Answer:
Explanation:
We have in this question the equilibrium
X ( g ) + Y ( g ) ⇆ Z ( g )
With the equilibrium contant Kp = pZ/(pX x pY)
The moment we change the concentration of Y, we are changing effectively the partial pressure of Y since pressure and concentration are directly proportional
pV = nRT ⇒ p = nRT/V and n/V is molarity.
Therefore we can calculate the reaction quotient Q
Qp = pZ/(pX x pY) = 1/ 1 x 0.5 atm = 2
Since Qp is greater than Kp the system proceeds from right to left.
We could also arrive to the same conclusion by applying LeChatelier´s principle which states that any disturbance in the equilibrium, the system will react in such a way to counteract the change to restore the equilibrium. Therefore, by having reduced the pressure of Y the system will react favoring the reactants side increasing some of the y pressure until restoring the equilibrium Kp = 1.
Element at Extreme Left In Periodic Table:
The elements of Group I-A (1) are present at extreme left of the periodic table. They are called as Alkali Metals. Alkali Metals are strong metals. These elements can easily loose their valence electron. The valence shell electronic configuration of these elements is,
ns¹
where n is principle quantum number, which shows main energy level or shell. These metals can gain Noble gas configuration (stable configuration) either by loosing one electron or by gaining seven or more electrons. As it is quite reasonable to loose one electron instead of gaining seven or more electrons so these element easily loose one electron to gain noble as configuration. The Metallic character decreases along the period from left to right. So Group II-A (2) are second most metallic elements and so on. These metals at extreme left mainly exist in solid form.
Element at Extreme Right In Periodic Table:
Elements present at extreme right of the periodic table lacks the properties of metallic character and act as non-Metals. They have almost complete outermost shell or have the deficiency of one or two electrons. They are not as hard as metallic elements and they exist with complete octet like in Noble gases, or deficient with one electron (Halogens) or two electrons (oxygen group). These elements tend to gain or accept electron if their valence shell is deficient with required number of elements. Like the valence electronic configuration of Halogens is,
ns², np⁵
So, Halogens readily accept one electron and attain noble gas configuration. Elements at extreme left exist mainly in gas phase.
2C_6H_14 + 19O_2 → 12CO_2 + 14H_2O
<em>Step 1</em>. Write the <em>condensed structural formula</em> for 2,3-dimethylbutane.
(CH_3)_2CHCH(CH_3)_2
<em>Step 2</em>. Write the <em>molecular formula</em>.
C_6H_14
<em>Step 3</em>. Write the <em>unbalanced chemical equation</em>.
C_6H_14 + O_2 → CO_2 + H_2O
<em>Step 4</em>. Pick the <em>most complicated-looking formula</em> (C_6H_14) and balance its atoms (C and H).
<em>1</em>C_6H_14 + O_2 → <em>6</em>CO_2 + <em>7</em>H_2O
<em>Step 5</em>. Balance the <em>remaining atoms</em> (O).
1C_6H_14 + (<em>19/2</em>)O_2 → 6CO_2 + 7H_2O
Oops! <em>Fractional coefficients</em>!
<em>Step 6</em>. <em>Multiply all coefficients by a number</em> (2) to give integer coeficients..
2C_6H_14 + 19O_2 → 12CO_2 + 14H_2O
