Answer:
71.7 L
Explanation:
Using the ideal gas equation;
PV = nRT
Where;
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/Kmol)
T = temperature (K)
According to the information provided in this question;
P = 1 atm (STP)
V = ?
n = 3.2mol
T = 273K (STP)
Using PV = nRT
V = nRT/P
V = 3.2 × 0.0821 × 273/1
V = 71.7 L
Answer:
A) During this procedure ( hypoventilation ) The CO2 in the arterial blood vessels and the lungs increases and this drives the PH level in the system lower, and the equilibrium will shift to the right. this is because the Blood-PH level is controlled by CO2 - bicarbonate buffer system
B) The blood PH may rise to 7.60 during Hyperventilation because the removal of CO2 from the lungs causes the increase in 
which is directly proportional to the increase in Blood PH levels
C) Hyper ventilation before a dash would be useful because it will remove excessive Hydrogen ions and and raise the Blood PH levels in preparedness of the production of acids like Lactic acid
Explanation:
A) During this procedure ( hypoventilation ) The CO2 in the arterial blood vessels and the lungs increases and this drives the PH level in the system lower, and the equilibrium will shift to the right. this is because the Blood-PH level is controlled by CO2 - bicarbonate buffer system
⇄ 
B) The blood PH may rise to 7.60 during Hyperventilation because the removal of CO2 from the lungs causes the increase in which is directly proportional to the increase in Blood PH levels
C) Hyper ventilation before a dash would be useful because it will remove excessive Hydrogen ions and and raise the Blood PH levels in preparedness of the production of acids like Lactic acid
First, we need the no.of moles of O2 = mass/molar mass of O2
= 55 g / 32 g/mol
= 1.72 mol
from the balanced equation of the reaction:
2H2 (g) + O2(g) → 2H2O(g)
we can see that the molar ratio between O2: H2O = 1: 2
So we can get the no.of moles of H2O = 2 * moles of O2
= 2 * 1.72 mol
= 3.44 mol
So by substitution by this value in ideal gas formula:
PV = nRT
when P = 12.4 atm & n H2O = 3.44 mol & R= 0.0821 & T = 85 + 273=358K
12.4 atm *V = 3.44 * 0.0821 * 358 = 8.15 L
∴ V ≈ 8.2 L
The answer is 0.59 M.
Molar mass (Mr) of MgCl₂ is the sum of the molar masses of its elements.
So, from the periodic table:
Mr(Mg) = 24.3 g/l
Mr(Cl) = 35.45 g/l
Mr(MgCl₂) = Mr(Mg) + 2Mr(Cl) = 24.3 + 2 · 35.45 = 24.3 + 70.9 = 95.2 g/l
So, 1 mol has 95.2 g/l.
Our solution contains 55.8g in 1 l of solution, which is 55.8 g/l
Now, we need to make a proportion:
1 mole has 95.2 g/l, how much moles will have 55.8 g/l:
1 M : 95.2 g/l = x : 55.8 g/l
x = 1 M · 55.8 g/l ÷ 95.2 g/l ≈ 0.59 M
