Answer:
A magnetic field is a picture that we use as a tool to describe how the magnetic force is distributed in the space around and within something magnetic.
Explanation:
<u>Answer:</u> C) be hypertonic to Tank B.
<u>Explanation: </u>
<u>
The ability of an extracellular solution to move water in or out of a cell by osmosis</u> is known as its tonicity. Additionally, the tonicity of a solution is related to its osmolarity, which is the <u>total concentration of all the solutes in the solution.
</u>
Three terms (hypothonic, isotonic and hypertonic) are used <u>to compare the osmolarity of a solution with respect to the osmolarity of the liquid that is found after the membrane</u>. When we use these terms, we only take into account solutes that can not cross the membrane, which in this case are minerals.
- If the liquid in tank A has a lower osmolarity (<u>lower concentration of solute</u>) than the liquid in tank B, the liquid in tank A would be hypotonic with respect to the latter.
- If the liquid in tank A has a greater osmolarity (<u>higher concentration of solute</u>) than the liquid in tank B, the liquid in tank A would be hypertonic with respect to the latter.
- If the liquid in tank A has the same osmolarity (<u>equal concentration of solute</u>) as the liquid in tank B, the liquid in tank A would be isotonic with respect to the latter.
In the case of the problem, option A is impossible because the minerals can not cross the membrane, since it is permeable to water only. There is no way that the concentration of minerals decreases in tank A, so <u>the solution in this tank can not be hypotonic with respect to the one in Tank B. </u>
Equally, both solutions can not be isotonic and neither we can say that the solution in tank A has more minerals that the one in tank B because the liquid present in tank B is purified water that should not have minerals. Therefore, <u>options B and D are also not correct.</u>
Finally, the correct option is C, since in the purification procedure the water is extracted from the solution in tank A to obtain a greater quantity of purified water in tank B. In this way, the solution in Tank A would be hypertonic to Tank B.
Answer:
8608.18 balloons
Explanation:
Hello! Let's solve this!
Data needed:
Enthalpy of propane formation: 103.85kJ / mol
Specific heat capacity of air: 1.009J · g ° C
Density of air at 100 ° C: 0.946kg / m3
Density of propane at 100 ° C: 1.440kg / m3
First we will calculate the propane heat (C3H8)
3000g * (1mol / 44g) * (103.85kJ / mol) * (1000J / 1kJ) = 7.08068 * 10 ^ 6 J
Then we can calculate the mass of the air with the heat formula
Q = mc delta T
m = Q / c delta T = (7.08068 * 10 ^ 6 J) / (1.009J / kg ° C * (100-25) ° C) =
m = 93566.96kg
We now calculate the volume of a balloon.
V = 4/3 * pi * r ^ 3 = 4/3 * 3.14 * 1.4m ^ 3 = 11.49m ^ 3
Now we calculate the mass of the balloon
mg = 0.946kg / m3 * 11.49m ^ 3 = 10.87kg
The amount of balloons is
93566.96kg / 10.87kg = 8608.18 balloons
The relation between density and mass and volume is

the dose required is 2.5 tsp
each tsp contain 5mL
So dose required in mL = 2.5 X 5 = 12.5 mL
the mass will be calculated using following formula


The mass of dose in grams will be 15.38 g
Answer:
The enthalpy of combustion of ethanol in kJ/mol is -1419.58 kJ/mol.
Explanation:
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change. Working with this equation, and assuming no heat is lost to the surroundings, we write
:
qcal= Ccal × ΔT= 490 J/K × 276.7 K= <u>135,583 J</u> = 135.58 kJ
Note we expressed the temperature change in K, because the heat capacity is written in K.
<u>
Now that we have the heat of combustion, we need to calculate the molar heat. </u>
Because qsystem = qrxn + qcal and qrxn = -qcal, the heat change of the reaction is -135.58 kJ.
This is the heat released by the combustion of 4.40 g of ethanol ; therefore, we can write the <u>conversion factor as 135.58 kJ/ 4.40 g</u>.
The molar mass of ethanol is 46.07 g, so the heat of combustion of 1 mole of ethanol is
:
molar heat of combustion= -135.58 kJ/4.40 g x 46.07 g/ 1 mol= -1419.58 kJ/mol
Therefore, the enthalpy of combustion of ethanol in kJ/mol is -1419.58 kJ/mol.