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seraphim [82]
3 years ago
10

How far apart must two point charges of 75.0 nC (typical of static electricity) be to have a force of 1.00 N between them? (answ

er in mm)
Chemistry
1 answer:
Jet001 [13]3 years ago
3 0

Answer:

7.12 mm

Explanation:

From coulomb's law,

F = kqq'/r².................... Equation 1

Where F = force, k = proportionality constant, q and q' = The two point charges, r = distance between the two charges.

Make r the subject of the equation,

r = √(kqq'/F).......................... Equation 2

Given: q = q' = 75.0 nC = 75×10⁻⁹ C, F = 1.00 N

Constant: k = 9.0×10⁹ Nm²/C².

Substitute into equation 2

r = √[ (75×10⁻⁹ )²9.0×10⁹/1]

r = 75×10⁻⁹.√(9.0×10⁹)

r = (75×10⁻⁹)(9.49×10⁴)

r = 711.75×10⁻⁵

r = 7.12×10⁻³ m

r = 7.12 mm

Hence the distance between the point charge = 7.12 mm

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3 years ago
Calculate the concentration of A bottle of wine contains 12.9% ethanol by volume. The density of ethanol (CH3OH) is 0.789 g/cm e
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Answer:

The mass percentage of the solution is 10.46%.

The molality of the solution is 2.5403 mol/kg.

Explanation:

A bottle of wine contains 12.9% ethanol by volume.

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Mass of the ethanol = m

Density of the ethanol ,d= 0.789 g/cm^3=0.789 g/mL

1 cm^3=1 mL

m=d\times v=0.798 g/ml\times 12.9 mL = 10.1781 g

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M=D\times V=1.00 g/ml\times 87.1 mL =87.1 g

Mass percent

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m=\frac{m}{\text{molar mass of ethanol}\times M(kg)}

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0.100 g Pb × 100 g Water/0.0012 g Pb = 8.3 × 10³ g Water

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Since the solution is diluted, we will assume the density of the sample is equal than the density of water (1 g/mL).

8.3 × 10³ g × 1 mL/1 g = 8.3 × 10³ mL

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