Can somebody please help me

List the number of

rjkz [21]

Answer:

beryllium: 2 valence electrons

nitrogen: 5 valence electrons

oxygen: 6 valence electrons

fluorine: 7 valence electrons

magnesium: 2+ valence electrons

phosphorus: 5 valence electrons

sulfur: 6 valence electrons

chlorine: 7 valence electrons

Explanation:

6 0

3 years ago

Please answer A S A P the send me the completed version

Murljashka [212]

Answer:

A DOCX file is a Microsoft Word document that typically contains text

4 0

3 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

Given the following reaction: \ce{Cu + 2AgNO3 -> 2Ag + Cu(NO3)2}Cu+2AgNO3 2Ag+Cu(NOX 3 )X 2 How many moles of \ce{Ag}

Alexandra [31]

Answer:

0.252 mol

Explanation:

Given the following reaction:

Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂

How many moles of Ag will be produced from 16.0 g Cu, assuming AgNO3 is available in excess.

First, we write the balanced equation.

Cu + 2 AgNO₃ → 2 Ag + Cu(NO₃)₂

We can establish the following relations.

The molar mass of Cu is 63.55 g/mol.
The molar ratio of Cu to Ag is 1:1.

The moles of Ag produced from 16.0 g of Cu are:

$16.0gCu.\frac{1molCu}{63.55gCu} .\frac{1molAg}{1molCu} =0.252 molAg$

6 0

3 years ago

Read 2 more answers

Classify the following as either solutions or colloids. If a colloid, name the type of colloid and identify both the dispersed a

aev [14]

Answer:

a. glucose in water( solution)

b. smoke in air (colloids)

c. carbon dioxide in air (solution)

d. milk( colloids)

Explanation:

A solution is said to be formed when a solute dissolves in a solvent to form a homogeneous mixture. The solute particles are less than 10^-9m in size. Familiar solutions are those where the solute are dissolved in a liquid solvent. When the liquid water, the solution is known as an aqueous solution. A typical example is (glucose in water). In some other cases, the apparent solution of a solute in a solvent is accompanied by a chemical reaction and this is often known as a chemical reaction. A typical example is (carbon dioxide in air).

Colloids are also known as false solutions. Here, the individual solute particles are larger than the particles of the true solution, but not large enough to be seen by the naked eye. When a light beam is placed beside a beaker containing a colloid, the light rays of the beam can be clearly seen. This shows that it exhibits the Tyndall effect while a solution dosent exhibit such.

In a colloid, the liquid solvent is more appropriately know as the DISPERSION medium while the solid solute particles constitute the DISPERSED substance. This can either be solid, liquid or gas.

For example:

--> smoke in air : Dispersion medium is gas while the dispersed substance is solid.

--> milk: Dispersion medium is liquid while the dispersed substance is liquid.

7 0

3 years ago