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o-na [289]
3 years ago
12

Determine the magnitude of the acceleration experienced by an electron in an electric field of 664 N/CN/C .

Chemistry
1 answer:
VARVARA [1.3K]3 years ago
5 0

Answer:

a=-1.17\times 10^{14} \hspace{3}\frac{m}{s^2}

Explanation:

A charged particle that is in a region where there is an electric field, experiences a force equal to the product of its charge by the intensity of the electric field:

F_e=q*E

If the electric field is uniform, the force is constant and so is the acceleration. Applying the equations of uniformly accelerated rectilinear motion, we obtain the velocity of the particle at any time or after having moved a certain distance:

a=\frac{qE}{m} (1)

Where:

E=Electric\hspace{3}field\hspace{3}strength=664\frac{N}{C} \\q=Electric\hspace{3}charge\hspace{3}of\hspace{3}the\hspace{3}particle\\m=Mass\hspace{3}of\hspace{3}the\hspace{3}particle

The electric charge and the mass of the electron are known constants:

q=-1.6\times 10^{-19} C\\m=9.1\times 10^{-31} kg

So, replacing this data in the equation (1) :

a=\frac{(-1.6\times 10^{-19})*(664) }{9.1\times 10^{-31} } =-1.167472527\times 10^{14} \approx -1.17\times 10^{14} \hspace{3}\frac{m}{s^2}

The minus sign is due to the fact that the charge is negative, therefore it experiences a force in the opposite direction to the field.

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Aspartamate is methyl ester of aspartic acid. It is used as an artificial sweetener in foods and beverages. The molecular formula of aspartamate is: C₁₄H₁₈N₂O₅. Molecular mass of aspartamate is 294.3 g/mol.

294.3 g of aspartamate contains 1 mol of aspartamate= 6.023 X 10²³ number of aspartamate molecules. So, 5 mg of aspartamate contains (5 X 6.023 X 10²³)/ (294.3 X 1000) = 1.023 X 10¹⁹ number of aspartamate molecules. Each molecule of aspartamate has 18 H-atoms, so, 5 mg i.e, 1.023 X 10¹⁹ number of aspartamate molecules contain 18 X 1.023 X 10¹⁹= 1.84 X 10²⁰ number of hydrogen atoms.

7 0
3 years ago
Calculate the experimental specific heat capacity of an object of mass 1.32 kg, given that the object releases 1.95 kJ of heat w
andreev551 [17]

Answer:

0.076 J/gºC.

Explanation:

From the question given above, the following data were obtained:

Mass (M) of object = 1.32 kg

Heat (Q) released = –1.95 kJ

Change in temperature (ΔT) = –19.5°C (decrease in temperature)

Specific heat capacity (C) =?

Next, we shall convert 1.32 kg to grams (g). This can be obtained as follow:

1 kg = 1000 g

Therefore,

1.32 kg = 1.32 kg × 1000 g / 1 kg

1.32 kg = 1320 g

Next, we shall convert –1.95 kJ to joule (J). This can be obtained as follow:

1 kJ = 1000 J

Therefore,

–1.95 kJ = –1.95 kJ × 1000 / 1 kJ

–1.95 kJ = –1950 J

Finally, we shall determine the specific heat capacity of the object. This can be obtained as follow:

Mass (M) of object = 1320 g

Heat (Q) released = –1950 J

Change in temperature (ΔT) = –19.5°C (decrease in temperature)

Specific heat capacity (C) =?

Q = MCΔT

–1950 = 1320 × C × –19.5

–1950 = –25740 × C

Divide both side by –25740

C = –1950 / –25740

C = 0.076 J/gºC

Thus, the specific heat capacity of the object is 0.076 J/gºC.

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3 years ago
Which factor would be most likely to shrink the size of an atom’s electron cloud?
Shalnov [3]

Answer:

A. Forming a positive ion!

Hope this helped :)

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Why does metal need to be extracted from an ore before it can be used?
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Metal ore has other elements in it as well. Also sediment and stone might cover the ore. We don't want to have a phone with sediment on it do we? thus these few reasons are why.
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What mass of aluminum chloride could be made from 8.1 g of aluminum and 4.2 L of chlorine at STP?
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In this problem Al metal is a limiting reactant as it is present in less amount as compared to chlorine gas, Hence, controls the formation of ALCl3. So, the amount of AlCl3 produced is 40.05 grams. Solution is as follow,

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