex]x = 2.928 \ J/mol [/tex]
So at the first temperature
[tex]J_{m} = [ \frac{8.314 * 298 }{2* 2.928 }] ^{\frac{1}{2} } - 0.5 [/tex]
=> [tex]J_{m} = 20 [/tex]
So at the second temperature
[tex]J_{m} = [ \frac{8.314 * 373 }{2* 2.928 }] ^{\frac{1}{2} } - 0.5 [/tex]
=> [tex]J_{m} = 22.5 [/tex]