What product has almost the same compounds as baking soda 10pts

A 29.05 gram sample of cobalt is heated in the presence of excess oxygen. A metal oxide is formed with a mass of 40.88 g. Determ

deff fn [24]

The empirical formula of the oxide is Co₂O₃.

Step 1. Calculate the mass of oxygen

Your reaction is

Cobalt + oxygen ⟶ cobalt oxide

29.05 g + x g ⟶ 40.88 g

According to the Law of Conservation of Mass, the total mass of the reactants must equal the total mass of the products. Thus,

29.05 g + x g ⟶ 40.88 g

x = 40.88 – 29.05 = 11.83

Step 2. Calculate the moles of each element

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

So, our job is to calculate the molar ratio of Co to O.

Moles of Co = 29.05 g Co × (1 mol Co /(58.93 g Co) = 0.492 96 mol Co

Moles of O = 11.83 g O × (1 mol O/16.00 g O) = 0.739 38 mol O

Step 3. Calculate the molar ratio of the elements

Divide each number by the smaller number of moles

Co:O = 0.429 26:0.739 38 = 1:1.4999

Step 4. Multiply each number by a factor that makes the ratio close to whole numbers

Multiply by 2. Then

Co:O = 2:2.998 ≈ 2:3

Step 5: Write the empirical formula

EF = Co₂O₃

4 0

3 years ago

When acid reacts with Base what result

elena55 [62]

What occurs is they neutralize both acid and base characteristics/features, usually producing a salt.

Hope this shells!

5 0

2 years ago

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What is the ground state electron configuration for magnesium?

Stella [2.4K]

Magnesium :

[Ne] 3s²

Answer A

hope this helps!

5 0

3 years ago

Jeff used an equal arm balance to weigh a 4.312 g sample of sodium chloride. Which of these measurements made by Jeff is the mos

Virty [35]

Answer:

4.3.

Explanation:

Measurements that are close to the known value are said to be accurate, whereas measurements that are close to each other are said to be precise.

The most closest measurement to the known value is 4.3 g.
So, the right choice is: 4.3 g.

3 0

4 years ago

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Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.210 M HClO(aq) with 0.210 M KOH(aq).

Degger [83]

a) before addition of any KOH :

when we use the Ka equation & Ka = 4 x 10^-8 :

Ka = [H+]^2 / [ HCIO]

by substitution:

4 x 10^-8 = [H+]^2 / 0.21

[H+]^2 = (4 x 10^-8) * 0.21

= 8.4 x 10^-9

[H+] = √(8.4 x 10^-9)

= 9.2 x 10^-5 M

when PH = -㏒[H+]

PH = -㏒(9.2 x 10^-5)

= 4

b)After addition of 25 mL of KOH: this produces a buffer solution

So, we will use Henderson-Hasselbalch equation to get PH:

PH = Pka +㏒[Salt]/[acid]

first, we have to get moles of HCIO= molarity * volume

=0.21M * 0.05L

= 0.0105 moles

then, moles of KOH = molarity * volume

= 0.21 * 0.025

=0.00525 moles

∴moles HCIO remaining = 0.0105 - 0.00525 = 0.00525

and when the total volume is = 0.05 L + 0.025 L = 0.075 L

So the molarity of HCIO = moles HCIO remaining / total volume

= 0.00525 / 0.075

=0.07 M

and molarity of KCIO = moles KCIO / total volume

= 0.00525 / 0.075

= 0.07 M

and when Ka = 4 x 10^-8

∴Pka =-㏒Ka

= -㏒(4 x 10^-8)

= 7.4

by substitution in H-H equation:

PH = 7.4 + ㏒(0.07/0.07)

∴PH = 7.4

c) after addition of 35 mL of KOH:

we will use the H-H equation again as we have a buffer solution:

PH = Pka + ㏒[salt/acid]

first, we have to get moles HCIO = molarity * volume

= 0.21 M * 0.05L

= 0.0105 moles

then moles KOH = molarity * volume
= 0.22 M* 0.035 L

=0.0077 moles

∴ moles of HCIO remaining = 0.0105 - 0.0077= 8 x 10^-5

when the total volume = 0.05L + 0.035L = 0.085 L

∴ the molarity of HCIO = moles HCIO remaining / total volume

= 8 x 10^-5 / 0.085

= 9.4 x 10^-4 M

and the molarity of KCIO = moles KCIO / total volume

= 0.0077M / 0.085L

= 0.09 M

by substitution:

PH = 7.4 + ㏒( 0.09 /9.4 x 10^-4)

∴PH = 8.38

D)After addition of 50 mL:

from the above solutions, we can see that 0.0105 mol HCIO reacting with 0.0105 mol KOH to produce 0.0105 mol KCIO which dissolve in 0.1 L (0.5L+0.5L) of the solution.

the molarity of KCIO = moles KCIO / total volume

= 0.0105mol / 0.1 L

= 0.105 M

when Ka = KW / Kb

∴Kb = 1 x 10^-14 / 4 x 10^-8

= 2.5 x 10^-7

by using Kb expression:

Kb = [CIO-] [OH-] / [KCIO]

when [CIO-] =[OH-] so we can substitute by [OH-] instead of [CIO-]

Kb = [OH-]^2 / [KCIO]

2.5 x 10^-7 = [OH-]^2 /0.105

∴[OH-] = 0.00016 M

POH = -㏒[OH-]

∴POH = -㏒0.00016

= 3.8
∴PH = 14- POH

=14 - 3.8

PH = 10.2

e) after addition 60 mL of KOH:

when KOH neutralized all the HCIO so, to get the molarity of KOH solution

M1*V1= M2*V2

when M1 is the molarity of KOH solution

V1 is the total volume = 0.05 + 0.06 = 0.11 L

M2 = 0.21 M

V2 is the excess volume added of KOH = 0.01L

so by substitution:

M1 * 0.11L = 0.21*0.01L

∴M1 =0.02 M

∴[KOH] = [OH-] = 0.02 M

∴POH = -㏒[OH-]

= -㏒0.02

= 1.7

∴PH = 14- POH

= 14- 1.7

= 12.3

8 0

3 years ago

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