2 nitrogen atoms, 4 hydrogen atoms and 3 oxygen atoms
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:
8.73
Explanation:
when you are adding or subtracting numbers, the sigfig (significant figure) is based on how many numbers after the decimal. with this info, we can see that 15.67 has 2 sigfigs and 6.943 has 3 sigfigs. when you subtract normally, you would get 8.727, which has 3 sigfigs, so you would round the last 7 up to get 8.73 with 2 sigfigs!
also it is to 2 sigfigs because we know that we go by the least number of sigfigs. hope this helped!
Answer is: molarity of hydrofluoric solution is 0.09 M.
Chemical reaction: HF(aq) + KOH(aq) → KF(aq) + H₂O(l).
V(HF) = 30.0 mL.
c(KOH) = 0.122 M.
V(KOH) = 22.15 mL:
c(HF) = ?.
From chemical reaction: n(HF) : n(KOH) = 1 : 1.
n(HF) = n(KOH).
c(HF) · V(HF) = c(KOH) · V(KOH).
c(HF) = c(KOH) · V(KOH) ÷ V(HF).
c(HF) = 0.122 M · 22.15 mL ÷ 30 mL:
c(HF) = 0.09 M.
Balanced chemical equation:
2 C2H2 + 5 O2 = 4 CO2 + 2 H2O
2 moles C2H2 ---------------- 5 moles O2
moles C2H2 ------------------ 84 moles O2
moles C2H2 = 84 * 2 / 5
molesC2H2 = 168 / 5 => 33.6 moles of C2H2