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quester [9]
2 years ago
5

2.3 x 10²¹ particles H = __________ mol H (round to the nearest thousandth)

Chemistry
1 answer:
Natalka [10]2 years ago
7 0

Taking into account the definition of avogadro's number, 3.82×10⁻³ moles of H are 2.3×10²¹ particles of H.

<h3>Avogadro's Number</h3>

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023×10²³ particles per mole. Avogadro's number applies to any substance.

<h3>This case</h3>

Then you can apply the following rule of three: if 6.023×10²³ particles are contained in 1 mole of H, then 2.3×10²¹ particles are contained in how many moles of H?

amount of moles of H= (2.3×10²¹ particles × 1 mole)÷ 6.023×10²³ particles

<u><em>amount of moles of H= 3.82×10⁻³ moles</em></u>

Finally, 3.82×10⁻³ moles of H are 2.3×10²¹ particles of H.

Learn more about Avogadro's Number:

<u>brainly.com/question/11907018?referrer=searchResults </u>

<u>brainly.com/question/1445383?referrer=searchResults </u>

<u>brainly.com/question/1528951?referrer=searchResults</u>

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Which of the following statements is true?
EastWind [94]
The first statement is False... as
For exothermic reaction :
A+B》 C+D + HEAT..(heat is considered as a product)... as for endo.. heat is a reactant.
So tjey can't be of the same energy...

2nd one...based on the
A+B》 C+D+HEAT...For exo reaction... the product have more Heat energy than potential...so its false
Recall...energy can nither be created nor destroyed but converted from one form to another....

The 4th one however is true for heat...the reactants have nore energy than the products..
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5 0
3 years ago
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
Read 2 more answers
A student measures a mass of an 8cm3 block of brown sugar to be 19.9G. what is the density of the brown sugar?
Naily [24]

Answer:

<h2>2.49 g/cm³</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question we have

density =  \frac{19.9}{8}  \\  = 2.4875

We have the final answer as

<h3>2.49 g/cm³</h3>

Hope this helps you

3 0
3 years ago
What pressure will be exerted by 0.675 moles of a gas at 25*C if it is in a 0.750-L container?
Westkost [7]
PV=nRT

P=nRT/V

P=[(0.650mol)(0.08206)(298K)]/(0.750L)=21.2atm



6 0
3 years ago
3) 2500 pounds to Kilograms<br><br> Would this be 1,250?
Softa [21]

Answer:

1134

Explanation:

(the ones place was rounded up). 1133.981 is the unrounded answer.

8 0
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