Answer:
Incomplete Q.
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Answer:
nearer to the north pole
Explanation:
because there is more water, which is basically gone everywhere else on Mars except for the south pole which has less water than the north pole
Answer:
437.5 kg of first solution and 812.5 kg of second solution should be mixed to get desired solution.
Explanation:
Let the mass of the first solution be x and second solution be y.
Amount solution required = 1250 kg
x + y = 1250 kg....[1]
Percentage of ethanol in required solution = 12% of 1250 kg
Percentage of ethanol in solution-1 = 5% of x
Percentage of ethanol in required solution = 25% of y
5% of x + 25% of y =12% of 1250 kg

x + 5y = 3000 kg...[2]
Solving [1] and [2] we :
x = 437.5 kg , y = 812.5 kg
437.5 kg of first solution and 812.5 kg of second solution should be mixed to get desired solution.
Answer:
Endothermic.
Explanation:
Hello there!
In this case, it is necessary to keep in mind that exothermic processes are characterized by the release of energy and the endothermic processes by the absorption of heat. In such a way, every process from solid to liquid or gas is endothermic as they require energy to separate the molecules and therefore turn out in the phase change. On the other hand, every process from gas to liquid or solid is exothermic as heat is released to rejoin the molecules and produce the phase change.
Therefore, since solid water molecules tend to be well-arranged, it is necessary to add heat to the system to produce the phase change until gas; in such a way, this process is endothermic as energy must be absorbed by the ice.
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We first need to convert grams of hydrogen chloride to moles.
molarity = moles of solute/liters of solution
To convert from grams to moles, we need to divide by it's molecular mass.
4.5 g HCl * (1 mole HCl/ 36.46 g HCl) = 0.123 moles HCl
0.12 moles/0.9 L = 0.13 M HCl
If they gave you the volume as 900 mL, then that would be 3 sig figs. And the sig figs would be limited to 2 because the mass.If, however, they gave you the volume as 0.9 L, then that would be 1 sig fig. And the final answer would be limited to to 1 sig fig.