D - density: 13,534 g/ml
m - mass: 10g
V - volume: ??
_____________
d = m/V
V = m/d
V = 10/13,534
V = 0,7389 ml
:•)
Answer:
Percent yield = 89.1%
Explanation:
Based on the equation:
Cl₂ + 2KI → 2KCl + I₂
<em>1 mole of Cl₂ reacts with 2 moles of KI to produce to moles of KCl</em>
<em />
To solve this quesiton we must find the moles of each reactant in order to find the limiting reactant. With the limiting reactant we can find the moles of KCl and the mass:
<em>Moles Cl₂:</em>
8x10²⁵ molecules * (1mol / 6.022x10²³ molecules) = 133 moles
<em>Moles KI -Molar mass: 166.0028g/mol-</em>
25g * (1mol / 166.0028g) = 0.15 moles
Here, clarely, the KI is the limiting reactant
As 2 moles of KI produce 2 moles of KCl, the moles of KCl produced are 0.15 moles. The theoretical mass is:
0.15 moles * (74.5513g / mol) =
11.2g KCl
Percent yield is: Actual yield (10.0g) / Theoretical yield (11.2g) * 100
<h3>Percent yield = 89.1%</h3>
Answer:
Pressure = 4313.43mmHg
Explanation:
P1 = ?
V1 = 0.335L
V2 = 1700mL =1700*10^-3L = 1.7L
P2 = 850mmhg
From Boyle's law, the volume of a fixed mass of gas is inversely proportional to its pressure provided that temperature remains constant.
P = k / v
K = pv. P1V1 = P2V2 = P3V3 =........=PnVn
P1V1 = P2V2
Solve for P1,
P1 = (P2*V2) / V1
P1 = (850 * 1.7) / 0.335
P1 = 4313.43mmHg
The pressure of the gas was 4313.43mmHg
<h3>Balanced equation :
2C₂H₆ (g) + 7O₂ (g) ⟶ 4CO₂ (g) + 6H₂O (ℓ)</h3><h3>Further explanation</h3>
Alkanes are saturated hydrocarbons that have single bonds in chains
General formula for alkanes :
Hydrocarbon combustion reactions (specifically alkanes)
So that the burning of ethane with air (oxygen):
2C₂H₆ (g) + 7O₂ (g) ⟶ 4CO₂ (g) + 6H₂O (ℓ)
or we can use mathematical equations to solve equilibrium chemical equations by giving the coefficients for each compound involved in the reaction
C₂H₆ (g) + aO₂ (g) ⟶ bCO₂ (g) + cH₂O (ℓ)
C : left 2, right b ⇒ b=2
H: left 6, right 2c⇒ 2c=6⇒ c= 3
O : left 2a, right 2b+c⇒ 2a=2b+c⇒2a=2.2+3⇒2a=7⇒a=7/2
