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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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Somebodyy help me asap plz i can’t do this
Solnce55 [7]
Negative charge is the answer
5 0
3 years ago
Just enough 0.500 M HCl is added to 30.0 mL of 2.5 M NH3 to reach the equivalence point. The Kb of NH3 = 1.8 X 10-5
algol13

a.NH₃+HCl⇒NH₄Cl

b.volume HCl=150 ml

c. pH=4.82

<h3>Further explanation</h3>

Reaction

NH₃+HCl⇒NH₄Cl

The equivalence point⇒mol NH₃=HCl

Titration formula :

M₁V₁n₁=M₂V₂n₂(n=acid base valence, NH₃=HCl=1)

mol NH₃

\tt 2.5\times 30=75~mlmol

mol HCl=75 mlmol

  • Volume HCl :

\tt \dfrac{75}{0.5}=150~ml

Volume total :

\tt 150+30=180~ml

  • molarity of salt(NH₄Cl)

mol NH₄Cl=mol NH₃=75 mlmol=0.075 mol

\tt M=\dfrac{0.075}{0.180}= 0.42

  • pH of solution

Dissociation of NH₄Cl at water to find [H₃O⁺]

\tt NH_4+H_2O\rightarrow NH_3+H_3O^+

ICE at equilibrium :

0.41-x            x        x

Ka(Kw:Kb)= 10⁻¹⁴ : 1.8.10⁻⁵=5.6.10⁻¹⁰

\tt Ka=\dfrac{NH_3.H_3O}{NH_4}=\dfrac{x^2}{0.41}

[H₃O⁺]=x :

\tt \sqrt{5.6.10^{-10}\times 0.41}=1.515.10^{-5}

pH=-log[H₃O⁺]

\tt pH=5-log~1.515=4.82

7 0
3 years ago
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First, let's write the chemical formula for each of the substances mentioned in the problem.

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Lithium Chloride: LiCl

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