SOMEONE PLZ HELP!!!!

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⁻).

Answer:

Final volume  = 0.44 L

Explanation:

Given data:

Initial volume of balloon = 1.1 L

Initial pressure = 0.80 atm

Final volume = ?

Final pressure = 2.0 atm

Solution:

The given problem will be solved through the Boly's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

0.80 atm × 1.1 L = 2.0 atm × V₂

V₂ = 0.88 atm. L/ 2.0 atm

V₂ = 0.44 L

Answer:

ARSENIC

Explanation:

It has an atomic number of 33

Answer:..

Explanation:

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]

Answer: Proton - positive charge (+)

Neutron - neutral charge (0)

Electron - negative charge (-)

Explanation:

Answer:

QC= [O2]^3/[F2]^10

Explanation:

Answer:

acidic because of electrical issues and the body of electrical equipment

Answer:

7.71 × 10⁻⁴ M/s

Explanation:

The initial rate of the reaction can be expressed by using the formula:

[tex]\dfrac{\Delta [O_2]}{\Delta t}[/tex]

where the number of moles of O₂ = [tex]\dfrac{PV}{RT}[/tex]

where;

Pressue P = 1.00 atm

Volume V =5.74mL =  (5.74 /1000) L

Rate R = 0.082 L atm/mol.K

Temperature = 298 K

[tex]= \dfrac{1.00 \ atm \times \dfrac{5.74 }{1000}L}{0.082 \ L \ atm/mol.K \times 298 K}[/tex]

= 2.35 × 10⁻⁴ mol

Δ[O₂] = [tex]\dfrac{moles \ produced - initial \ mole}{\dfrac{5.08 }{1000}L }[/tex]

Δ[O₂] = [tex]\dfrac{2.35 \times 10^{-4} M - 0 M}{\dfrac{5.08 }{1000}}[/tex]

Δ[O₂]  = 0.04626 M

The initial rate = [tex]\dfrac{\Delta [O_2]}{\Delta t}[/tex]

= [tex]\dfrac{0.04626}{60}[/tex]

= 7.71 × 10⁻⁴ M/s