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Ann [662]
3 years ago
13

I BEG YOU I NEED IT TILL 12 PLEASE ILL GIVE THE BRAINLEST AND HEARTS AND POINTS HELP PLEASEEEE !!!!!!!!!!!!!!!!!!!!!!!!!

Chemistry
2 answers:
Dennis_Churaev [7]3 years ago
6 0
I’ve never seen this in my life Jesus
Lubov Fominskaja [6]3 years ago
3 0
UHHHHHHHH IDRK TRY THE LINK ABOVE
You might be interested in
The electron configuration of a neutral atom is 1s22s22p63s2. Write a complete set of quantum numbers for each of the electrons.
Pavlova-9 [17]

Answer:

The four quantum number for each electron will be:

1s^{2}

n=1;l=0;m=0;s=+\frac{1}{2}/-\frac{1}{2}

2s^{2}

n=2;l=0;m=0;s=+\frac{1}{2}/-\frac{1}{2}

2p^{6}

n=1;l=1;\\m=+1,0.-1 \\s=+\frac{1}{2}/-\frac{1}{2}

3s^{2}

n=3;l=0;m=0;s=+\frac{1}{2}/-\frac{1}{2}

Explanation:

As the element is neutral, the number of protons will be equal to number of electrons which will be the atomic number of the element.

Number of electrons =12

Atomic number = 12

Element : Magnesium

The principal shell is represented by "n"

i) For "s" subshell the value of l =0 (azimuthal quantum number) thus m (magnetic quantum number)= 0

The two electrons in s subshell will have either plus half or minus half spin quantum number

ii) for "p" subshell the value for l =1

thus m = 0 or +1 or -1

The two electrons in each orbital will have either plus half or minus half spin quantum number

8 0
3 years ago
!
11111nata11111 [884]
The correct answer is B. Homeostasis
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
Solve the quadratic equation 2x^2+13x=15 by method of completing the square<br>​
mafiozo [28]

Answer:

x = 1, -7.5

Explanation:

2x² + 13x = 15

Divide the left side of the equation by 2

2(x² + 6.5x) = 15

Divide 6.5 by 2 and square the quotient

6.5/2 = 3.25

3.25² = 10.5625

Add 10.5625 to the left side

2(x² + 6.5x + 10.5625) = 15

Since you have a 2 outside the parentheses, you will be adding 10.5625•2 to the right side.

10.5625 • 2 = 21.125

2(x² + 6.5x + 10.5625) = 36.125

To factor (x² + 6.5x + 10.5625), add b/2 to x

b/2 = 6.5/2 = 3.25

2(x + 3.25)² = 36.125

Divide by 2

(x + 3.25)² = 18.0625

Square root.

(x + 3.25) = √18.0625

x + 3.25 = ±4.25

Subtract 3.25.

x = 4.25 - 3.25 = 1

x = -4.25 - 3.25 = -7.5

x = 1, -7.5

4 0
3 years ago
Balanced equations illustrate what law?
Klio2033 [76]
Conservation of mass
8 0
3 years ago
Read 2 more answers
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