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NARA [144]
3 years ago
7

Help me with this chemistry question please.​

Chemistry
1 answer:
rodikova [14]3 years ago
4 0

Answer:

B) Because you are adding positively charged protons, which pull in the outer electrons.

Explanation:

Since the increase in protons have a greater effect then the increase in electrons the protons pull the electrons in closer resulting the radius of an atom to decrease from left to right across a period.

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Write a ground state electron configuration for each neutral atom
Gre4nikov [31]

Answer:

Pb[lead] [Xe]4f^145d^106s^26p^2

U[uranium] 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

This notation can be written in core notation or noble gas notation by replacing the

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

with the noble gas [Rn].

[Rn]7s25f4

N[nitrogen] The full electron configuration for nitrogen is 1s^2 2s^2 2p^3.

Ti[titanium] Ti2+:[Ar]3d^2

Ti:1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2

1s^2 2s^2 2p^6 3s^2 3p^5 = 17 electrons

(1) electron gain will result to a

negative charge (−), and

(2) electron loss will result to a positive charge (+),

1s^2 2s^2 2p^6 3s^2 3p^6 = 18 electrons

Hg[mercury] You should then find its atomic number is 80. It has a Xe core, so in shorthand notation, you can include [Xe]instead of

1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6,

for 54 electrons. For the 6th row of the periodic table, we introduce the 4f orbitals, and proceed to atoms having occupied 5d orbitals. We, as usual, have the ns orbitals, and n=? for the 6th period?

Mercury has a regular electron configuration. It becomes:

[Xe]4f145d106s2

Explanation:

socratic.org helped me! I'm really sorry if this is wrong!

6 0
2 years ago
Calculate the concentration of all species in a 0.160 m solution of h2co3.
enot [183]

             H₂CO₃ ⇔       HCO₃⁻   +     H⁺

I            0.160               0                 0

C            -x                  +x               +x

E          0.160-x          +x                +x

Ka1 = [HCO₃⁻][H⁺] / [H₂CO₃]

4.3 x 10⁻⁷ = x² / (0.160-x)   (x is neglected in 0.160-x = 0.160)

x² = 6.88 x 10⁻⁸

x = 2.62 x 10⁻⁴

             HCO₃⁻    ⇔            CO₃⁻²    +   H⁺

I          2.62 x 10⁻⁴               0                2.62 x 10⁻⁴

C          -x                            +x              +x

E       2.62 x 10⁻⁴ - x           +x              2.62 x 10⁻⁴ + x

Ka2 = [CO₃⁻²][H⁺] / [HCO₃⁻]

5.6 x 10⁻¹¹ = x(2.62 x 10⁻⁴ + x) / (2.62 x 10⁻⁴ - x)

x = 5.6 x 10⁻¹¹

Thus,

[H₂CO₃] = 0.160 - (2.62 x 10⁻⁴) = 0.16 M

[HCO₃⁻] = 2.62 x 10⁻⁴ - ( 5.6 x 10⁻¹¹) = 2.6 x 10⁻⁴ M

[CO₃⁻²] = 5.6 x 10⁻¹¹ M

[H₃O⁺] = 2.62 x 10⁻⁴ + 5.6 x 10⁻¹¹ = 2.6 x 10⁻⁴ M

[OH⁻] = 3.8 x 10⁻¹¹

8 0
3 years ago
6. Which part leads the blood to the heart?
Dmitriy789 [7]

Answer:

Your answer should be D

7 0
2 years ago
Read 2 more answers
Calculation of Molar Ratios of Conjugate Base to Weak Acid from pH For a weak acid with a pKa of 6.0, calculate the ratio of con
Free_Kalibri [48]

Explanation:

According to the Handerson equation,  

          pH = pK_{a} + log \frac{\text{salt}}{\text{acid}}

or,      pH = pK_{a} + log \frac{\text{conjugate base}}{\text{acid}}

Putting the given values into the above equation as follows.

     pH = pK_{a} + log \frac{\text{conjugate base}}{\text{acid}}

       5.0 = 6.0 + log \frac{\text{conjugate base}}{\text{acid}}[/tex]

      log \frac{\text{conjugate base}}{\text{acid}} = -1.0

or,      \frac{\text{conjugate base}}{\text{acid}} = 10^{-1.0}

                            = 0.1

Therefore, we can conclude that molar ratios of conjugate base to weak acid for given solution is 0.1.

7 0
3 years ago
Calculate the ph of 0,24 m of kch3coo.? ​
Galina-37 [17]

Answer:

Correct option is A)

[H

+

]=

KaC

=

1.8×10

−6

=1.34×10

−3

pH=−log[H

+

]

=2.88

Explanation:

here is your answer if you like my answer please follow

6 0
3 years ago
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