To insert a thermometer into an adapter, use <u>mineral oil</u> to prepare the thermometer. Then, hold the thermometer <u>close to</u> the adapter and<u> slowly turn</u> the thermometer into the adapter.
The term "temperature" refers to a measurement of how cold or hot an actual physical object is. It is measured with a thermometer, which gives readings in Celsius, Kelvin, and Fahrenheit (°C, K, and °F).
The average kinetic energy of the particles in a given substance is often measured by temperature. A thermometer is a tool used to gauge a substance's or a body's temperature (degree of hotness or coolness). It is a bulb-shaped piece of thin glass that usually contains either coloured alcohol or mercury.
In order to get readings throughout the distillation process, a thermometer adapter is used with a temperature probe. Use mineral oil to prepare or make the thermometer suitable before inserting it into the adapter. After that, slowly insert the thermometer into the adaptor while holding it close to it.
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Bro honestly I don’t understand either
Glucose is a simple sugar with the molecular formula C6H12O6. It is a carbohydrate.
Answer: The volume of 0.640 grams of 
gas at Standard Temperature and Pressure (STP) is 0.449 L.
Explanation:
Given: Mass of gas = 0.640 g
Pressure = 1.0 atm
Temperature = 273 K
As number of moles is the mass of substance divided by its molar mass.
So, moles of (molar mass = 32.0 g/mol) is as follows.
Now, ideal gas equation is used to calculate the volume as follows.
PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.
Thus, we can conclude that the volume of 0.640 grams of gas at Standard Temperature and Pressure (STP) is 0.449 L.
Answer:
There is 54.29 % sample left after 12.6 days
Explanation:
Step 1: Data given
Half life time = 14.3 days
Time left = 12.6 days
Suppose the original amount is 100.00 grams
Step 2: Calculate the percentage left
X = 100 / 2^n
⇒ with X = The amount of sample after 12.6 days
⇒ with n = (time passed / half-life time) = (12.6/14.3)
X = 100 / 2^(12.6/14.3)
X = 54.29
There is 54.29 % sample left after 12.6 days
