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3 years ago

Do Neutros and Protons affect the element?

Chemistry

1 answer:

noname [10]3 years ago

4 0

Answer:

Protons identify the element and add to the mass of the element. The mass of a proton is about 1. Neutrons keep the nucleus (the place where neutrons and protons are) together. they also add to the mass The mass of a neutron is also about 1

Explanation:

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Hydrogen has three isotopes. The most common one, protium, has no neutrons. Deuterium, the second isotope, has one neutron. Trit

Korvikt [17]

Atomic number is the number of protons in an atom. Since these are all Hydrogen, the number of protons remains the same, and the Atomic Number for each is 1. Mass number is the number of Protons and Neutrons in an atom. So, Protium is 1, Deuterium is 2, Tritium is 3. The number of neutrons is stated in the question (Protium 0, Deuterium 1, Tritium 2). Hope this helps, if you wouldn’t mind marking brandies if it did!

3 0

4 years ago

Which statement describes a digital signal used to store information

Sladkaya [172]

Answer:

it is 1,4

Explanation:

Just took the test hope you get it right. c:

8 0

3 years ago

Read 2 more answers

An element that has the electron configuration 1s22s22p 4 has how many electrons?

Tpy6a [65]

Electrons ruminate around the nucleus in particular circular path known as orbit called shell.

Electronic configuration is the dispense of electrons.

Electronic configuration is also known as electronic structure.

Also termed as spatial methodize.

More than one orbital in the electron shell is known as subshell, they have similar energy level.

In Electronic configuration there are four shell present namely k,l,m,n and there are 4 sub-shell namely s,p,d,f.

As we know that every subshell fills gradually s max 2 electrons p - 6 , d - 10 and f - 14.

So,

According to questions,

Thge element has configuration 1s2 2s2 2p4 hence 8 electrons.

To know more about Electronic Configuration here :

brainly.com/question/14283892?referrer=searchResults

#SPJ4

7 0

2 years ago

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s )

Fiesta28 [93]

This is an incomplete question, here is a complete question.

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.

$M_3O_4(s)\rightleftharpoons 3M(s)+2O_2(g)$

Substance ΔG°f (kJ/mol)

M₃O₄ -9.50

M(s) 0

O₂(g) 0

What is the standard change in Gibbs energy for the reaction, as written, in the forward direction? delta G°rxn = kJ / mol.

What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K?

What is the equilibrium pressure of O₂(g) over M(s) at 298 K?

Answer :

The Gibbs energy of reaction is, 9.50 kJ/mol

The equilibrium constant of this reaction is, 0.0216

The equilibrium pressure of O₂(g) is, 0.147 atm

Explanation :

The given chemical reaction is:

$PCl_3(l)\rightarrow PCl_3(g)$

First we have to calculate the Gibbs energy of reaction $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{M(s)}\times \Delta G^0_{(M(s))}+n_{O_2(g)}\times \Delta G^0_{(O_2(g))}]-[n_{M_3O_4(s)}\times \Delta G^0_{(M_3O_4(s))}]$

where,

$\Delta G^o$ = Gibbs energy of reaction = ?

n = number of moles

Now put all the given values in this expression, we get:

$\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (0kJ/mol)]-[1mole\times (-9.50kJ/K.mol)]$

$\Delta G^o=9.50kJ/mol$

The Gibbs energy of reaction is, 9.50 kJ/mol

Now we have to calculate the equilibrium constant of this reaction.

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs free energy = 9.50kJ/mol = 9500 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = 298 K

K = equilibrium constant = ?

$9500J/mol=-(8.314J/K.mol)\times (298K)\times \ln (K)$

$K=0.0216$

The equilibrium constant of this reaction is, 0.0216

Now we have to calculate the equilibrium pressure of O₂(g).

The expression of equilibrium constant is:

$K=(P_{O_2})^2$

$0.0216=(P_{O_2})^2$

$P_{O_2}=0.147atm$

The equilibrium pressure of O₂(g) is, 0.147 atm

5 0

4 years ago

is 0.0410 M−1 s −1 . We start with 0.105 mol C2F4 in a 4.00-liter container, with no C4F8 initially present. What will be the co

Jet001 [13]

Answer:

After three hours, concentration of C₂F₄ is 0.00208M

Explanation:

The rate constant of the reaction:

2 C2F4 → C4F8 is 0.0410M⁻¹s⁻¹

As the units are M⁻¹s⁻¹, this reaction is of second order. The integrated law of a second-order reaction is:

$\frac{1}{[A]} =\frac{1}{[A]_0} +Kt$

<em>Where [A] and [A]₀ represents initial and final concentrations of the reactant (C₂F₄), K is rate constant (0.0410M⁻¹s⁻¹) and t is time of the reaction (In seconds). </em>

3.00 hours are in seconds:

3 hours ₓ (3600 seconds / 1 hour) = 10800 seconds

Initial concentration of C2F4 is:

0.105mol / 4.00L = 0.02625M

Replacing in the integrated law:

$\frac{1}{[A]_0}= \frac{1}{0.02625} +0.0410M^{-1}s^{-1}*10800s\\\frac{1}{[A]_0}=480.9M^{-1}$

[A] = 0.00208M

<h3>After three hours, concentration of C₂F₄ is 0.00208M</h3>

5 0

3 years ago