The correct answer is B) Basic. Hope this helps.
Answer:
Percent error = 25%
Explanation:
Given data:
Measured density of water = 1.25 g/mL
Accepted density value of water = 1 g/mL
Percent error = ?
Solution:
Formula:
Percent error = (measured value - accepted value / accepted value) × 100
Now we will put the values in formula:
Percent error = (1.25 g/mL - 1 g/mL /1 g/mL )× 100
Percent error = (0.25 g/mL /1 g/mL )× 100
Percent error = 0.25 × 100
Percent error = 25%
<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:
Heat given off was -34.34kJ
Explanation:
Mass of iron bar = 869g
Initial temperature (T1) = 94°C
Final temperature (T2) = 5°C
Specific heat capacity of iron (c) = 0.444J/g°C
Heat energy (Q) = Mc∇T
Q = heat energy
c = specific heat capacity
∇T = change in temperature
M = mass of the substance
Q = mc∇T
∇T = T2 - T1
Q = Mc(T2 -T1)
Q = 869 * 0.444 * (5 - 94)
Q = 385.836 * -89
Q = -34339.404J
Q = -34.34kJ
The heat given of was -34.34kJ
The last row going across
