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Serjik [45]
3 years ago
7

2. 85.6 GRAMS of RbBr

Chemistry
1 answer:
Georgia [21]3 years ago
4 0

Explanation:

1 mole is equal to 1 moles RbBr, or 165.3718 grams.

You might be interested in
How many atoms of lead are contained in 0.532mol?​
Colt1911 [192]

Answer:

3.2043 x 10²³

Explanation:

No. of Mole of lead (Pb) = 0.532 mol

No. of atoms of lead = ?

Solution:

Formula Used to calculate

no. of moles = numbers of particles (ions, molecules, atoms) /Avogadro's number

Avogadro's no. = 6.023 x10²³

So,

The formula could be written as

no. of atoms of lead Pb = no. of moles x 6.023 x10²³

Put the values in above formula

no. of atoms of lead Pb = 0.532 mol x 6.023 x10²³

no. of atoms of lead Pb = 3.2043 x 10²³

so 3.2043 x 10²³ atoms of lead are contained in 0.532 mole.

6 0
3 years ago
1. To become stable, the following element:
Delicious77 [7]

Answer:

A

Explanation:

4 0
2 years ago
Write a balanced equation for the oxidation-reduction reaction that occurs when hydrogen peroxide reacts with ferrous ion
Alex17521 [72]
H₂O₂ + 2FeSO₄ + H₂SO₄ → Fe₂(SO₄)₃ + 2H₂O

H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O  k=1
Fe²⁺ → Fe³⁺ + e⁻                 k=2

H₂O₂ + 2H⁺ + 2Fe²⁺ → 2H₂O + 2Fe³⁺


7 0
3 years ago
Question 38 (1 point)
Fiesta28 [93]

Answer: Protons because they have a positive charge.

Explanation:

8 0
3 years ago
The equilibrium concentrations of the reactants and products are [HA]=0.280 M, [H+]=4.00×10−4 M, and [A−]=4.00×10−4 M. Calculate
Tems11 [23]

Answer:

6.24

Explanation:

The following data were obtained from the question:

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

pKa =.?

Next, we shall write the balanced equation for the reaction. This is given below:

HA <===> H+ + A-

Next, we shall determine the equilibrium constant Ka for the reaction. This can be obtained as follow:

Equilibrium constant for a reaction is simply the ratio of concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant for the above equation is given below:

Ka = [H+] [A−] /[HA]

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

Equilibrium constant (Ka) =

Ka = (4×10¯⁴ × 4×10¯⁴) / 0.280

Ka = 1.6×10¯⁷/ 0.280

Ka = 5.71×10¯⁷

Therefore, the equilibrium constant for the reaction is 5.71×10¯⁷

Finally, we shall determine the pka for the reaction as follow:

Equilibrium constant, Ka = 5.71×10¯⁷

pKa =?

pKa = – Log Ka

pKa = – Log 5.71×10¯⁷

pKa = 6.24

Therefore, the pka for the reaction is 6.24.

6 0
3 years ago
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