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27 days ago

Suppose that one of millikan's oil drops has a charge of −4.8×10−19

Chemistry

2 answers:

eduard [2.5K]27 days ago

8 0

Answer: The drop has 3 excess electrons.

Explanation:

Provided:

Millikan's oil drop charge = $4.8\times 10^{-19}C$

To determine the number of extra electrons, we divide the charge of the oil drop by the charge of a single electron.

$\text{Excess electrons}=\frac{\text{Charge on millikan's oil drop}}{\text{Charge on 1 electron}}$

It is known that:

Charge of one electron = $1.60\times 10^{-19}C$

Substituting the values into the equation gives:

$\text{Excess electrons}=\frac{4.8\times 10^{-19}C}{1.60\times 10^{-19}C}\\\\\text{Excess electrons}=3$

Consequently, the drop contains 3 excess electrons.

alisha [2.7K]27 days ago

8 0

The correct choice is c.

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How many moles of nitrogen gas are there in 6.8 liters at room temperature and pressure (293 K and 100 kPa)?

Alekssandra [2719]

Utilize the ideal gas law:

n = PV / RT

P = 100kPa = 100 x 1000 x (9.8 x 10^{-6}) = 0.98 atm
Convert kPa to atm, where 1 Pa = 9.8 x 10^{-6} atm.
T = 293 K
V = 6.8 L
R = 1/12
Substituting all values leads to:
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4 0

1 month ago

What is the percentage by mass of solution formed by dissolving 36.0 grams of HCl in 98.0 grams of water?

eduard [2520]

Convert HCl and H2O to moles.

36.0 g of HCl = 0.987 moles HCl

98.0 g of H2O = 5.44 moles H2O

Based on the stoichiometric ratio for HCl,
there are 0.987 moles of H and 0.987 moles of Cl.

For H₂O, according to the stoichiometric ratio, you have 10.88 moles of H and 5.44 moles of O.

Combining them:
11.867 moles H
0.987 moles Cl
5.44 moles O

Revert the moles back to grams, then divide by the total mass and multiply by 100 for the percentage by mass.

11.867 moles H = 11.96 g H
0.987 moles Cl = 34.99 g Cl
5.44 moles O = 87.03 g O

11.96/(36.0+98.0)(100) = 8.93% for H
34.99/(36.0+98.0)(100) = 26.11% for Cl
87.03/(36.0+98.0)(100) = 64.96% for O.

4 0

26 days ago

Read 2 more answers

A scientist measures the speed of sound in a monatomic gas to be 449 m/s at 20∘C. What is the molar mass of this gas?

Tems11 [2403]

Answer:

The molar mass of the gas is 36.25 g/mol.

Explanation:

To determine this, we utilize the mathematical relationship:

ν = $\sqrt{3RT/M}$

Where, ν represents the speed of light in a gas (ν = 449 m/s),

R denotes the universal gas constant (R = 8.314 J/mol.K),

T stands for the temperature of the gas in Kelvin (T = 20 °C + 273 = 293 K),

M is the molar mass of the gas in (Kg/mol).

ν = $\sqrt{3RT/M}$

(449 m/s) = √(3(8.314 J/mol.K)(293 K)/M,

by squaring both sides:

(449 m/s)² = (3(8.314 J/mol.K)(293 K))/M,

thus M = (3(8.314 J/mol.K)(293 K)/(449 m/s)² = 7308.006/201601 = 0.03625 Kg/mol.

Thus, the molar mass of the gas is 36.25 g/mol.

7 0

8 days ago

The elements X and Y combine in different ratios to form four different types of compounds: XY, XY2, XY3, and XY4. Consider that

eduard [2520]

Answer:

The ratios arranged in ascending order are; The ratio of the mass of Y to X in XY2 divided by the mass of Y to X in XY, The ratio of the mass of Y to X in XY3 divided by the mass of Y to X in XY, The ratio of the mass of Y to X in XY4 divided by the mass of Y to X in XY

1) Mass ratio = 3

2) Mass ratio = 2

3) Mass ratio = 4

Explanation:

Comprehensive calculations are displayed in the attachment.

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11 days ago

If you pump air into cycle tyre a slight warming effect is noticed at valve stem why

castortr0y [2743]

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27 days ago