Which of these elements has two valence electrons? hydrogen (H) barium (Ba) nitrogen (N) krypton (Kr) bromine (Br)

For molecules with similar molecular weights, the dispersion forces become stronger as the molecules become more polarizable. fo

sweet [91]

True ncifjfjedkxkcnnfenskxkcjcndnenejskkxcfenen

8 0

3 years ago

How is a heterogeneous mixture different from a homogeneous mixture?

MArishka [77]

A homogeneous mixture has the same uniform appearance and composition throughout. Many homogeneous mixtures are commonly referred to as solutions. A heterogeneous mixture consists of visibly different substances or phases. The three phases or states of matter are gas, liquid, and solid. Now this is my answer. So revise it a little. You dont want plagiarism

6 0

4 years ago

Hydrogen gas and solid gold(III) sulfide are reacted. Pure solid gold metal is formed and gaseous hydrogen sulfide is released.

alexdok [17]

The reactants of the reaction will be solid gold (III) sulfide and hydrogen gas, while the products will be pure solid gold and hydrogen sulfide gas.

<h3>Chemical reactions</h3>

Reactants react together during reactions to arrive at products.

In this case, solid gold (III) sulfide and hydrogen gas react according to the following equation:

$Au_2S_3(s) + 3 H_2 (g)-- > 2 Au(s) + 3H_2S(g)$

Reactants = solid gold (III) sulfide, hydrogen gas

Products: pure solid gold, hydrogen sulfide gas

More on chemical reactions can be found here: brainly.com/question/1689737

8 0

2 years ago

An acid and a base react to form salt and water. Which of the following best describes this type of chemical reaction? Redox Neu

I am Lyosha [343]

Answer:

Neutralization

Explanation:

When an acid react with base it form the salt and water. The reaction is also called neutralization reaction because both neutralize each other.

In neutralization reaction equal amount of acid and base react to neutralize each other and equal amount of water and salt are formed. When pH does not reach to 7 its means there is less amount of one of reactant which is not fully neutralize.

Neutralization reactions are also used as first aid. For example when someone is dealing with HCl for cleaning purpose of toilet and get touched. It is advised to neutralize it with soap, milk or egg white.

Example:

Hydrochloric acid when react with the sodium hydroxide, a salt sodium chloride and water are formed.

Chemical equation:

HCl + NaOH → NaCl + H₂O

Titration:

Neutralization reactions are also used to determine the concentration of solution. Titration is a quantitative technique in which acid or base is gradually added into the solution whose concentration is to be determine until the neutral point is reached.

3 0

4 years ago

Read 2 more answers

Using the following portion of the activity series for oxidation half-reactions, determine which combination of reactants will r

dusya [7]

The question is incomplete, the complete question is:

Using the following portion of the activity series for oxidation half-reactions, determine which combination of reactants will result in a reaction.

$Li(s)\rightarrow Li^+(aq)+e^-$

$Al(s)\rightarrow Al^{3+}(aq)+3e^-$

A) Li(s) with Al(s)

B) Li(s) with $Al^{3+}$ (aq)

C) $Li^+$ (aq) with Al(s)

D) $Li^+$ (aq) with $Al^{3+}$ (aq)

Answer: The correct option is B): Li(s) with $Al^{3+}$ (aq)

Explanation:

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction.

The chemical species will undergo a reduction reaction if the value of standard reduction potential is more positive or less negative.

For the given half-reactions:

$Li(s)\rightarrow Li^+(aq)+e^-;E^o_{Li^+/Li}=-3.04V$

$Al(s)\rightarrow Al^{3+}(aq)+3e^-;E^o_{Al^{3+}/Al}=-1.662V$

As the value of standard reduction potential of aluminium is less negative. Thus, it undergoes reduction reaction and lithium will undergo oxidation reaction.

The half-reaction follows:

Oxidation half-reaction: $Li(s)\rightarrow Li^+(aq)+e^-$ ( × 3)

Reduction half-reaction: $Al^{3+}(aq)+3e^-\rightarrow Al(s)$

Overall cell-reaction: $3Li(s)+Al^{3+}(aq)\rightarrow 3Li^+(aq)+Al(s)$

Hence, the correct option is B): Li(s) with $Al^{3+}$ (aq)

5 0

3 years ago