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3 years ago

What is the potential at a distance of 5.0 x 10-10 m from a nucleus of charge 50e? question options:?

1 answer:

6 0

Electricpotential V=keQ/r
r = 5.0*10-10 m,
Q = 50e = 50*1.60*10-19
C = 80*10-19 C
V=9.0*109 Nm2C-2* 80*10-19 C / 5.0*10-10

m = 144V

I hope it helped:)

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Temka [501]

Answer: a

Explanation:

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3 years ago

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a gas has a volume of 4.54 L at 1.65 atm and 75. I decrease celsius. At what pressure would the volume of the gas be increased t

Vikentia [17]

Answer:

The answer to your question is P2 = 1.52 atm

Explanation:

Data

Volume 1 = V1 = 4.54 l

Pressure 1 = P1 = 1.65 atm

Temperature 1 = T1 = 75°C

Volume 2= V2 = 5.33 l

Pressure 2 = P2 = ?

Temperature 2 = 103°C

Process

1.- Convert temperature to °K

Temperature 1 = 75 + 273 = 348°K

Temperature 2 = 103 + 273 = 376°K

2.- Use the combine gas law to find the final pressure

P1V1/T1 = P2V2/T2

-Solve for P2

P2 = P1V1T2 / T1V2

-Substitution

P2 = (1.65 x 4.54 x 376) / (348 x 5.33)

-Simplification

P2 = 2816.62 / 1854.84

-Result

P2 = 1.52 atm

5 0

3 years ago

H2(g) + F2(g)2HF(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.20 moles

abruzzese [7]

<u>Answer:</u> The value of $\Delta S^o$ for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$H_2(g)+F_2(g)\rightarrow 2HF(g)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(HF(g))})]-[(1\times \Delta S^o_{(H_2(g))})+(1\times \Delta S^o_{(F_2(g))})]$

We are given:

$\Delta S^o_{(HF(g))}=173.78J/K.mol\\\Delta S^o_{(H_2)}=130.68J/K.mol\\\Delta S^o_{(F_2)}=202.78J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (173.78))]-[(1\times (130.68))+(1\times (202.78))]\\\\\Delta S^o_{rxn}=14.1J/K$

Entropy change of the surrounding = - (Entropy change of the system) = -(14.1) J/K = -14.1 J/K

We are given:

Moles of hydrogen gas reacted = 2.20 moles

By Stoichiometry of the reaction:

When 1 mole of hydrogen gas is reacted, the entropy change of the surrounding will be -14.1 J/K

So, when 2.20 moles of hydrogen gas is reacted, the entropy change of the surrounding will be = $\frac{-14.1}{1}\times 2.20=-31.02J/K$

Hence, the value of $\Delta S^o$ for the surrounding when given amount of hydrogen gas is reacted is -31.02 J/K

7 0

3 years ago

What is the molar mass of CuCIO3

Genrish500 [490]

<h3>Answer:</h3>

147.05 g/mol

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<h3>Explanation:</h3>

<u>Step 1: Define</u>

CuClO₃

Molar Mass of Cu - 63.55 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CuClO₃ - 63.55 + 35.45 + 3(16.00) = 147.05 g/mol

5 0

3 years ago

Describe the process of diffusion. How does diffusion help cells survive?

lubasha [3.4K]

Answer:

Diffusion is the movement of a substance from an area of high concentration to an area of low concentration.

Diffusion happens in liquids and gases because their particles move randomly from place to place.

Diffusion is an important process for living things; it is how substances move in and out of cells.

6 0

3 years ago