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Shtirlitz [24]
3 years ago
11

How is the element’s period related to the number of energy levels over which its electrons are spread?

Chemistry
1 answer:
Tanzania [10]3 years ago
7 0

Answer:

my turn.:). i think that the more dense the state of an element is the more energy it will have.

Explanation:

if somthing is tightly packed electricity can pass through it very easy but is somthing like oxegen is spread out it can not pass through very easily. hope i helped. :)

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Chameleons have the ability to change their coloring to help them blend in with their environmerit.
Andre45 [30]

Answer:

B

Explanation:

trust me

3 0
3 years ago
Read 2 more answers
er solutions can be produced by mixing a weak acid with its conjugate base or by mixing a weak base with its conjugate acid. The
liubo4ka [24]

Answer:

You need to add 19,5 mmol of acetates

Explanation:

Using the Henderson-Hasselbalch equation:

pH = pKa + log₁₀ [base]/[acid]

For the buffer of acetates:

pH = pKa + log₁₀ [CH₃COO⁻]/[CH₃COOH]

As pH you want is 5,03, pka is 4,74 and milimoles of acetic acid are 10:

5,03 = 4,74 + log₁₀ [CH₃COO⁻]/[10]

1,95 = [CH₃COO⁻]/[10]

<em>[CH₃COO⁻] = 19,5 milimoles</em>

Thus, to produce an acetate buffer of 5,03 having 10 mmol of acetic acid, you need to add 19,5 mmol of acetates.

I hope it helps!

7 0
4 years ago
The quantity (current times voltage ) is
maksim [4K]

Answer:

Electrical power

Explanation:

An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.

Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.

The electrical power of an electric circuit can be defined as a measure of the rate at which energy is either produced or absorbed in the circuit.

<em>Mathematically, electrical power is given by the formula;</em>

Electrical \; power = current * voltage

This ultimately implies that, the quantity (current times voltage ) is electrical power and it is measured (S.I units) in Watt (W).

7 0
3 years ago
Muiltplying 2.5 x 10^10 by 3.5 x 10^-7
andrey2020 [161]

<u>Answer: </u>

The value of \left(2.5 \times 10^{10}\right) \times\left(3.5 \times 10^{-7}\right) is 8750

<u>Solution: </u>

\left(2.5 \times 10^{10}\right) \times\left(3.5 \times 10^{-7}\right)

\Rightarrow\left(2.5 \times 10^{10} \times 3.5 \times 10^{-7}\right)

\Rightarrow\left(2.5 \times 3.5 \times\left(10^{10} \times 10^{-7}\right)\right)

\Rightarrow(2.5 \times 3.5) \times\left(10^{10+(-7)}\right)

\Rightarrow(2.5 \times 3.5) \times\left(10^{10+(-7)}\right)

\Rightarrow(2.5 \times 3.5) \times 10^{3}

\Rightarrow 8.75 \times 10^{3}

\Rightarrow 8.75 \times 1000=8750

6 0
3 years ago
g Using the complex based titration system: 50.00 mL 0.00250 M Ca2 titrated with 0.0050 M EDTA, buffered at pH 11.0 determine (i
kobusy [5.1K]

Answer:

Explanation:

a).

conc of Ca²⁺ =0.0025 M

pCa = -log(0.0025) = 2.6

logK,= 10.65 So lc = 4.47 x 10.

Formation constant of Ca(EDTA)]-z= 4.47 x 10¹⁰ At pH = 11, the fraction of EDTA that exists Y⁻⁴ is  \alpha_{Y^{-4}} =0.81

So the Conditional Formation constant= K_f =0.81x 4.47 x10¹⁰

=3.62x10¹⁰

b)

At Equivalence point:

Ca²⁺ forms 1:1 complex with EDTA At equivalence point,

Number of moles of Ca²⁺= Number of moles of EDTA Number of moles of Ca²⁺ = M×V = 0.00250 M × 50.00 mL = 0.125 mol

Number of moles of EDTA= 0.125 mol

Volume of EDTA required = moles/Molarity = 0.125 mol / 0.0050 M = 25.00 mL  

V e= 25.00 mL  

At equivalence point, all Ca²⁺ is converted to [CaY²⁻] complex. So the concentration of Ca²⁺ is determined by the dissociation of [CaY²⁻] complex.  

[CaY^{2-}] = \frac{Initial,moles,of, Ca^{2+}}{Total,Volume} = \frac{0.125mol}{(50.00+25.00)mL} = 0.001667M

                                                            {K^'}_f

                       Ca²⁺      +      Y⁴          ⇄     CaY²⁻

Initial                 0                  0                      0.001667

change             +x                  +x                     -x

equilibrium        x                    x                    0.001667 - x

{K^'}_f = \frac{[CaY^{2-}]}{[Ca^{2+}][Y^4]}=\frac{0.001667-x}{x.x} =\frac{0.001667-x}{x^2}\\\\x^2 = \frac{0.001667-x}{{K^'}_f}\\ \\

x^2=\frac{0.001667}{3.62\times10^{10}}=4.61\exp-14

x = 2.15×10⁻⁷

[Ca+2] = 2.15x10⁻⁷ M  

pca = —log(2 15x101= 6.7

3 0
4 years ago
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