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Rom4ik [11]
2 years ago
12

While looking at xenon (Xe) on the periodic table, a student needs to find an element with a smaller atomic mass in the same gro

up. Where should the student look?
directly up

directly down

directly to the left

directly to the right
Chemistry
2 answers:
Andru [333]2 years ago
6 0
Groups are the columns which go down. (The rows are periods).

As you can see in the periodic table the atomic number increases as you go down the periodic table.

Therefore the answer is to look directly up :)
zvonat [6]2 years ago
3 0

Answer is: directly up.

Neon (Ne) is noble gas with atomic number 10 and atomic mass around 20.

1) directly up is noble gas helium (He) with atomic mass around 4.

2) directly down is noble gas argon (Ar) with atomic mass around 40.

3) directly to the left is florine (F), with smaller atomic mass, but also from different group.

4) directly to the right there no elements, because noble gases are far right in 18. group of Periodic table.

You might be interested in
Question 4: Which members of an ecosystem are part of the energy flow?
Vanyuwa [196]

Answer:

The answer is A (number 1)

5 0
2 years ago
Hydrazine (N2H4) and dinitrogen tetroxide (N2O4) form a self-igniting mixture that has been used as a rocket propellant. The rea
Alexxx [7]

Answer:

Explanation:

An oxidizing accepts an electron and becomes reduced while a reducing agent loses an electron and become oxidized.

Chemical equation:

1) 2 N₂H₄ + N₂O₄ → 3 N₂ + 4 H₂O

2) Hydrazine ( N₂H₄)  is being oxidized

Dinitrogen tetroxide N₂O₄ is being reduced

3) The reducing agent is Hydrazine ( N₂H₄) and the oxidizing agent is dinitrogen tetroxide (N₂O₄)

5 0
2 years ago
_________is a process in which o2 is released as a by-product of oxidation-reduction reactions
Kisachek [45]

Answer : The combustion is a process in which oxygen is released as a by-product of oxidation-reduction reactions.

Explanation :

Combustion reaction : It is defined as the reactions in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide and water.

The chemical equation of combustion reaction is:

CH_4+2O_2\rightarrow CO_2+2H_2O

The combustion reaction is also a redox reaction.

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

The combustion reaction is also a redox reaction in which the carbon shows oxidation by the addition of oxygen or removal of hydrogen and oxygen shows reduction by the addition of hydrogen or removal of oxygen.

Hence, the combustion is a process in which oxygen is released as a by-product of oxidation-reduction reactions.

6 0
2 years ago
You have 49.8 g of O2 gas in a container with twice the volume as one with CO2 gas. The pressure and temperature of both contain
IrinaVladis [17]

Answer:

34.2 g is the mass of carbon dioxide gas one have in the container.

Explanation:

Moles of O_2:-

Mass = 49.8 g

Molar mass of oxygen gas = 32 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{49.8\ g}{32\ g/mol}

Moles_{O_2}= 1.55625\ mol

Since pressure and volume are constant, we can use the Avogadro's law  as:-

\frac {V_1}{n_1}=\frac {V_2}{n_2}

Given ,  

V₂ is twice the volume of V₁

V₂ = 2V₁

n₁ = ?

n₂ = 1.55625 mol

Using above equation as:

\frac {V_1}{n_1}=\frac {V_2}{n_2}

\frac {V_1}{n_1}=\frac {2\times V_1}{1.55625}

n₁ = 0.778125 moles

Moles of carbon dioxide = 0.778125 moles

Molar mass of CO_2 = 44.0 g/mol

Mass of CO_2 = Moles × Molar mass = 0.778125 × 44.0 g = 34.2 g

<u>34.2 g is the mass of carbon dioxide gas one have in the container.</u>

5 0
2 years ago
Explain the effects of nh3 and hcl on the cuso4 solution in terms of le chatelier's principle
Fittoniya [83]

The Principle of Le Chatelier states that if a system in equilibrium is subjected to a disturbance, the system will react in such a way that it will diminish the effect of that disturbance. Thus, when the concentration of one of the substances in an equilibrium system is changed, the equilibrium varies in such a way that it can compensate for this change.

For example, if the concentration of one of the reactants is increased, the equilibrium shifts to the right or to the side of the products. Also, if you add more reagents, the reaction will move even more to the right until the balance is re-established again, increasing the quantity of products.

In this way, adding HCl to a solution of CuSO4 will produce the following reaction:

CuSO4 (aq) + 2HCl (aq) ⇔ CuCl2 (aq) + H2SO4 (aq)

Initially the solution of CuSO4 in water will be blue, but when adding HCl the solution will change color to green, since the aqueous solutions of CuCl2 are green. By adding more HCl this color will intensify as the balance shifts to the right, producing more CuCl2 and H2SO4.

On the other hand, adding NH3 to a solution of CuSO4 will produce the following reaction

CuSO4 (aq) + 4NH3 (aq) ⇔ [Cu(NH3)4] SO4 (s)

Thus, by adding NH3 to the CuSO4 solution we will observe the formation of a precipitate corresponding to [Cu(NH3) 4] SO4. <u>When adding more NH3, the formation of more precipitate will be observed as the equilibrium moves to the right, producing a greater quantity of [Cu (NH3) 4] SO4.</u>

6 0
2 years ago
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