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anyanavicka [17]
3 years ago
12

-------,-------,--------,and---------- are the four types of main waves

Chemistry
1 answer:
myrzilka [38]3 years ago
4 0

Answer:

As far as i know there are only three types of waves: transverse waves, longitudinal waves, and surface waves. Though maybe I'm incorrect.

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At 25 ∘C only 0.0180 mol of the generic salt AB is soluble (at equilibrium) in 1.00 L of water.What is the value for Ksp for thi
Ghella [55]

Answer:

Ksp = 3.24 x 10⁻⁴

Explanation:

The dissociation equilibrium for a generic salt AB is:

AB(s) ⇄ A⁺(aq) + B⁻(aq)

              s            s

For instance, the expression for the Ksp constant is:

Ksp = [A⁺] [B⁻] = s x s = s²

According to the problem, 0.0180 mol of the salt is soluble in 1.00 L os water. That means that the solubility of the salt (s) is equal to 0.0180 mol per liter.

s = moles of solute/L of solution = 0.0180 mol/L

Thus, we calculate Ksp from the s value as follows:

Ksp = s² = (0.0180)² = 3.24 x 10⁻⁴

6 0
3 years ago
Sodiun atom, potassium atom, and cesium atoms have the same ?
Tpy6a [65]
Sodium atom , potassium atom and cesium atom have the same group number which is group 1
3 0
3 years ago
Read 2 more answers
Which statement accurately describes the rock layers
ipn [44]

Answer:

Layer 8 is older than layer 1

Explanation:

7 0
3 years ago
An atom has radius of 227 pm and crystallizes in a body-centered cubic unit cell. What is the volume of the unit cell
sdas [7]

We will see that the volume of the unit cell is 144,070,699.06 pm^3

<h3>How to get the volume of a body-centered cubic unit cell?</h3>

In a body-centered cubic unit cell, the side length of the cube is given as:

S = \frac{4}{\sqrt{3} } *R

Where R is the radius of the atom.

And the volume of a cube is the side length cubed, then we can see that the volume of our cube will be:

V = S^3 = (\frac{4}{\sqrt{3} }*227pm)^3

Solving that we get:

V  = (\frac{4}{\sqrt{3} }*227pm)^3 = 144,070,699.06pm^3

This is the approximated volume of the unit cell.

If you want to learn more about unit cell structures, you can read:

brainly.com/question/13110055

5 0
2 years ago
An acidified solution was electrolyzed using copper electrodes. A constant current of 1.18 A caused the anode to lose 0.584 g af
Alexxx [7]

Answer:

\boxed{\text{(a) 209 mL; (b) } 6.09 \times 10^{23}}

Explanation:

(a) Gas produced at cathode.

(i). Identity

The only species known to be present are Cu, H⁺, and H₂O.

Only the H⁺ and H₂O can be reduced.

The corresponding reduction half reactions are:

(1) 2H₂O + 2e⁻ ⇌ H₂ + 2OH⁻;     E° = -0.8277 V

(2) 2H⁺ +2e⁻ ⇌ H₂;                     E° =  0.0000 V

Two important points to remember when using a table of standard reduction potentials:

  • The higher up a species is on the right-hand side, the more readily it will lose electrons (be oxidized).
  • The lower down a species is on the left-hand side, the more readily it will accept electrons (be reduced}.

H⁺ is below H₂O, so H⁺ is reduced to H₂.

The cathode reaction is 2H⁺ +2e⁻ ⇌ H₂, and the gas produced at the cathode is hydrogen.

(ii) Volume

a. Anode reaction

The only species that can be oxidized are Cu and H₂O.

The corresponding half reactions  are:

(3) Cu²⁺ + 2e⁻ ⇌ Cu;                E° =  0.3419 V

(4) O₂ + 4H⁺ + 4e⁻ ⇌ 2H₂O     E° =   1.229   V

Cu is above H₂O, so Cu is more easily oxidized.

The anode reaction is Cu ⇌ Cu²⁺ + 2e⁻.

b. Overall reaction:

Cu           ⇌ Cu²⁺ + 2e⁻

<u>2H⁺ +2e⁻ ⇌ H₂            </u>        

Cu + 2H⁺ ⇌ Cu²⁺ + H₂

c. Moles of Cu lost

n_{\text{Cu}} = \text{0.584 g } \times \dfrac{\text{1 mol}}{\text{63.55 g}} = 9.190 \times 10^{-3}\text{ mol Cu}

d. Moles of H₂ formed

n_{\text{H}_{2}}} = 9.190 \times 10^{-3}\text{ mol Cu} \times \dfrac{\text{1 mol H}_{2}}{\text{1 mol Cu}} =9.190 \times 10^{-3}\text{ mol H}_{2}

e. Volume of H₂ formed

Volume of 1 mol at STP (0 °C and  1 bar) = 22.71 mL

V = 9.190 \times 10^{-3}\text{ mol}\times \dfrac{\text{22.71 L}}{\text{1 mol}}  = \text{0.209 L} = \boxed{\textbf{209 mL}}

(b) Avogadro's number

(i) Moles of electrons transferred

\text{Moles of electrons} = 9.190 \times 10^{-3}\text{ mol Cu}\times \dfrac{\text{2 mol electrons}}{\text{1 mol Cu}}\\\\\\= \text{0.018 38 mol electrons}

(ii) Number of coulombs

Q  = It  

Q = \text{1.18 C/s} \times 1.52 \times 10^{3} \text{ s} = 1794 C

(iii). Number of electrons

n = \text{ 1794 C} \times \dfrac{\text{1 electron}}{1.6022 \times 10^{-19} \text{ C}} = 1.119 \times 10^{22} \text{ electrons}

(iv) Avogadro's number

N_{\text{A}} = \dfrac{1.119 \times 10^{22} \text{ electrons}}{\text{0.018 38 mol}} = \boxed{6.09 \times 10^{23} \textbf{ electrons/mol}}

6 0
3 years ago
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