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Aleksandr-060686 [28]
3 years ago
5

Visible light, X Rays infrared radiation and radio waves all have the same

Chemistry
1 answer:
ryzh [129]3 years ago
8 0

The answer is speed of light. All kinds of light waves travel at 3.00 m/s.

You might be interested in
How much heat is required to take a 150 g sample of water from 10.0 ℃ to 95.0 ℃? cs,water = 4.184 J/g*℃
oee [108]

Answer don't know at all but i will find out by tomorrow

Explanation: sorry

6 0
3 years ago
You made up a saturated solution of calcium sulfate (CaSO4). The temperature is 25°C. You then add 5.00 × 10−3 M sodium sulfate
BARSIC [14]

Answer:

Ca^{+2}=5.13x10^{-3}M\\SO_4^{-2}=0.010 M

Explanation:

Hello,

At first, the answer is wrong, consider the following procedure:

- the pKs, leads to a Ks of:

pKs=-Log(Ks)

Ks=10^{-4.58}=2.63x10^{-5}

- now, stating the equilibrium for the calcium sulfate (whereas x is the change during the equilibrium):

Ks=[Ca^{+2}]^{eq}[SO_{4}^{-2}]^{eq}\\Ks=[x]^{eq}[x]^{eq}\\Ks=x^{2}\\x=\sqrt{2.63x10^{-5}}\\x=5.13x10^{-3} M

- x becomes the concentration of calcium ions and one of the contributions for the total concentration of sulfate ions, thus:

[Ca^{+2}]^{eq}=5.13x10^{-3} M

- finally, since the sodium sulfate is totally dissolved, we just simply add its concentration (equal to the sulfate ions yielded by itself) to the outgoing sulfate concentration from the calcium sulfate:

[SO_4^{-2}]^{total}=5x10^{-3}M+5.13x10^{-3}M\\[SO_4^{-2}]^{total}=0.010 M

Bibliography:https://books.google.com.co/books?id=LQwyfFc3pXsC&lpg=PA243&ots=zIJN3nSFWE&dq=pKs%20chemistry&pg=PA244#v=onepage&q=pKs%20chemistry&f=false page 243

Best regards!

7 0
3 years ago
Read 2 more answers
Where is the only place the weak necular force can be seen
kykrilka [37]
I’m not that smart sorry
3 0
4 years ago
Read 2 more answers
CH3COOH  CH3COO– + H+
Oxana [17]

(a)

pH = 4.77

; (b)

[

H

3

O

+

]

=

1.00

×

10

-4

l

mol/dm

3

; (c)

[

A

-

]

=

0.16 mol⋅dm

-3

Explanation:

(a) pH of aspirin solution

Let's write the chemical equation as

m

m

m

m

m

m

m

m

l

HA

m

+

m

H

2

O

⇌

H

3

O

+

m

+

m

l

A

-

I/mol⋅dm

-3

:

m

m

0.05

m

m

m

m

m

m

m

m

l

0

m

m

m

m

m

l

l

0

C/mol⋅dm

-3

:

m

m

l

-

x

m

m

m

m

m

m

m

m

+

x

m

l

m

m

m

l

+

x

E/mol⋅dm

-3

:

m

0.05 -

l

x

m

m

m

m

m

m

m

l

x

m

m

x

m

m

m

x

K

a

=

[

H

3

O

+

]

[

A

-

]

[

HA

]

=

x

2

0.05 -

l

x

=

3.27

×

10

-4

Check for negligibility

0.05

3.27

×

10

-4

=

153

<

400

∴

x

is not less than 5 % of the initial concentration of

[

HA

]

.

We cannot ignore it in comparison with 0.05, so we must solve a quadratic.

