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2 years ago

If the sun was 1 meter (about 3 feet) in size the earth would be __________ meters away

2 answers:

8 0

The Sun is 864,400 miles (1,391,000 kilometers) across. This is about 109 times the diameter of Earth. The Sun weighs about 333,000 times as much as Earth. It is so large that about 1,300,000 planet Earths can fit inside of it.Answer:

Explanation:

Vera_Pavlovna [14]2 years ago

7 0

Answer:

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You decide to establish a new temperature scale on which the melting point of ammonia (-77.75 ∘c) is 0∘a, and the boiling point

Hoochie [10]

First we need to know that the boiling point of water in C is 100 and we just need to solve for x in the equation:

-33.75-(-77.75) / 100 = 100-(-77.75) / x
44.4/100 = 177.75 / x
x = 177.75*100/44.4 = 400.33

The boiling point of water in ∘a would be 400.33∘a.

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3 years ago

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What is bulimia? can someone explain it to me.

Charra [1.4K]

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4 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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Leave the answer blank if no precipitate will form. (Express your answer as a chemical formula.) Formula of precipitate ZnCl2(aq

Rasek [7]

Answer:

1. Zn(OH)₂ (s)

2. Ag₂CO₃ (s)

3. Ni₃(PO₄)₂(s)

4. No reaction

5. (NH₄)₂CO₃(s)

Explanation:

Let's state the equations and we analyse some solubility and precipitation information:

ZnCl₂(aq) + 2KOH(aq) → Zn(OH)₂ (s) + 2KCl (aq)

All the salts from the halogens with group 1, are soluble.

The OH⁻ reacts to Zn cation in order to produce a precipitate. This is ok, but if the base is in excess, the Zn(OH)₂ will be soluble

K₂CO₃(aq) + 2AgNO₃(aq) → Ag₂CO₃ (s) ↓ + 2KNO₃(aq)

All salts from nitrate are soluble

All salts from carbonates are insoluble

2(NH₄)₃PO₄(aq) + 3Ni(NO₃)₂(aq) → Ni₃(PO₄)₂(s) ↓ + 6NH₄NO₃(aq)

Salts from phosphates are insoluble

All salts from nitrate are soluble

NaCl(aq) + KNO3(aq) → NO REACTION

All salts from nitrate are soluble

All the salts from the halogens with group 1, are soluble

Na₂CO₃(aq) + 2NH₄Cl(aq) → 2NaCl(aq) + (NH₄)₂CO₃(s) ↓

All salts from carbonates are insoluble

All the salts from the halogens with group 1, are soluble

7 0

3 years ago

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What are the reasons for existence of large number of organic compounds in nature.Explain

Law Incorporation [45]

Answer:

because of catenation of carbon.

Explanation:

Catenation is the binding of an element to its self through covalent bonds to form chain or ring molecules. carbon is able to form continuous links with other carbon atoms which is the reason for the existence of a large number of organic compounds.

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3 years ago