All categories

3 years ago

The ph of a 0.55 m aqueous solution of hypobromous acid, hbro, at 25.0 Â°c is 4.48. what is the value of ka for hbro?

1 answer:

6 0

Hbro dissociate as follows
HBro---> H+ + BrO-
Ka= (H+)(BrO-) / HBro
PH = -log (H+)
therefore (H+) = 10^-4.48= 3.31 x10^-5
ka is therefore= ( 3.31 x 10^-5)^2/0.55=1.99 x10^-9

You might be interested in

If 5.57 g of Ag2O is sealed in a 75.0-mL tube filled with 760 torr of N2 gas at 28 ∘C, and the tube is heated to 310 ∘C, the Ag2

Snezhnost [94]

Answer: Total pressure = 7293.2 torr or 9.60 atm

Explanation:

Total pressure = partial pressure of nitrogen + partial pressure of oxygen

The partial pressure due to nitrogen is determined using the equation of Gay-Lussac's law: P₁/T₁=P₂/T₂

P₁ = 760 torr = 1atm, T₁ = 28∘C = (273+28)K = 301k, P₂ = ?, T₂ = 310∘C =(310+273)K = 583K

P₂ = P₁ T₂/ T₁

P₂ = 760 * 583 / 301 = 1472.03 torr

The pressure due to Oxygen gas produced is calculated thus:

Balanced equation of the decomposition of Ag₂O at s.t.p. is as follows;

2Ag₂O ----> 4Ag + O₂(g)

2 moles of Ag₂O produces 1 mole of O₂

molar mass of Ag₂O = (2*108 + 16)g = 232g/mol

molar volume of gas at s.t.p. = 22.4L

2*232g i.e. 464g of Ag₂O produces 22.4L of O₂

5.57g of Ag₂O will produce 5.57g*22.4L/464g = 0.269L or 269mL of O₂

Using the General gas equation P₁V₁/T₁=P₂V₂/T₂

P₁ = 1atm = 760 torr, V₁ = 269mL, T₁=273K, P₂ = ?, V₂= 75mL, T₂ = 583K

P₂ = P₁V₁T₂/V₂T₁

P₂ = 760*269*583 / 75*273

P₂ = 5821.17 torr

Total pressure = (1472.03 + 5821.17) torr

Total pressure = 7293.2 torr or 9.60 atm

7 0

3 years ago

Dr. I. M. A. Brightguy adds 0.1727 g of an unknown gas to a 125-mL flask. If Dr. B finds the pressure to be 736 torr at 20.0°C,

AlladinOne [14]

Answer:

The gas that Dr. Brightguy added was O₂

Explanation:

Ideal Gases Law to solve this:

P . V = n . R . T

Firstly, let's convert 736 Torr in atm

736 Torr is atmospheric pressure = 1 atm

20°C = 273 + 20 = 293 T°K

125 mL = 0.125L

0.125 L . 1 atm = n . 0.082 L.atm / mol.K . 293K

(0.125L .1atm) / (0.082 mol.K /L.atm . 293K) = n

5.20x10⁻³ mol = n

mass / mol = molar mass

0.1727 g / 5.20x10⁻³ mol = 33.2 g/m

This molar mass corresponds nearly to O₂

7 0

3 years ago

Scientists use what properties to help them identify and classify matter

Volgvan

Scientists use the physical and chemical properties to help them identify and classify matter. These physical and chemical properties are in a macro-perspective, in which these matter contains compounds, elements and atoms. Hence, matter can be classified in various ways, 
1. Atomic number either atomic mass each element has
2. By substance of that matter either pure substance or mixed substance 
3. If they cannot reduce a certain substance into a much smaller quantified atomic structure then they they’ll use (2) to identify and classify it.

5 0

3 years ago

Which orbitals form a pi bond?

bekas [8.4K]

Answer:

p and d orbitals

Explanation:

A π bond forms when two orbitals overlap side-on.

The most common types are formed by the overlap of p orbitals (Fig. 1).

However, d orbitals can also overlap sideways with each other to form dπ-dπ bonds and with p orbitals to form dπ-pπ bonds (Fig.2).

8 0

3 years ago

One mole of water weights?

Gemiola [76]

Answer:

c 18.0ml

Explanation:

The average mass of one H2O molecule is 18.02 amu. The number of atoms is an exact number, the number of mole is an exact number; they do not affect the number of significant figures. The average mass of one mole of H2O is 18.02 grams. This is stated: the molar mass of water is 18.02 g/mol.

8 0

2 years ago