Answer: A mutation because a change in DNA that is natural is a mutation. :)
Answer:
A. AN INCREASE IN BLOOD ACIDITY NEAR THE TISSUES
B. AN INCREASE IN BLOOD TEMPERATURE NEAR THE TISSUES.
C. THE PRESENCE OF A PRESSURE GRADIENT FOR OXYGEN.
Explanation:
Metabolically active tissues need more oxygen to carry out theirs functions. They are involved during excercise and other active phsiological conditions.
There is the reduction in the amount of oxygen reaching these tissues resulting in carbon IV oxide build up, lactic acid formation and temperature increases.
The acidity of the blood near the tissues is increased due to the accumulation of carbon IV oxide in the tissues resulting into a decreased pH. This reduces the affinity of heamoglobin to oxygen in the blood near the metabollically active tissues.
There is also the increase in temperature causing rapid offload of oxygen from oxy-heamoglobin molecules.
The partial pressure of oxygen gradient also affects the rate of oxygen offload by the blood. In metabollically active tissues, the partial pressure of oxygen is reduced in the tissues causing a direct offloading of oxygen to the tissues.
Answer:
280
Explanation:
total pressure=partial pressure (He)+partial pressure (N2)+partial pressure (H2)
Answer:
9.52 × 10³ J
Explanation:
There is some info missing but I found it on the web.
<em>Calculate the energy required to heat 179g of ethanol from -2.2 °C to 19.6 °C. Assume the specific heat capacity of ethanol under these conditions is 2.44 J/g.°C. Be sure your answer has the correct number of significant digits.</em>
Step 1: Given data
- Mass of ethanol (m): 179 g
- Initial temperature: -2.2°C
- Final temperature: 19.6 °C
- Specific heat capacity of ethanol (c): 2.44 J/g.°C
Step 2: Calculate the temperature change
ΔT = 19.6 °C - (-2.2 °C) = 21.8 °C
Step 3: Calculate the energy required (Q)
We will use the following expression.
Q = c × m × ΔT
Q = 2.44 J/g.°C × 179 g × 21.8 °C
Q = 9.52 × 10³ J
Answer:
-285.4 J/K
Explanation:
Let's consider the following balanced equation.
HCl(g) + NH₃(g) ⇒ NH₄Cl(s)
We can calculate the standard entropy change for the reaction (ΔS°r) using the following expression.
ΔS°r = 1 mol × S°(NH₄Cl(s)) - 1 mol × S°(HCl(g)) - 1 mol × S°(NH₃(g))
ΔS°r = 1 mol × 94.6 J/K.mol - 1 mol × 187 J/K.mol - 1 mol × 193 J/K.mol
ΔS°r = -285.4 J/K
