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3 years ago

What is the ending of the electron configuration of each element in group 4?

1 answer:

6 0

<span>because p6 will be the group 8. You have to count the 2 electrons from the "s" block that are Group I and Group II Group I s1 Group II s2 Group III s2 p1 Group IV s2 p2 Group V s2 p3 Group VI s2 p4 Group VII s2 p5 Group VIII s2 p6</span>

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Li = # of atom <br> O= # of atom

defon

answer:

O= 2 atoms

Li= 3 atoms

5 0

2 years ago

Consider this reaction at equilibrium at a total pressure P1: 2SO2(g) + O2(g) â†’ 2SO3(g) Suppose the volume of this system is c

oksian1 [2.3K]

Answer:

The new equilibrium total pressure will be increased to one-half to initial total pressure.

Explanation:

From the information given :

The equation of the reaction can be represented as;

$2SO_{2(g)}+O_{2(g)} \to2SO_{3(g)}$

From above equation:

2 moles of sulphur dioxide reacts with 1 mole of oxygen (i.e 2 moles +1 mole =3 moles ) to give 2 moles of sulphur trioxide

So; suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished.

So if this process takes place ; the equilibrium will definitely shift to the side with fewer moles , thus the equilibrium will shift to the right. As such; there is increase in pressure.

Let the total pressure at the initial equilibrium be $P_1$

and the total pressure at the final equilibrium be $P_2$

According to Boyle's Law; Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.

Thus;

P ∝ 1/V

P = K/V

PV = K

where K = constant

So;

PV = constant

Hence;

$P_1V_1 = P_2V_2$

From the foregoing; since the volume is decreased to one- half to initial Volume; then ,

$V_2 = \dfrac{V_1}{\dfrac{3}{2}} ----- (1)$

also;

Thus ;

$P_1V_1 = P_2( \dfrac{V_1}{\frac{3}{2}})$

$P_1V_1 = P_2 * 2 \dfrac{V_1}{3}$

$3 P_1 V_1 = 2 P_2 V_1$

Dividing both sides by $V_1$

$3P_1 = 2P_2$

$P_2 =P_1 \dfrac{3}{2} ----- (2)$

From ;

$P_1V_1 = P_2V_2$

$P_2 V_2 = P_1 * \dfrac{3}{2}* \dfrac{V_1}{\frac{3}{2}}$

$P_2 V_2 = P_1 * \dfrac{3}{2}* \dfrac{2 }{3}}*V_1$

$P_2 V_2 = P_1 V_1$

Thus; The new equilibrium total pressure will be increased to one-half to initial total pressure.

7 0

3 years ago

What else is produced during the combustion of butane, C4H10?

denpristay [2]

Another product: CO₂

<h3>Further explanation</h3>

Given

Reaction

2C₄H₁₀ + 13O₂⇒ 8__+ 10H₂O

Required

product compound

Solution

In the combustion of hydrocarbons there can be 2 kinds of products

If there is excess Oxygen, you will get Carbon dioxide(CO₂) and water in the product

If Oxygen is low, you'll get Carbon monoxide(CO) and water

Or in other ways, we can use the principle of the law of conservation of mass which is also related to the number of atoms in the reactants and in the products

if we look at the reaction above, there are C atoms on the left (reactants), so that in the product there will also be C atoms with the same number of C atoms on the left

2C₄H₁₀ + 13O₂⇒ 8CO₂+ 10H₂O

5 0

2 years ago

Read 2 more answers

A person's _____ changes as he travels from the earth to space, but his _____ remains the same.

EleoNora [17]

A persons weight changes as he travels from earth to space but his mass remains the same

4 0

3 years ago

Read 2 more answers

Which correctly describes the reaction between potassium and excess water?

nika2105 [10]

Balanced chemical reaction: 2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g).

KOH is inorganic compound p<span>otassium hydroxide, a strong base.
H</span>₂ is hydrogen gas.
In balanced chemical reaction number of atoms on both side of chemical reaction must be same. There are two potassium atoms, four hydrogen atoms and two oxygen atoms on both side of reaction.

7 0

3 years ago