<h3>
Answer:</h3>
0.111 J/g°C
<h3>
Explanation:</h3>
We are given;
- Mass of the unknown metal sample as 58.932 g
- Initial temperature of the metal sample as 101°C
- Final temperature of metal is 23.68 °C
- Volume of pure water = 45.2 mL
But, density of pure water = 1 g/mL
- Therefore; mass of pure water is 45.2 g
- Initial temperature of water = 21°C
- Final temperature of water is 23.68 °C
- Specific heat capacity of water = 4.184 J/g°C
We are required to determine the specific heat of the metal;
<h3>Step 1: Calculate the amount of heat gained by pure water</h3>
Q = m × c × ΔT
For water, ΔT = 23.68 °C - 21° C
= 2.68 °C
Thus;
Q = 45.2 g × 4.184 J/g°C × 2.68°C
= 506.833 Joules
<h3>Step 2: Heat released by the unknown metal sample</h3>
We know that, Q = m × c × ΔT
For the unknown metal, ΔT = 101° C - 23.68 °C
= 77.32°C
Assuming the specific heat capacity of the unknown metal is c
Then;
Q = 58.932 g × c × 77.32°C
= 4556.62c Joules
<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
- We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
4556.62c Joules = 506.833 Joules
c = 506.833 ÷4556.62
= 0.111 J/g°C
Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C
Maalox is the trade name for an antacid and antigas medication used for relief of heartburn, bloating, and acid indigestion in which
4 ml contains
= 320mg of aluminum hydroxide
= 320mg of magnesium hydroxide
= 32mg of simethicone
recommended doses = 4 times * 2 tea spoon = 8 tea spoon/ day
given = 1 tea spoon = 5 ml
8 tea spoon = 40 ml
hence,
amount of aluminum hydroxide = 320/4 * 40 = 3200mg = 3.2 g
amount of magnesium hydroxide = 320/4 * 40 = 3.2 g
amount of simethicone = 32/4 * 40 = 320 mg = 0.32g
To know more about antacid visit :
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Answer:
Joe Mama Formula
Explanation:
JoeMamic Acid is the answer
Answer: is A
Explanation:
how becuase i just take the test
Both of you are overlooking a pretty big component of the question...the Group I cation isn't being dissociated into water. We're testing the solubility of the cation when mixed with HCl. And this IS a legitimate question, seeing as our lab manual is the one asking.
<span>By the way, the answer you're looking for is "Because Group I cations have insoluble chlorides". </span>
<span>"In order...to distinguish cation Group I, one adds HCl to a sample. If a Group I cation is present in the sample, a precipitate will form." </span>
