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2 months ago

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A metallic object holds a charge of −4.8 × 10−6 C. What total number of electrons does this represent? (e = 1.6 × 10−19 C is the

magnitude of the electronic charge.)

1 answer:

Anarel [2.9K]2 months ago

5 0

Answer:

$n=3.0\times 10^{13}$

Explanation:

Charge of one electron = $-1.6\times 10^{-19}\ C$

The formula for calculating charge is:

$Charge=n\times q_e$

Given that: Charge = $-4.8\times 10^{-6}\ C$

$-4.8\times 10^{-6}=n\times (-1.6\times 10^{-19})$

$n=\frac{4.8\times 10^{-6}}{1.6\times 10^{-19}}=3.0\times 10^{13}$

Total electrons, n = $3.0\times 10^{13}$

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Answer:

The integer value of x in the hydrate is 10.

Explanation:

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Molar concentration of the solution = 0.0366 M

Volume of the solution = 5.00 L

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By solving for x, we arrive at:

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1 month ago

Grignard reagents are air-and moisture-sensitive. List at least threereactants, solvents, and/or techniques that were utilized i

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Diethyl ether (DTH) and Tetrahydrofuran (THF).

Clarification:

Grignard reactions react with water, resulting in the formation of alkanes. The presence of water leads to rapid decomposition of the reagent.

Therefore, solvents like anhydrous diethyl ether or tetrahydrofuran (THF), as well as poly(tetramethylene ether) glycol (PTMG), are used in experimental procedures to limit the exposure of Grignard reagents to air and moisture.

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1 month ago

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1) Drift velocity: $3.32\cdot 10^{-4}m/s$

2. $5.6\cdot 10^{13}$ electrons per individual

Explanation:

1)

In a conducting material with an electric current, the drift velocity of electrons can be calculated using this equation:

$v_d=\frac{I}{neA}$

where

I stands for current

n represents the density of free electrons

$e=1.6\cdot 10^{-19}C$ indicates the charge of an electron

A signifies the wire's cross-sectional area

The wire's cross-sectional area can be determined as

$A=\pi r^2$

where r denotes the wire's radius. Thus, the equation transforms to

$v_d=\frac{I}{ne\pi r^2}$

In this scenario, we have:

I = 8.0 A as the current

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d = 1.5 mm is the diameter, making the radius

r = 1.5/2 = 0.75 mm = $0.75\cdot 10^{-3}m$

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2)

The entire length of the cord is

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And the cross-sectional area is

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$N'=8\cdot 10^9$

Hence, the number of electrons distributed to each person is:

$N_e = \frac{N}{N'}=\frac{4.5\cdot 10^{23}}{8\cdot 10^9}=5.6\cdot 10^{13}$

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Answer:

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1 month ago

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