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lukranit [14]
4 years ago
10

The charge per unit length on a glass rod is 0.00500 C/m. If the rod is 1 mm long, how many electrons have been removed from the

glass rod?
Chemistry
1 answer:
ehidna [41]4 years ago
5 0

Answer:

3.125 × 10¹³ Electrons

Explanation:

Data provided in the question:

Charge per unit length on rod = 0.00500 C/m

Length of the rod = 1 mm = 1 × 10⁻³ m

Therefore, the total charge on the rod

= Charge per unit length on rod × Length of the rod

= 0.00500 C/m × ( 1 × 10⁻³ m)

= 5  × 10⁻⁶ C

Thus,

Number of electrons removed

=  total charge on the rod ÷ Charge of an electron

= 5  × 10⁻⁶ ÷ (1.6 × 10⁻¹⁹)

= 3.125 × 10¹³ Electrons

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The radius of an atom is closest in size to a
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3 years ago
Calculate how many times more soluble Mg(OH)2 is in pure water Based on the given value of the Ksp, 5.61×10−11, calculate the ra
maks197457 [2]

Answer:

molar solubility in water = 2.412 * 10^-4  mol/L

molar solubility of NaOH in 0.130M = 3.32 * 10^-9 mol/L

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

Explanation:

The Ksp refers to the partial solubilization of a mostly insoluble salt. This is an equilibrium process.

 

The equation for the solubilization reaction of Mg(OH)2 can be given as:

 

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

 Ksp can then be given as followed:

Ksp = [Mg^2+][OH^–]²  

<u>Step 2:</u> Calculate the solubility in water

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X of OH-

The concentration at equilibrium will be XM Mg^2+ and 2X OH-

Ksp = [Mg^2+][OH^–]²  

5.61*10^-11 = X * (2X)² = X *4X² = 4X³

 X = <u>2.412 * 10^-4 mol/L = solubility in water</u>

<u>Step 3</u>: Calculate solubility in 0.130 M NaOH

The initial concentration of Mg^2+  = 0 M

The initial concentration of OH- = 0.130 M

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X +0.130 for OH-

The concentration at equilibrium will be XM Mg^2+ and 0.130 + 2X OH-

The value of "[OH–] + 2X" is, because the very small value of X, equal to the value of [OH–] .

Let's consider:

[Mg+2] = X

[OH] = 0.130

Ksp = [Mg^2+][OH^–]²  

5.61*10^-11 = X *(0.130)²  

5.61*10^-11 = X * (0.130)^2

X = <u>3.32*10^-9 = solubility in 0.130 M NaOH </u>

<u>Step 4:</u> Calculate how many times Mg(OH)2 is better soluble in pure water.

(2.412*10^-4)/ (3.32*10^-9) = 0.73 * 10^5

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

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Serga [27]

Answer:

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Explanation:

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I mole of photons contains Avogadro's number of particles, 6.02 × 10^23

Therefore, the energy per mile of photon is 3.84 × 10^-22 KJ × 6.02 × 10^23

Energy per mile of photon = 2.31 × 10^2 KJ/mol

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