Neutrons released in a fission reaction can strike other nuclei and cause

How did john dalton explain a chemical reaction using his atomic theory?

Archy [21]

Hi! Dalton said that the atoms of reactants are rearranged to form new substances as products.

7 0

3 years ago

N the reaction Mg + 2HCl -> MgCl2 + H2 which element’s oxidation number does not change?

BartSMP [9]

Answer:- Oxidation number of Cl does not change as it is -1 on both sides.

Explanations:- oxidation number of Mg on reactant side is 0 as it is in its elemental form(not combined with another element).

Oxidation number of hydrogen in its compounds is +1, so if H is +1 in HCl the oxidation number of Cl is -1 as the sum has to be zero.

On product side, Mg oxidation number is +2 as the oxidation number of alkaline earth metals in their compounds is +2.

Two Cl are present in magnesium chloride, so if Mg is +2 then Cl is -1.

Oxidation number of H on product side is 0 as it is present in its elemental for, $H_2$ ,

So, it is only chlorine(Cl) whose oxidation number does not change for the given equation.

4 0

3 years ago

In solids, the particles of matter are not in constant motion.

Vsevolod [243]

True because in a solid particles dont move so it wouldn't be in a constant motion.

5 0

3 years ago

Read 2 more answers

Below are the reduction half reactions for chemolithoautotrophic nitrification, where ammonia is a source of electrons and energ

SSSSS [86.1K]

Answer: The molecules of oxygen gas that will be reduced to water are 42 molecules

Explanation:

We are given:

$E^o_{(NO_2^-/NH_4)}=-0.41V\\E^o_{(O_2/H_2O)}=0.82V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, oxygen will undergo reduction reaction will get reduced.

$NH_4$ will undergo oxidation reaction and will get oxidized.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

The half reactions follows:

Oxidation half reaction: $NH_4\rightarrow NO_2^-+6e^-$ ( × 4)

Reduction half reaction: $O_2+4e^-\rightarrow 2H_2O$ ( × 6)

Overall reaction: $4NH_4+6O_2\rightarrow 4NO_2^-+12H_2O$

We are given:

Molecules of $NH_4$ = 28

By Stoichiometry of the reaction:

4 molecules of $NH_4$ reacts with 6 molecules of oxygen gas

So, 28 molecules of $NH_4$ will react with = $\frac{6}{4}\times 28=42$ molecules of oxygen gas

Hence, the molecules of oxygen gas that will be reduced to water are 42 molecules

4 0

3 years ago

Bond order of n2 bond order of N2

snow_lady [41]

Answer:

The MO method for N2+ gives the bond order equal to 2.5. But first, we look at the diagram of molecular orbitals for N2 (the bond order for the nitrogen molecule is 3). the N2+ molecule). That is, the bond order for N2+ is 2.5.

4 0

3 years ago