Answer:
Cp = 0.093 J.g⁻¹.°C⁻¹
Solution:
The equation used for this problem is as follow,
Q = m Cp ΔT ----- (1)
Where;
Q = Heat = 300 J
m = mass = 267 g
Cp = Specific Heat Capacity = ??
ΔT = Change in Temperature = 12 °C
Solving eq. 1 for Cp,
Cp = Q / m ΔT
Putting values,
Cp = 300 J / (267 g × 12 °C)
Cp = 0.093 J.g⁻¹.°C⁻¹
Answer:
4.34atm
Explanation:
The following data were obtained from the question:
P1 = 3.50 atm
T1 = 24°C = 24 +273 = 297K
T2 = 95°C = 368K
P2 =?
Using P1 /T1 = P2 /T2, we can obtain the new pressure as follows:
P1 /T1 = P2 /T2
3.5 / 297 = P2 / 368
Cross multiply to express in linear form:
297 x P2 = 3.5 x 368
Divide both side by the 297
P2 = (3.5 x 368) /297
P2 = 4.34atm
Therefore, the gas pressure in the container at 95°C will be 4.34atm
Answer:
V = 22.34 L
Explanation:
Given data:
Volume of O₂ needed = ?
Temperature and pressure = standard
Number of molecules of water produced = 6.0× 10²³
Solution:
Chemical equation:
2H₂ + O₂ → 2H₂O
Number of moles of water:
1 mole contain 6.022× 10²³ molecules
6.0× 10²³ molecules × 1 mole / 6.022× 10²³ molecules
0.99 mole
Now we will compare the moles of oxygen and water.
H₂O : O₂
2 : 1
0.996 : 0.996
Volume of oxygen needed:
PV = nRT
V = nRT/P
V = 0.996 mol × 0.0821 atm.L/mol.K × 273.15 K / 1 atm
V = 22.34 L
Answer: A) monosaccharides are as well simple sugars
