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2 years ago

Which particle has no charge? 1.) electron 2.) neutron 3.) positron 4.) proton

Chemistry

2 answers:

Marianna [84]2 years ago

4 0

The neutron has no charge because it is a neutral particle.

drek231 [11]2 years ago

3 0

It could be proton or neutron

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Which of the following represents positron emission? Select one: a. 3015P ---> 3014Si + ___ b. 23892U ---> 23490Th + ___ c

xxTIMURxx [149]

The simple trick which one can consider in such problem where it is asked for positron emission is :
When the atomic number goes DOWN by one and mass number remains unchanged, then a positron is emitted.

a. $P^{30}_{15} ----\ \textgreater \ Si^{30}_{14} + e^{0}_{1}^+$

Here the atomic number decreases by one.

Similarly, options b and d are eliminated.

Option c is also not the answer.
For c, Count the atomic number on left side and compare it with right side. You will see it is 9 on left and 8 on right. Atomic no. did go down by 1. But the atomic mass is changed as well.

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3 years ago

Is this sentence true or false ? hydrogen is considered to be a metal

victus00 [196]

It is false, hope this helps!

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3 years ago

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A neutral solution has a pH of: a. 1 b. 5 c. 7 d. 14

zavuch27 [327]

The answer is 7 my guy

3 0

2 years ago

Where does the formation of rocks began

bekas [8.4K]

Hi!

All rocks are connected in a cycle of creation, change, and destruction called the Rock Cycle. The rock cycle begins with molten rock (magma below ground, lava above ground), which cools and hardens to form igneous rock.

Hope this helps!

~CoCo

7 0

3 years ago

What is the balanced form of the following equation? Br 2 + S 2 O 3 2– + H 2 O → Br 1– + SO 4 2– + H +

nikitadnepr [17]

Answer:

4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

Explanation:

Br₂ + S₂O₃²⁻ + H₂O → Br⁻ + SO₄²⁻ + H⁺

This is a redox reaction:

Br₂ changes the oxidation state from 0 to -1, so it was reduced

In the S₂O₃⁻² anion S changes the oxidation state from +2 to +6 in sulfate anion. (S₂O₃⁻², it is called thiosulfate)

We have protons in the main equation, so we assume we are in acidic medium:

Br₂ + 2e⁻ → 2Br⁻ Reduction

We balanced the bromide with 2, so the bromine has gained 2 electrons.

5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8e- Oxidation

First of all, we add 2 to the sulfate anion in the product side, in order to balance the S.

As we have 8 O in right side, and 3 O in left side, we must add 5 O. We add 5 water in the place where the O are lower (reactant side).

Now, we have 10 H, in the reactant side, so we balance the product side with protons (10 H⁺).

Sulfur changed the oxidation state from +2 to +6, so it released 4 electrons, but, if you see thiosulfate anion you have 2 sulfurs so finally it has released 8 electrons.

Electrons are unbalanced so we multiply reduction x4, and oxidation x1.

(Br₂ + 2e⁻ → 2Br⁻) . 4 = 4Br₂ + 8e⁻ → 8Br⁻

(5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8e-) . 1 = STAYS THE SAME.

We sum both half reactions, to cancel the elecetrons:

4Br₂ + 8e⁻ + 5H₂O + S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8e- + 8Br⁻

Finally the balanced reaction is: 4Br₂+ 5H₂O+ S₂O₃²⁻ → 2SO₄²⁻ + 10H⁺ + 8Br⁻

5 0

3 years ago