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:
A net ionic equation shows only the chemical species that are involved in a reaction, while a complete ionic equation also includes the spectator ions.
Brainlist pls!
Answer:
6m/s
Explanation:
Data obtained from the question include:
Mass = 0.5kg
Momentum = 3 kg.m/s
Velocity =.?
Momentum is simply the product of mass and velocity as shown by the equation below:
Momentum = Mass x Velocity
Velocity = Momentum /mass
Velocity = 3kg.m/s / 0.5kg
Velocity = 6m/s
Therefore, the velocity of the ball is 6m/s
Answer:
152 pm
Explanation:
According to the question, we can estimate the bond length from the given values of the atomic radii. This now is the upper limit of the bond length for the molecule.
Since we have that;
Atomic radius of H= 37.0 pm
Atomic radius of Br = 115.0 pm
Bond length = Atomic radius of H + Atomic radius of Br
Bond length = 37.0 pm + 115.0 pm
Bond length = 152 pm
The percent yield of the reaction between ammonia gas with oxygen gas is 90.52%.
A chemical reaction between ammonia gas (NH3) with oxygen gas (O2)
NH₃ + O₂ → NO₂ + H₂O
The balanced reaction 4NH₃ + 7O₂ → 4NO₂ + 6H₂O
Calculate the number of moles from the reactant
- Ammonia gas
Molar mass N = 14 gr/mol
Molar mass H = 1 gr/mol
Molar mass NH₃ = 14 + (3 × 1) = 14 + 3 = 17 gr/mol
mass = 28.5 grams
n = m ÷ molar mass = 28.5 ÷ 17 = 1.68 mol - Oxygen gas
Molar mass O = 16 gr/mol
Molar mass O₂ = 16 × 2 = 32 gr/mol
mass = 83.4 grams
n = m ÷ molar mass = 83.4 ÷ 32 = 2.61 mol - n O₂ ÷ coefficient O₂ = 2.61 ÷ 7 = 0.37
n NH₃ ÷ coefficient NH₃ = 1.68 ÷ 4 = 0.42
0.42 > 0.37 it means that the ammonia gas is in excess and the O₂ is limiting.
According to stoichiometry, the number of moles NO₂ with the number of moles O₂ has the ratio with the coefficient in reaction.
- Theoretically the number moles of NO₂
n O₂ : n NO₂ = 7 : 4
2.61 : n NO₂ = 7 : 4
n NO₂ = 4 x 2.61 : 7 = 1.49 mol - The actual number of moles NO₂
Molar mas NO₂ = 14 + (16 × 2) = 14 + 32 = 46 gr/mol
n NO₂ = m ÷ molar mass = 61.9 ÷ 46 = 1.35 mol
The percent yield NO₂ is the ratio of the actual number of moles NO₂ with the theoretical number of moles NO₂ times 100%.
P = (1.35 ÷ 1.49) × 100%
P = 0.9052 × 100%
P = 90.52%
Learn more about stoichiometry here: brainly.com/question/13691565
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