Answer:
68325.8 g/mol
Explanation:
Given that:
Pressure = 743 torr
Temperature = 37 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (37 + 273.15) K = 310.15 K
T = 310.15 K
Volume = 2.30 mL = 0.00230 L ( 1 mL = 0.001 L)
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 62.364 L.torr/K.mol
Applying the equation as:
743 torr × 0.00230 L = n × 62.364 L.torr/K.mol × 310.15 K
⇒n = 8.84 × 10⁻⁵ moles
Since, given that:-
4 moles of oxygen gas is transferred by 1 mole of haemoglobin.
8.84 × 10⁻⁵ moles of oxygen gas is transferred by 8.84 × 10⁻⁵/4 moles of haemoglobin.
Moles of haemoglobin = 2.21 × 10⁻⁵ moles
Mass = 1.51 g
The formula for the calculation of moles is shown below:
Thus,
Answer:
D
Explanation:
If you add two or more thermochemical equations to give a final equation, then you also add the heats of reaction to give the final heat of reaction
Answer:
89.3 mL
Explanation:
First we <u>convert 20.0 °C and 54.0 °C to K</u>:
- 20.0 °C + 273.16 = 293.16 K
- 54.0 °C + 273.16 = 327.16 K
With the absolute temperatures we can solve this problem by using <em>Charles' law</em>:
Where in this case:
We <u>input the data</u>:
- 293.16 K * V₂ = 327.16 K * 80.0 mL
And <u>solve for V₂</u>: