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katen-ka-za [31]
3 years ago
9

Name two gases that are found in air?

Chemistry
2 answers:
Diano4ka-milaya [45]3 years ago
7 0

Answer:

Oxygen and Nitrogen

hope this helps

Norma-Jean [14]3 years ago
5 0

Answer:

oxygen and carbon dioxide

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In a closed system containing three different gases as shown, X2 + Y2 2XY. Choose the TWO things that MUST happen as the volume
Kitty [74]

Answer:

3) The relative concentrations of each gas must remain constant.

4)The concentration of each gas will not change.

Explanation:

  • For the equilibrium system:

<em>X₂ + Y₂ ⇄ 2XY,</em>

The no. of moles of gases in each side is constant; there is 2 moles of gases at reactants side and 2 moles of gases at products side.

So, changing the volume will not affect on the equilibrium system.

<em>So, the right choice is:</em>

3) The relative concentrations of each gas must remain constant.

4)The concentration of each gas will not change.

5 0
3 years ago
If a compound contains a poly atomic ion then what type of compound is it A. Convalent Compound B. Ionic Compound
den301095 [7]

Answer:

B. Ionic Compound

Explanation:

An ionic compound is that compound which contains a positively charged ion called CATION and a negatively charged ion called ANION. The cation loses or transfers electrons to the anion, hence, making the former (cation) positive and the latter (anion) negative.

A polyatomic ion is an ion that contains more than one type of atom e.g OH-, NO3²-, CO3²- etc. A polyatomic ion usually has an overall charge formed from the charges of the individual atoms that makes it up. For example, in OH-, the overall charge is -1.

Since a polyatomic ion can have an overall positive or negative charge, it must enter a reaction with another ion that complements it i.e. a negative polyatomic ion will react with a positive ion to neutralize its charge. Hence, this forms an IONIC COMPOUND. This is why most compounds with polyatomic ions are IONIC COMPOUNDS.

For example, CaCO3 is an ionic compound formed when Ca²+ (cation) reacts with the polyatomic anion: CO3²-

5 0
2 years ago
WRITE THE ELECTRONIC CONFIGURATION OF THE FOLLOWING ELEMENT
Sunny_sXe [5.5K]
1 Hydrogen 1s1
2 Helium 1s2
3 Lithium 2s1
5 0
2 years ago
Read 2 more answers
Sodium tert-butoxide (NaOC(CH3)3) is classified as bulky and acts as Bronsted Lowry base in the reaction. It is reacted with 2-c
IceJOKER [234]

Answer:but-1-ene

Explanation:This is an E2 elimination reaction .

Kindly refer the attachment for complete reaction and products.

Sodium tert-butoxide is a bulky base and hence cannot approach the substrate 2-chlorobutane from the more substituted end and hence major product formed here would not be following zaitsev rule of elimination reaction.

Sodium tert-butoxide would approach from the less hindered side that is through the primary centre and hence would lead to the formation of 1-butene .The major product formed in this reaction would be 1-butene .

As the mechanism of the reaction is E-2 so it will be a concerted mechanism and as sodium tert-butoxide will start abstracting the primary hydrogen through the less hindered side simultaneously chlorine will start leaving. As the steric repulsion in this case is less hence the transition state is relatively stabilised and leads to the formation of a kinetic product 1-butene.

Kinetic product are formed when reactions are dependent upon rate and not on thermodynamical stability.

2-butene is more thermodynamically6 stable as compared to 1-butene  

The major product formed does not follow the zaitsev rule of forming a more substituted alkene as sodium tert-butoxide cannot approach to abstract the secondary proton due to steric hindrance.

5 0
3 years ago
Write a net ionic equation for the overall reaction that occurs when aqueous solutions of sodium hydroxide and hydrosulfuric aci
Helga [31]

Answer:

2OH-(aq) + 2H+(aq) → 2H2O(l)

Explanation:

Step 1: Data given

sodium hydroxide = NaOH

hydrosulfuric acid = H2S

Step 2: The unbalanced equation

NaOH(aq) + H2S(aq) → Na2S(aq) + H2O(l)

Step 3: Balancing the equation

On the left side we have 1x Na (in NaOH), on the right side we have 2x Na (in Na2S). To balance the amount of Na on both sides, we have to multiply NaOH on the left side by 2.

2NaOH(aq) + H2S(aq) → Na2S(aq) + H2O(l)

On the left side we have 4x H (2x in NaOH and 2x in H2S), on the right side we have 2x H (in H2O). To balance the amount of H on both sides, we have to multiply H2O on the right side by 2. Now the equation is balanced.

2NaOH(aq) + H2S(aq) → Na2S(aq) + 2H2O(l)

Step 4: The net ionic equation

2Na+(aq) + 2OH-(aq) + 2H+(aq) + S^2-(aq) → 2Na+(aq) + S^2-(aq) + 2H2O(l)

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

2OH-(aq) + 2H+(aq) → 2H2O(l)

8 0
3 years ago
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