Answer:
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Explanation:
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Mass of the water : 2.23 g
<h3>Furter explanation</h3>
Heat
Q = m.c.Δt
m= mass, g
c = heat capacity, for water : 4.18 J/g° C.
ΔT = temperature
Q= 140 J
Δt = 75 - 60 = 15
mass of the water :
<span> First you need to know how many isotopes there are of silicon, and its average atomic units (look at periodic table). Then make up a system of equations to solve for it. Theres 3 stable silicon isotopes (28, 29, 30) so you will need to have 3 equations. You must be given the percent abundance of at least one of the isotopes to solve because here I can only see 2 equations (numbered down below) set x = percent abundance of si-28 y = percent abundance of si-29 z = percent abundance of si-30 since all of silicon atoms account for 100% of all silicon: x + y + z = 100% = 1 therefore: 1) x = 1 - y - z You also have 2) 28x + 29y + 30z = average atomic mass you can substitute x so that equation becomes: 28 (1 - y - z) + 29y + 30z = average atomic mass See how you have 2 variables here? You cant go on until you know the value of one isotope already or you have given a clue which you can derive the third equation</span>
<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.
30% should be the percentage of oxygen if the total mass of fe2o3 is 160.
