The unit is the Kelvin, but most of the time 273 is subtracted from the Kelvin temperature and the new number is called "degrees Celsius".
Answer:
pH ≅ 4.80
Explanation:
Given that:
the volume of HN₃ = 25 mL = 0.025 L
Molarity of HN₃ = 0.150 M
number of moles of HN₃ = 0.025 × 0.150
number of moles of HN₃ = 0.00375 mol
Molarity of NaOH = 0.150 M
the volume of NaOH = 13.3 mL = 0.0133
number of moles of NaOH = 0.0133× 0.150
number of moles of NaOH = 0.001995 mol
The chemical equation for the reaction of this process can be written as:
1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water
thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol
Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L
Concentration of 
= 
= 0.0458 M
Concentration of 
= 
= 0.0521 M
GIven that :
Ka = 
Thus; it's pKa = 4.72
pH ≅ 4.80
Answer:
A model is developed for predicting oxygen uptake, muscle blood flow, and blood chemistry changes under exercise conditions. In this model, the working muscle mass system is analyzed. The conservation of matter principle is applied to the oxygen in a unit mass of working muscle under transient exercise conditions. This principle is used to relate the inflow of oxygen carried with the blood to the outflow carried with blood, the rate of change of oxygen stored in the muscle myoglobin, and the uptake by the muscle. Standard blood chemistry relations are incorporated to evaluate venous levels of oxygen, pH, and carbon dioxide.
Explanation:
Answer:
KBr is limiting reactant.
Explanation:
Given data:
Mass of KBr =4g
Mass of Cl₂ = 6 g
Limiting reactant = ?
Solution:
Chemical equation:
2KBr + Cl₂ → 2KCl + Br₂
Number of moles of KBr:
Number of moles = mass/molar mass
Number of moles = 4 g/ 119 gmol
Number of moles = 0.03 mol
Number of moles of Cl₂:
Number of moles = mass/molar mass
Number of moles = 6 g/ 70 gmol
Number of moles = 0.09 mol
Now we will compare the moles of reactant with product.
KBr : KCl
2 : 2
0.03 : 0.03
KBr : Br₂
2 : 1
0.03 : 1/2×0.03= 0.015
Cl₂ : KCl
1 : 2
0.09 : 2/1×0.09 = 0.18
Cl₂ : Br₂
1 : 1
0.09 : 0.09
Less number of moles of product are formed by the KBr thus it will act as limiting reactant while Cl₂ is present in excess.
