Answer:
3.) 51.7 L
Explanation:
To find the volume, you need to use the Ideal Gas Law:
PV = nRT
In the equation,
-----> P = pressure (kPa)
-----> V = volume (L)
-----> n = moles
-----> R = Ideal Gas constant (8.314 kPa*L/mol*K)
-----> T = temperature (K)
First, you need to convert the temperature from Celsius to Kelvin. Then, you can plug the given values into the equation and simplify to find "V".
P = 68.0 kPa R = 8.314 kPa*L/mol*K
V = ? L T = 27.00 °C + 273 = 300 K
n = 1.41 moles
PV = nRT <----- Ideal Gas Law
(68.0 kPa)V = (1.41 moles)(8.314 kPa*L/mol*K)(300 K) <----- Insert values
(68.0 kPa)V = 3516.822 <----- Multiply right side
V = 51.7 <----- Divide both sides by 68.0
Taking credit for a someone else's research or written work
Answer:
16. Option c.
Explanation:
The reaction is:
C₈H₁₈ + O₂ → CO₂ + H₂O
Let's count, we have 8 C, 18 H and 2 O in reactant side
We have 1 C, 2 H and 3 O in product side
In order to balance the C, we add 8 to CO₂ and to balance the H, we add a 9 to H₂O
Now we have 8 C on both sides and 18 H On both sides. Finally we have 25 O on product side.
C₈H₁₈ + O₂ → 8CO₂ + 9H₂O
To balance the O in product side we must add 25/2, but as it is a rational number, we must multiply x2 to get an integer number (x2 in all the stoichiometry). The balanced reaction is :
2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O
with 16 C, 36 H and 50 O, on both sides
Answer:
(upper right) corner of the periodic table to the bottom left corner
Explanation:
<em>Potas</em><em>sium</em><em> </em><em>has</em><em> </em><em>an</em><em> </em><em>atomi</em><em>c</em><em> </em><em>radii</em><em> </em><em>that</em><em> </em><em>is</em><em> </em><em>greate</em><em>r</em><em> </em><em>t</em><em>han</em><em> </em><em>t</em><em>hat</em><em> </em><em>of</em><em> </em><em>lithiu</em><em>m</em><em> </em><em>th</em><em>at</em><em> is</em><em> </em><em>why</em><em> </em><em>pot</em><em>assium</em><em> </em><em>is</em><em> </em><em>more</em><em> </em><em>reactiv</em><em>e</em><em> </em><em>than</em><em> </em><em>lithi</em><em>um</em><em>.</em>
<em>tHx</em><em> </em><em>fOr</em><em> pOinTs</em><em>.</em><em>.</em><em>.</em>
