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2 years ago

What quantum numbers specify these subshells? 2S, 6P, and 3D. (The answer is n= and L=)

Chemistry

1 answer:

statuscvo [17]2 years ago

4 0

N = 1
l = from 0 to (n-1)
ml = -1... + 1
ms = 1/2 or -1/2

eg = 2s
n = 2, m = 0, n = 0
s = 1/2, -1/2

hope this help

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Determinación de pH Expresa las siguientes concentraciones de [H+ ] en función del pH • [H+] = 0.001 M • [H+] = 0.002 M • [H+] =

Pavel [41]

Answer:

• pH = 3.0

• pH = 2.70

• pH = 3.61

• pH = 8.28

• pH = 1.40

Explanation:

El pH es una medida en química usada para determinar el grado de acidez o basicidad en una solución.

Se define como:

pH = -log₁₀ [H⁺]

El - logaritmo de la concentración molar de H⁺

Para las concentraciones de H⁺ dadas:

• [H+] = 0.001 M

pH = -log (0.001M) = 3

pH = 3.0

• [H+] = 0.002 M

pH = -log (0.002M)

pH = 2.70

• [H+] = 2.45X10-4 M

pH = -log (2.45X10-4 M )

pH = 3.61

• [H+] = 5.2X10-9 M

pH = -log (5.2X10-9 M)

pH = 8.28

• [H+] = 0.04 M

pH = -log (0.04M)

pH = 1.40

8 0

2 years ago

If 36.9 mL of B2H6 reacted with excess oxygen gas, determine the actual yield of B2O3 if the percent yield of B2O3 was 75.7%. (T

zhuklara [117]

Answer: The actual yield of $B_2O_3$ is 60.0 g

Explanation:-

The balanced chemical reaction :

$B_2H_6(l)+3O_2(g)\rightarrow B_2O_3(s)+3H_2O(l)$

Mass of $B_2H_6$ = $Density\times Volume=1.131g/ml\times 36.9ml=41.7g$

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} B_2H_6=\frac{41.7g}{27.668g/mol}=1.51moles$

According to stoichiometry:

1 mole of $B_2H_6$ gives = 1 mole of $B_2O_3$

1.51 moles of $B_2H_6$ gives = $\frac{1}{1}\times 1.51=1.51$ moles of $B_2O_3$

Theoretical yield of $B_2O_3=moles\times {Molar mass}}=1.14mol\times 69.62g/mol=79.3g$

Percent yield of $B_2O_3$ = $75.7\%$

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

$75.7\%=\frac{\text{Actual yield}}{79.3}\times 100$

${\text{Actual yield}}=60.0g$

Thus the actual yield of $B_2O_3$ is 60.0 g

7 0

2 years ago

How many grams are in 2.5 pound sample

julsineya [31]

About 2,500 grams Ans balkfdoaks;

5 0

2 years ago

What is the pH of a 0.050 M triethylamine, (C2H5)3N, solution? Kb for triethylamine is 5.3 ´ 10-4. Question options: 1) 11.69 2)

blsea [12.9K]

the pH of a 0.050 M triethylamine, is 11.70

For triehtylamine, $(C_{2}H_{5})_{3}N$ , the reaction will be
$(C_{2}H_{5})_{3}N + H_{2}O ---\ \textgreater \ ( C_{2}H_{5})_{3}NH^{+} + OH^{-}$

and we know, pH = -log[H+] and pOH = -log[OH-]

Also, pOH + pH = 14

Now, the Kb value = 5.3 x 10^-4

And $kb = \frac{( [( C_{2}H_{5})_{3}NH^{+} ]* OH^{-} )}{[( C_{2}H_{5})_{3}N]}$

thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050

=0.00516 M

Thus, pOH = 2.30
pH = 14 - pOH = 11.7

6 0

2 years ago

Acetaldehyde shows two UV bands, one with a lmax of 289 nm ( 5 12) and one with a lmax of 182 nm ( 5 10,000). Which is the n p*

UkoKoshka [18]

Answer:

The bands are due to:

λmax = 289 nm n→π* transition (E = 12)

λmax = 182 nm π→π* transition (E=10000)

Explanation:

The two types of acetaldehyde transition are as follows:

n→π* and π→π*

From the attached diagram we have to:

ΔEn→π* < ΔEπ→π*

ΔEα(1/λ)

Thus:

λn→π* > λπ→π*

In n→π* spin forbidden, the intensity is low. Thus, the molar extinction E for n→π* is very low.

The same way, for π→π* spin allowed the intensity is high. Thus, the molar extinction coefficient E for π→π* is high too.

The bands are due to:

λmax = 289 nm n→π* transition (E = 12)

λmax = 182 nm π→π* transition (E=10000)

5 0

2 years ago