What is an orderly, three dimensional arrangement formed by ions called?

A 100.0 mLflask is filled with 0.065 moles of A and allowed to react to form B according to the reaction below. The following ex

kodGreya [7K]

The question is incomplete. The complete question is :

A 100.0 mL flask is filled with 0.065 moles of A and allowed to react to form B according to the reaction below. The following experimental data are obtained for the amount of A as the reaction proceeds. What is the average rate of appearance of B in units of M/s between t = 10 min. and t = 30 min.? Assume that the volume of the flask is constant. A(g) → B(g)

Time 0.0 10.0 20.0 30.0 40.0

Moles of A 0.065 0.051 0.042 0.036 0.031

Solution :

Consider the following reaction as follows :

$A \rightarrow B$

The experiment data is given as follows :

Time (min) : 0.0 10.0 20.0 30.0 40.0

Moles of A : 0.065 0.051 0.042 0.036 0.031

According to the rate of reaction concept, the rate can be expressed as a consumption of the reactant and formation of the product as follows :

Average rate : $= -\frac{d[A]}{dt} = \frac{d[B]}{dt}$

Now we have to calculate the average rate between 10.0 to 30.0 min w.r.t. A as follows :

Rate $=-\frac{(0.051-0.036) mol \times \frac{1}{0.1 \ L}}{(30.0-10.0) mol \times \frac{60 \ s}{1 \ min}}$

$=\frac{0.15 \ M}{20 \ min \times \frac{60 \ s}{1 \ min}}$

$= 1.25 \times 10^{-4 }\ M/s$

Therefore, the rate = $= 1.3 \times 10^{-4 }\ M/s$

5 0

3 years ago

Please help! ill give medal

weeeeeb [17]

An example of a general formula of an acid is
<span>HCl

This is called as hydrochloric acid or hydrogen chloride.

H represents for the atom of Hydrogen
and Cl represents for the atom of Chlorine

Since their charges are -1, and +1, it's ratio is 1:1</span>

3 0

3 years ago

A red sports drink contains Red 40 dye. A 5.4 mL aliquot of this sports drink was diluted to 25.0 mL with deionized water in a v

miskamm [114]

Answer:

84.0 ppm is the concentration of Red 40 dy in the original sports drink.

Explanation:

Concentration of red dye in sport drink before dilution $C_1=?$

Volume of the sport drink before dilution $V_1=5.4 mL$

Concentration of red dye in sport drink after dilution $C_2=18.1 ppm$

Volume of the sport drink after dilution $V_2=25.0 mL$

$C_1V_1=C_2V_2$ ( dilution )

$C_1=\frac{C_2V_2}{V_1}=\frac{18.1 ppm\times 25.0 mL}{5.4 mL}$

$C_1=83.80 ppm\approx 84.0ppm$

84.0 ppm is the concentration of Red 40 dy in the original sports drink.

6 0

3 years ago

Calculate the pH for each case in the titration of 50.0 mL of 0.230 M HClO ( aq ) 0.230 M HClO(aq) with 0.230 M KOH ( aq ) . 0.2

Vikentia [17]

Answer:

pH before addition of KOH = 4.03

pH after addition of 25 ml KOH = 7.40

pH after addition of 30 ml KOH = 7.57

pH after addition of 40 ml KOH = 8.00

pH after addition of 50 ml KOH = 10.22

pH after addition 0f 60 ml KOH = 12.3

Explanation:

pH of each case in the titration given below

(6) After addition of 60 ml KOH

Since addition of 10 ml extra KOH is added after netralisation point.

Concentration of solution after addition 60 ml KOH is calculated by

M₁V₁ = M₂V₂

or, 0.23 x 10 = (50 + 60)ml x M₂

or M₂ = 0.03 Molar

so, concentration of KOH = 0.03 molar

[OH⁻] = 0.03 molar

pOH = 0.657

pH = 14 - 0.657 = 13.34

5 0

3 years ago

Question

AnnyKZ [126]

Blank #1: Polyatomic

Blank #2: 2 (see explanation)

Blank #3: 1 (see explanation)

<h3>Explanation</h3>

Both the ammonium ion $\text{NH}_4^{+}$ and the sulfate ion $\text{SO}_4^{2-}$ contain more than one atom in each ion. The two species are thus <em>polyatomic</em>. The chloride ion $\text{Cl}^{-}$ , for example, is <em>monoatomic</em>.

Superscripts above formulae of the ions indicate their charge. Each ammonium ion carries a positive one (+1) charge. Each sulfate ion carries a charge of negative two (-2).

Ammonium sulfate is an ionic compound. A sample of this compound contain myriads of ammonium ions and sulfate ions. The ions are packed in three-dimensional lattices. Thus unlike water, ammonium sulfate does not exist as molecules in nature.

Assuming that the second and third blanks refers to a formula unit, rather than a molecule, of ammonium sulfate. The empirical formula of ammonium sulfate gives the minimum whole-number ratio between the two ions in a sample.

Charges shall balance between the two ions. Ammonium ions are of charge +1. Sulfate ions are of charge -2. The sample shall thus contain two ammonium ions for every one sulfate ion.

The empirical formula of ammonium sulfate is therefore $(\text{NH}_4)_2\text{SO}_4$ .

There are thus two ammonium ions $\text{NH}_4^{+}$ and one sulfate ion $\text{SO}_4^{2-}$ in each formula unit of ammonium sulfate.

8 0

3 years ago