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3 years ago

Which subatomic particle is electrically neutral and found in the nucleus of an atom?

2 answers:

8 0

There are primarily three subatomic particles in an atom;
Protons
Electrons
Neutrons
----------------
Protons are positively charged subatomic particles in an atom. These particles give an element it's identity, based on it's 'Atomic Number', or how many protons are contained in the atom.

Electrons are negatively charged subatomic particles in an atom. These particles give an element it's reactivity, and are a major part in finding where an element lies on the Periodic Table by valence electrons. We can discuss more of this in PM if you'd like to know more about this. Electrons are also a major part of finding an element's 'atomic mass', or how many neutrons it has.

Neutrons are subatomic particles in an atom that have a neutral or 'no' charge. These are inside of the nucleus of an atom, as opposed to protons and electrons, which are farther outward of the center of an atom.

To answer your question, and reading from the text above, 'Neutrons' is your answer.

I hope this helps!

Valentin [98]3 years ago

3 0

The answer is neutron

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Select the answer that CANNOT be used to fill in the blank in the following sentence:

8_murik_8 [283]

Answer:

the answer is ur mom

Explanation:

8 0

3 years ago

g At elevated temperatures, molecular hydrogen and molecular bromine react to partially form hydrogen bromide: H 2 (g) Br 2 (g)

oksano4ka [1.4K]

Answer: At equilibrium , there are 0.274 moles of $Br_2$

Explanation:

Moles of $H_2$ = 0.682 mole

Moles of $Br_2$ = 0.440 mole

Volume of solution = 2.00 L

Initial concentration of $H_2$ = $\frac{0.682}{2.00}=0.341 M$

Initial concentration of $Br_2$ = $\frac{0.440}{2.00}=0.220 M$

Equilibrium concentration of $H_2$ = $\frac{0.516}{2.00}=0.258 M$

The given balanced equilibrium reaction is,

$H_2(g)+Br_2(g)\rightleftharpoons 2HBr(g)$

Initial conc. 0.341 M 0.220 M 0 M

At eqm. conc. (0.341-x) M (0.220-x) M (2x) M

Given : (0.341-x) M = 0.258 M

x= 0.083 M

Thus equilibrium concentartion of $Br_2$ = (0.220-0.083) M = 0.137 M

Thus moles of $Br_2$ at equilibrium = $0.137M\times 2.00L=0.274mol$

At equilibrium , there are 0.274 moles of $Br_2$

8 0

3 years ago

A chemical reaction in a lab experiment actually produces 1.23mol of BeBr2. The theoretical yield is 220g. To two decimal places

Mashutka [201]

Answer:

Percent yield = 94.62%

Explanation:

Given data:

Actual yield of BeBr₂ = 1.23 mol

Theoretical yield of BeBr₂ = 220 g

Percent yield = ?

Solution:

Theoretical yield in moles:

Number of moles = mass/molar mass

Number of moles = 220 g/168.82 g/mol

Number of moles = 1.30 mol

Percent yield:

Percent tiled = actual yield / theoretical yield × 100

by putting values,

Percent yield = 1.23 mol / 1.30 mol × 100

Percent yield = 0.9462 × 100

Percent yield = 94.62%

5 0

3 years ago

What are gases that are molecules

Vinvika [58]

Answer:

there are elemental molecules, made from one type of atom (pure), example: oxygen and there are compound molecules, made from more then one type of atom, for example carbon dioxide

7 0

3 years ago

Write the equilibrium constant expression Kc for the following reaction.2H20(g) ---------> O2(g) + 2H2(g) <---------Do not

Komok [63]

Answer:

$K_c\text{ = }\frac{[O_2][H_2]\placeholder{⬚}^2}{[H_2O]\placeholder{⬚}^2}$

Explanation:

Here, we want to write the equilibrium constant expression

To write this, we raise the concentrations of the reactants and products to the coefficient before them. These concentrations are represented by square brackets in which the chemical formula of the compound is placed

We place the representation of the products over that of the reactants

We have the expression written as follows:

$K_c\text{ = }\frac{[H_2]\placeholder{⬚}^2[O_2]}{[H_2O]\placeholder{⬚}^2}$

7 0

1 year ago