Answer:
0.0734 grams LiOH
Explanation:
The molarity formula looks like this:
Molarity (M) = moles / volume (L)
You are given the molarity and volume (in mL). Therefore, you can plug these values into the equation and solve for moles (after you convert from mL to L).
25 mL / 1,000 = 0.025 L
0.125 M = moles / 0.025 L
(0.125 M) x (0.025 L) = moles
0.003125 = moles
After using the molarity formula, we know that 0.003125 moles LiOH are needed to satisfy these conditions. Now, we can convert this to grams using the molar mass of LiOH. This can be determined using the values from the periodic table. Remember to write your conversion in a way that allows for the units to cancel out.
Molar Mass (LiOH) = 6.941 g/mol + 16.00 g/mol + 1.008 g/mol
Molar Mass (LiOH) = 23.499 g/mol
0.003125 moles 23.499 g
------------------------ x ------------------- = 0.0734 grams LiOH
1 mole
Answer:
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- <u><em>2)NH</em></u><em><u>₃</u></em>
Explanation:
The answer choices of this question are:
<h2 /><h2>Solution</h2>
The key to answer this question is to realize that <em>NH₃</em> is a gas at the given temperatures.
The solutilibty of all gases always decreases when the temperature increases. Thus, the solubility of NH₃ will decrease when the temperature rises from 10ºC to 50ºC.
The reason for this behavior of gases is that the temperature and the kinetic energy of the particles are directly proportional. Thus, as the temperature incrases the kinetic energy of the particles increases.
As the kinetic energy of the molecules of gas in the liquid solution increases, their speeds also increase, meaning that more molecules will escape from the solution to the gas phase, leaving the soluton with less dissolved gas molecules.
Answer:
Saturated solution
We should raise the temperature to increase the amount of glucose in the solution without adding more glucose.
Explanation:
Step 1: Calculate the mass of water
The density of water at 30°C is 0.996 g/mL. We use this data to calculate the mass corresponding to 400 mL.
Step 2: Calculate the mass of glucose per 100 g of water
550 g of glucose were added to 398 g of water. Let's calculate the mass of glucose per 100 g of water.
Step 3: Classify the solution
The solubility represents the maximum amount of solute that can be dissolved per 100 g of water. Since the solubility of glucose is 125 g Glucose/100 g of water and we attempt to dissolve 138 g of Glucose/100 g of water, some of the Glucose will not be dissolved. The solution will have the maximum amount of solute possible so it would be saturated. We could increase the amount of glucose in the solution by raising the temperature to increase the solubility of glucose in water.
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