274 mL H3 O+ and fully neutralized
It will take one teaspoon of Mg(OH)2 to completely neutralize 2.00×10^2mL of H3O+.
<h3>What is the purpose of milk of magnesia?</h3>
- For a brief period of time, this medicine is used to relieve sporadic constipation.
- It is an osmotic laxative, which means that it works by drawing water into the intestines, which aids in causing bowel movement.
<h3>What dosage of milk of magnesia is recommended for constipation?</h3>
- Take Milk of Magnesia once day, preferably before bed, in divided doses, or as prescribed by a physician.
- suggested dosage: 30 mL to 60 mL for adults and kids 12 years of age and older. 15 mL to 30 mL for children aged 6 to 11 years.
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the question you are looking for is
People often take milk of magnesia to reduce the discomfort associated with acid stomach or heartburn. The recommended dose is 1 teaspoon, which contains 4.00x 10^{2} mg of Mg(OH)_2. What volume of an HCl solution with a pH of 1.3 can be neutralized by one dose of milk of magnesia? If the stomach contains 2.00x10^{2}mL of pH 1.3 solution, is all the acid neutralized? If not, what fraction is neutralized?
Answer: Hello i am confused are you asking a question?
Explanation:
Answer:
ΔS = -661.0J/mol is the entropy change for the system
ΔS = -842J/mol.K is the entropy change for the surroundings
Explanation:
From the relationship between ΔG, T, ΔH and ΔS,
Mathematically, ΔG = ΔH - TΔS
TΔS = ΔH - ΔS
ΔS = ΔH - ΔS / T
but ΔG = -54 kJ/mol, ΔH = -251 kJ/mol and T = 25 °C (298 K)
plugging into the equation,
ΔS = -251 kJ/mol - ( -54 kJ/mol) / 298
ΔS = -0.6610KJ/mol or in J.mol
ΔS = -661.0J/mol is the entropy change for the system
- For entropy change for the surroundings = ΔS = ΔH/T
- ΔS = -0.84KJ/mol.K or -842J/mol.K is the entropy change for the surroundings
<u>Answer:</u>
<u>read below</u>
<u>Explanation:</u>
<u>When the zebra eats grass, it gets energy from the sun that has been stored in the grass. Similarly, the lion obtains energy stored in the zebra. The zebra and lion both obtain the sun's energy indirectly from the energy that the grass obtained through photosynthesis. </u>
