Answer:
1.06 mL of toluene will be needed.
Explanation:
Equi-molar mixture means equal moles of all the components.
as given the volume of ethyl acetate = 1mL
Density of ethyl acetate = 0.898 g/mL
The relation between density, mass and volume is :
mass=volumeXdensity
mass of ethyl acetate present = 1mL X 0.898g/mL = 0.598 grams
the moles are related to mass as:
For ethyl acetate molar mass = 4X12+8X1+2X16= 88g/mol
moles of ethyl acetate will be:
So we need 0.01 moles of toluene also
For 0.01 moles the mass of toluene required = 0.01 X molar mass of toluene
mass required = 0.01 X 92=0.92grams
for 0.92 grams of toluene volume required will be:
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.
First, we write the reaction equation:
NiCO₃ + 2HBr → NiBr₂ + H₂CO₃
Now, writing this in ionic form:
NiCO₃ + 2H⁺ + 2Br⁻ → NiBr₂ + 2H⁺ + CO₃⁻²
(NiCO₃ is insoluble so it does not dissociate in to ions very readily)
Overall equation:
NiCO₃ + 2Br⁻ → NiBr₂ + CO₃⁻²
The correct answer is 1. Lose electrons and become positive ions.
I hope my answer was beneficial to you! c:
Answer:
0.12693 mg/L
Explanation:
First we <u>calculate the concentration of compound X in the standard prior to dilution</u>:
- 10.751 mg / 100 mL = 0.10751 mg/mL
Then we <u>calculate the concentration of compound X in the standard after dilution</u>:
- 0.10751 mg/mL * 5 mL / 25 mL = 0.021502 mg/L
Now we calculate the<u> concentration of compound X in the sample</u>, using the <em>known concentration of standard and the given areas</em>:
- 2582 * 0.021502 mg/L ÷ 4374 = 0.012693 mg/L
Finally we <u>calculate the concentration of X in the sample prior to dilution</u>:
- 0.012693 mg/L * 50 mL / 5 mL = 0.12693 mg/L
