Answer: 94.13 L
Explanation: In STP in an ideal gas there is a standard value for both temperature and pressure. At STP,pressure is equal to 1atm and the temperature at 0°C is equal to 273.15K. This problem is an ideal gas so we use PV=nRT where R is a constant R= 0.08205 L.atm/mol.K.
To find volume, derive the equation, it becomes V=nRT/P. Substitute the values. V= 4.20 mol( 0.08205L.atm/mol.K)(273.15K) / 1 atm = 94.13 L. The mole units, atm and K will be cancelled out and L will be the remaining unit which is for volume.
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
Answer:
Hi there I think the awnser is B. or A. lol but im 90% its A.
Explanation:
xoxo hope this helps
hugs and kisses also your pfp is hella cute:)
Ur answer is going to be francium
Mg + Cl₂ = MgCl₂M(Mg) = 24г/моль m 1 моль Mg = 24 г.По условию задачи дано 12г. Mg Количество вещества n(Mg) =12÷24=0,5 мольРассуждаем: по уравнению реакции с 1 моль магния реагирует 1 моль хлора, следовательно с 0,5 моль будет реагировать 0,5 моль хлора.1 моль хлора при н.у. занимает объем 22,4л. , тогда 0,5 моль хлора займет:0,5х22,4л.= 11,2л. Ответ: Для взаимодействии 12 г. магния потребуется 11,2 л. хлора.