Then

x

2

0.05

−

x

=

3.27

×

10

-4

x

2

=

3.27

×

10

-4

(

0.05

−

x

)

=

1.635

×

10

-5

−

3.27

×

10

-4

x

x

2

+

3.27

×

10

-4

x

−

1.635

×

10

-5

=

0

x

=

1.68

×

10

-5

[

H

3

O

+

]

=

x

l

mol/L

=

1.68

×

10

-5

l

mol/L

pH

=

-log

[

H

3

O

+

]

=

-log

(

1.68

×

10

-5

)

=

4.77

(b)

[

H

3

O

+

]

at pH 4

[

H

3

O

+

]

=

10

-pH

l

mol/L

=

1.00

×

10

-4

l

mol/L

(c) Concentration of

A

-

in the buffer

We can now use the Henderson-Hasselbalch equation to calculate the

[

A

-

]

.

pH

=

p

K

a

+

log

(

[

A

-

]

[

HA

]

)

4.00

=

−

log

(

3.27

×

10

-4

)

+

log

(

[

A

-

]

0.05

)

=

3.49

+

log

(

[

A

-

]

0.05

)

log

(

[

A

-

]

0.05

)

=

4.00 - 3.49

=

0.51

[

A

-

]

0.05

=

10

0.51

=

3.24

[

A

-

]

=

0.05

×

3.24

=

0.16

The concentration of

A

-

in the buffer is 0.16 mol/L.

hope this helps :)

6 0
2 years ago
For a hydrogen- like atom, classify these electron transitions by whether they result in the absorption or emission of light: n=
Nataly_w [17]

The electron transfer from n = 1 to n = 3 occurs the greatest energy change

<h3>Further explanation</h3>

In an atom, there are energy levels in the skin and sub skin.

This energy level is expressed in terms of electron configurations.

Writing the electron configuration starts from the lowest to the highest subshell's energy level. There are 4 sub-shells in an atom's shell, namely s, p, d, and f. The maximum number of electrons for each subshell is

  • s: 2 electrons
  • p: 6 electrons
  • d: 10 electrons and
  • f: 14 electrons

The filling of electrons uses the following sequence:

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc.

Each sub-shell also has orbitals drawn in the form of a square box in which there are electrons symbolized by half arrows.

Each orbital in an atom consists of 4 quantum numbers, namely the main quantum number (n), the azimuth quantum number (l), and the magnetic quantum number (m) and the spin quantum number (s)

  • Value of n: positive integer
  • value of l: between 0 to n-1
  • m value: between -l to + l
  • value s: +1/2 or -1/2

Determination of electron configurations based on principles:

  • 1. Aufbau: Electrons occupy orbitals of the lowest energy level
  • 2 Hund: electron fills orbitals with the same energy level
  • 3. Pauli: there are no electrons that have 4 equal quantum numbers

Bohr's atomic model has shown the energy levels of electrons in the path of the atomic shell

The greater the value of n (the atomic shell, the main quantum number), the greater the energy level

In normal circumstances, electrons fill the skin at the lowest energy level starting from the skin K, L M and then N

When an atom gets energy from outside, the electrons will absorb energy so it moves to higher energy. This situation is called excited

Electrons will return to the original path or a lower energy level because the excited state is unstable. In this condition, the electron will release energy

The electron energy at the nth path can be formulated:

En = -Rh / n²

Rh = constant 2.179.10⁻¹⁸ J

So the electron transfer energy (delta E)

delta E = E end - E start

From the electron transfer available, because the value of the Rh constant is the same, the effect is the value of n (skin) ---> 1/n²

  • 1. n = 3 to n = 5

then the value of E is

delta E = -1/25 +1/9 = 0.07

  • 2. n = 1 to n = 3,

then the value of E is

delta E = -1/9 + 1/1 = 0.89

  • 3.n = 3 to n = 2,

then the value of E is

delta E = -1 / 9 + 1/16 = 0.14

  • 4. n = 2 to n = 1

then the value of E is

delta E = -1/4 + 1 = 0.75

These 4 values ​​indicate that regardless of the sign that there is an electron transfer from n = 1 to n = 3 the greatest energy change occurs

<h3>Learn more</h3>

 statement about electrons and atomic orbitals

brainly.com/question/1832385

Effective nuclear charge

brainly.com/question/5441986

statement about subatomic particles is true

brainly.com/question/3176193

Keywords: electron transfer, quantum number, electron shell

6 0
4 years ago
Read 2 more answers
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