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White raven [17]
3 years ago
13

Compared to the charge of a proton the charge of an electron has

Chemistry
1 answer:
mart [117]3 years ago
5 0

Answer:

Compared to the charge of proton the charge of an electron is negative.

Explanation:

An atom consist of electron, protons and neutrons. Protons and neutrons are present with in nucleus while the electrons are present out side the nucleus.

All these three subatomic particles construct an atom. A neutral atom have equal number of proton and electron. In other words we can say that negative and positive charges are equal in magnitude and cancel the each other. For example if neutral atom has 6 protons than it must have 6 electrons. The sum of neutrons and protons is the mass number of an atom while the number of protons are number of electrons is the atomic number of an atom.

For example

The carbon have six protons and six neutrons so its atomic mass is 12 amu and atomic number is six.

Electron:

The electron is subatomic particle that revolve around outside the nucleus and has negligible mass. It has a negative charge.

Symbol= e-

Mass= 9.10938356×10-31 Kg

It was discovered by j. j. Thomson in 1897 during the study of cathode ray properties.

Neutron and proton

While neutron and proton are present inside the nucleus. Proton has positive charge while neutron is electrically neutral. Proton is discovered by Rutherford while neutron is discovered by James Chadwick in 1932.

Symbol of proton= P+  

Symbol of neutron= n⁰  

Mass of proton=1.672623×10-27 Kg

Mass of neutron=1.674929×10-27 Kg

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. An inflated balloon has a volume of 6.0 L at sea level (1.0 atm) and is allowed to ascend in altitude until the pressure is 0.
Tju [1.3M]

Answer:

The answer to your question is Volume = 11.4 L

Explanation:

Data

Volume 1 = V1 = 6 L

Pressure 1 = P1 = 1 atm

Temperature 1 = T1 = 22°C

Volume 2 = V2 = ?

Pressure 2 = 0.45 atm

Temperature 2 = -21°C

Process

1.- Convert temperature (°C) to °K

T1 = 273 + 22 = 295°K

T2 = 273 + (-21) = 252°K

2.- Use the combined gas law to solve this problem

                   P1V1 / T1 = P2V2 / T2

-Solve for V2

                   V2 = P1V1T2 / T1P2

-Substitution

                   V2 = (6)(1)(252) / (295)(0.45)

- Simplification

                   V2 = 1512 / 132.75

- Result

                   V2 = 11.38 L

3 0
3 years ago
Why did we have you determine the reaction rate using reactant consumed per unit of time instead of product formed per unit of t
Vsevolod [243]

Answer:Rate of reaction can be determined in terms of concentration of reactants consumed or concentration of product formed per unit time

Explanation: For the reaction below

A ===>B

The reactant is A while the product given is B.

Reaction rate = Δ[B]/Δt = -Δ[A]/Δt

The concentration of A will decrease with time while the concentration of B will increases with time.

The negative sign in -Δ[A]/Δt is to convert the expression to positive since the change will always be negative (decreases)

7 0
3 years ago
Why are scientific models important in the study of science? A. They are useful only for teaching students about science. B. The
madreJ [45]
The correct answer should be B.
The correct answer would be B.
4 0
3 years ago
Read 2 more answers
El agua del mar contiene aproximadamente un 3,0 % m/v de sal (NaCl, 58,44 g/mol), (asuma que es la única fuente de cloruros) si
Alchen [17]

Answer:

s = 4.41 g/L.

Explanation:

¡Hola!

En este caso, considerando el escenario dado, se hace necesario para nosotros saber que la posible reacción de disociación la experimenta el cloruro de plomo (II) como se muestra a continuación:

PbCl_2(s)\rightleftharpoons 2Cl^-(aq)+Pb^{2+}(aq)

Lo cual hace que la expresión de equilibrio se calcule como:

Ksp=[Pb^{2+}][Cl^-]^2

Y que en términos de la solubilidad molar, s, se resuelve como:

1.6x10^{-5}=s(2s)^2\\\\1.6x10^{-5}=4s^3\\\\s=\sqrt[3]{\frac{1.6x10^{-5}}{4} } \\\\s=0.0159molPbCl_2/L

Ahora, convertimos este valor a g/L al multiplicarlo por la masa molar del cloruro de plomo (II):

s=0.0159molPbCl_2/L*\frac{278.1gmolPbCl_2}{1molmolPbCl_2} \\\\s=4.41g/L

¡Saludos!

7 0
3 years ago
Alguien sabe el nombre del compuesto etilisopropilacetileno?​
Paladinen [302]
Noooooooooooooooooooooo
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4 years ago
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