Answer: NH3 and H2O
Explanation:
NH4OH(l) --> NH3(g) + H2O(g)
Ammonia liquor, NH4OH is an highly unstable hydroxide of ammonium, hence it decomposes on heating to produce ammonia gas, NH3 which is liberated to the surroundings and water, H2O evaporated as steam. In a reverse reaction, ammoniacal liquor is formed by dissolving gaseous ammonia in cold water.
Thus, the products of NH4OH break down are NH3 and H2O
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.
![400 mL \times \frac{0.996g}{1mL} =398g](https://tex.z-dn.net/?f=400%20mL%20%5Ctimes%20%5Cfrac%7B0.996g%7D%7B1mL%7D%20%3D398g)
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.
![100gH_2O \times \frac{550gGlucose}{398gH_2O} = 138 gGlucose](https://tex.z-dn.net/?f=100gH_2O%20%5Ctimes%20%5Cfrac%7B550gGlucose%7D%7B398gH_2O%7D%20%3D%20138%20gGlucose)
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.
The overlapping of two s atomic orbitals produces two molecular orbitals, <span>one bonding molecular orbital and one anti-bonding molecular orbital.
Whenever two atomic orbitals overlap according to molecular orbital theory, it will produce one bonding and one anti bonding molecular orbital. Molecular orbital theory is also a way for determining molecular shape wherein electrons are not assigned to character bonds between atoms, however are dealt with as transferring underneath the effect of the nuclei within the entire molecule.</span>
Answer:
The molarity of the HCl solution is 4M.
Explanation:
![Molarity=\frac{number\:of\:moles}{Volume\:of\:solution\:in\:L}\\ \\The\:volume\:of\:solution=250mL=0.25L\\\\Number\:of\:moles=\frac{weight\:of\:HCl\:in\:solution}{molecular\:weight\:of\:HCl}\\ \\Molecular\:weight\:of\:HCl=36.5g\\\\The\:weight\:of\:HCl\:in\:solution=36.5g\\\\Number\:of\:moles\:of\:HCl=\frac{36.5}{36.5}=1\:mole\\\\The\:molarity\:of\:HCl\:solution=\frac{1}{0.25}=4M](https://tex.z-dn.net/?f=Molarity%3D%5Cfrac%7Bnumber%5C%3Aof%5C%3Amoles%7D%7BVolume%5C%3Aof%5C%3Asolution%5C%3Ain%5C%3AL%7D%5C%5C%20%5C%5CThe%5C%3Avolume%5C%3Aof%5C%3Asolution%3D250mL%3D0.25L%5C%5C%5C%5CNumber%5C%3Aof%5C%3Amoles%3D%5Cfrac%7Bweight%5C%3Aof%5C%3AHCl%5C%3Ain%5C%3Asolution%7D%7Bmolecular%5C%3Aweight%5C%3Aof%5C%3AHCl%7D%5C%5C%20%5C%5CMolecular%5C%3Aweight%5C%3Aof%5C%3AHCl%3D36.5g%5C%5C%5C%5CThe%5C%3Aweight%5C%3Aof%5C%3AHCl%5C%3Ain%5C%3Asolution%3D36.5g%5C%5C%5C%5CNumber%5C%3Aof%5C%3Amoles%5C%3Aof%5C%3AHCl%3D%5Cfrac%7B36.5%7D%7B36.5%7D%3D1%5C%3Amole%5C%5C%5C%5CThe%5C%3Amolarity%5C%3Aof%5C%3AHCl%5C%3Asolution%3D%5Cfrac%7B1%7D%7B0.25%7D%3D4M)
Hence, the molarity of the HCl solution = 4 M
Hey! Let me help you!
C)
<span>What
additional volume of 10.0 M HCl would be needed to exhaust the
remaining capacity of the buffer after the reaction described in Part B?
In other words, how much more of this HCl solution is required to
consume the remaining Tris in the buffer?
let x = how much more solution needed
x = 1.8 mL <=========
Your answer is 1.8 mL as it is needed for more of this HCI solution is requires to consume the remaining Tris in the buffer!
Have an awesome day! :D
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