Answer:
The number of moles of the gas is 9.295 moles or 9.30 moles
Explanation:
We use PV = nRT
Where P = 4.87 atm;
V = 67.54 L
R= 0.0821Latm/molK
T = 158 C = 158 +273 K = 431 K
the number of moles can be obtained by substituting the values in the respective columns and solve for n
n = PV / RT
n = 4.87 * 67.54 / 0.0821 * 431
n = 328.9198 / 35.3851
n = 9.295moles
The number of moles is approximately 9.30moles.
Answer:
-245.7°C es la temperatura del gas bajo 100 torr
Explanation:
Para resolver esta pregunta debemos hacer uso de la ley de Boyle que establece que la presión de un gas es directamente proporcional a la temperatura de este cuando el volumen permanece constante. La ecuación es:
P1T2 = P2T1
<em>Donde P es presión y T temperatura absoluta del estado inicial, 1, y final, 2.</em>
<em />
Reemplazando:
P1 = 1000torr
T2 = ? -Incógnita-
P2 = 100torr
T1 = 273K -Temperatura del hielo fundido = 0°C = 273K
1000torrT2 = 100torr273
T2 = 27.3K
27.3K - 273 =
<h3>-245.7°C es la temperatura del gas bajo 100 torr</h3>
<em />
Answer:
a) = 0.704%
b) = 1.30%
c) = 2.60%
Explanation:
Given that:
= 
For Part A; where Concentration of A = 0.270 M
Percentage Ionization(∝) 



percentage% (∝) = 
= 0.704%
For Part B; where Concentration of B =
M



percentage% (∝) = 0.0130 × 100%
= 1.30%
For Part C; where Concentration of C= 



percentage% (∝) = 0.02608 × 100%
= 2.60%
This is not a phospholipid as it does not contain a phosphate group at the end of the chain, and is not a triglyceride as there is no glyceryl moiety. Each carbon bonded to hydrogens makes the maximum number of C-H bonds possible, therefore there are no multiple bonds between carbons and the lipid is saturated. Therefore the answer is A.
Hope this helps!
Answer:
The answer would either be Carbon or Silicon.
Explanation: