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2 years ago

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

Chemistry

2 answers:

svlad2 [7]2 years ago

8 0

The best and correct option is c

Vanyuwa [196]2 years ago

8 0

Answer: The number of excess electrons that the drop contains are 3.

Explanation:

We are given:

Charge on millikan's oil drops = $4.8\times 10^{-19}C$

To calculate the number of excess electrons, we divide the charge on millikan's oil drop by the change on 1 electron.

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

We know that:

Charge on 1 electron = $1.60\times 10^{-19}C$

Putting values in above equation, we get:

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

Hence, the number of excess electrons that the drop contains are 3.

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A chef wants to make 1 gallon (128 ounces) of a 15% vinegar-to-oil salad dressing. He only has pure vinegar and a mild 4% vinega

noname [10]

Answer:

Pure vinegar: 12.2 ounces, 4% mix salad: 115.8 ounces

Explanation:

The chef wants to male a 15% vinegar-to-oil salad.

If we call:

$v$ the amount of vinegar in the salad

$o$ the amount of oil in the salad

This means that

$\frac{v}{o}=\frac{15}{100}$

In order to get this salad, the chef has to mix:

- Pure vinegar, which has 100% concentration of vinegar

- A 4% vinegar-to-oil salad: this means that the amount of vinegar in this salad is $v=0.04o$ (4% of the amount of oil)

This means that the total amount of vinegar in the final salad will be:

$v'=v+0.04o$ (1a)

Where v is the amount of pure vinegar added and $o$ is the amount of oil in the 4% salad

While the amount of oil needed is $o'=o$

So we have, since the final salad has 15% concentration of vinegar to oil:

$\frac{v+0.04o}{o}=\frac{15}{100}$ (1)

Moreover, we know that the final volume must be 128 ounces, so

$v'+o=128$ (2)

From eq.(2) and (1a) we get

$v+0.04o+o=128\\v=128-1.04o$

Substituting into (1) and solving for $o$ ,

$\frac{128-1.04o+0.04o}{o}=\frac{15}{100}\\100(128-o)=15o\\20(128-o)=3o\\2560-20o=3o\\o=\frac{2560}{23}=111.3$

Therefore the amount of vinegar must be

$v=128-1.04o=128-(1.04)(111.3)=12.2$

So, the amount of each that should be added is:

- Pure vinegar: 12.2 ounces

- 4% vinegar-to-oil mix: $1.04o=(1.04)(111.3)=115.8$ ounces

5 0

2 years ago

Be sure to answer all parts. Consider the formation of ammonia in two experiments. (a) To a 1.00−L container at 727°C, 1.30 mol

Sonbull [250]

Answer: The value of $K_c$ for $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ reaction is $5.13\times 10^2$

Explanation:

We are given:

Initial moles of nitrogen gas = 1.30 moles

Initial moles of hydrogen gas = 1.65 moles

Equilibrium moles of ammonia = 0.100 moles

Volume of the container = 1.00 L

For the given chemical equation:

$N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$

Initial: 1.30 1.65

At eqllm: 1.30-x 1.65-3x 2x

Evaluating the value of 'x'

$\Rightarrow 2x=0.100\\\\\Rightarrow x=0.050mol$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NH_3]^2}{[N_2]\times [H_2]^3}$

Equilibrium moles of nitrogen gas = $(1.30-x)=(1.30-0.05)=1.25mol$

Equilibrium moles of hydrogen gas = $(1.65-x)=(1.65-0.05)=1.60mol$

Putting values in above expression, we get:

$K_c=\frac{(0.100)^2}{1.25\times (1.60)^3}\\\\K_c=1.95\times 10^{-3}$

Calculating the $K_c'$ for the given chemical equation:

$2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$

$K_c'=\frac{1}{K_c}\\\\K_c'=\frac{1}{1.95\times 10^{-3}}=5.13\times 10^2$

Hence, the value of $K_c$ for $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ reaction is $5.13\times 10^2$

8 0

1 year ago

Read 2 more answers

The reaction N2 + 3 H2 → 2 NH3 is used to produce ammonia. When 450. g of hydrogen was reacted with nitrogen, 1575 g of ammonia

tiny-mole [99]

Answer:

The percent yield of this reaction is 70%

Explanation:

The reaction is: N₂ + 3H₂ → 2NH₃

We only have the mass of H₂, so we assume that N₂ is in excess

We convert the mass to moles, to work with the reaction:

450 g . 1mol / 2 g = 225 moles

Ratio is 2:3. 3 moles of H₂ can produce 2 moles of ammonia

Therefore 225 moles of H₂ will produce (225 .2)/ 3 = 150 moles

This is the 100% yield reaction → We convert the moles of NH₃ to mass

150 mol . 17g /1mol = 2550 g

Percent yield = (Produced yield/Theoretical yield) .100

Percent yield = (1575g/2550g) . 100 = 70%

7 0

2 years ago

Be sure to answer all parts. provide stepwise mechanisms illustrating how each product is formed by choosing, in order, the step

SpyIntel [72]

Answer:

Explanation:

find the solution below

8 0

2 years ago

What is the volume in liters of a cube whose edge is 4.33 cm long?

andreyandreev [35.5K]

Assume that the side length of the cube is m.
Since all sides of the cube are equal, therefore, the volume of the cube can be measured using the following rule:
volume of cube = m^3

Substitute in this equation with the length given:
volume = (4.33)^3 = 81.1827 cm^3

From the standards of units conversions, 1 cm^3 = 0.001 liters
Therefore:
volume of cube = 81.1827 cm^3 = 0.0811827 liters

5 0

2 years ago