Answer:
I think it's D. The expansion of water as it freezes increases the amount of nutrients that can dissolve in liquid water, but it could cause fluid in cells to dissolve more harmful substances.
Explanation:
I know when water freezes, it expands and between the two answers that discuss the expansion of water, D sounds the most logical to me lol.
<u>Answer:</u>
<em>1. A NaCl solution with a concentration of 50g/100mL of water at 40°C:</em> The NaCl solution with a given concentration is saturated at this temperature .As the temperature increases the solution will more dissolves.
<em>2. A sugar solution with a concentration of 200g/100mL of water at 40°C: </em>The sugar solution with a given concentration is saturated at this temperature. As the temperature increases the solution will more dissolves.
<em>3. A sugar solution with a concentration of 240g/100mL of water at 40°C:</em> The sugar solution with a given concentration is saturated at given temperature.
Answer:
The change of the volume of the device during this cooling is 
Explanation:
Given that,
Mass of oxygen = 10 g
Pressure = 20 kPa
Initial temperature = 110°C
Final temperature = 0°C
We need to calculate the change of the volume of the device during this cooling
Using formula of change volume
Put the value into the formula
Hence, The change of the volume of the device during this cooling is
Answer:
The resultant force would (still) be zero.
Explanation:
Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.
In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.
By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.
When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.
However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.
