The density of water is 1.0 grams per milliliter then it will be sink in water
Density is a word we use to describe how much space an object or substance takes up in the volume and in relation to the amount of matter in that object or substance its mass) and another way to put it is that density is the amount of mass per unit of volume
Here given density is 1.0 grams per milliliter and it will be sink in water because the density of water in 1.0 g/ml and this object is more dense than water and the density of an object determines whether it will float or sink in another substance and an object will float if it is less dense than the liquid it is placed in and an object will sink if it is more dense than the liquid it is placed
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The initial report of the team formed by PURE may contain findings similar to the processes used in the disinfection of the water. Usual cases of ammonia presence in water are due to chloramine disinfection process. Also, they should also look into the storage of their facility and the end user which may also have a window of contamination.
Yes
Methane also is the chief constituent of natural gas, which contains from 50 to 90 percent methane (depending on the source), and occurs as a component of firedamp (flammable gas) along coal seams.
Answer:
Lime water Ca(OH)2 is a base
Potassium oxide K2O is a basic oxide
Copper sulphate CuSO4 is a salt
And vinegar is an acid
Explanation:
Answer:
See explanation.
Explanation:
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In this case, we consider the questions:
a. Ideal gas at:
i. 273.15 K and 22.414 L.
ii. 500 K and 100 cm³.
b. Van der Waals gas at:
i. 273.15 K and 22.414 L.
ii. 500 K and 100 cm³.
Thus, we define the ideal gas equation and the van der Waals one as shown below:

Whereas b and a for hydrogen sulfide are 0.0434 L/mol and 4.484 L²*atm / mol² respectively, therefore, we proceed as follows:
a.
i. 273.15 K and 22.414 L.

ii. 500 K and 100 cm³ (0.1 L).

b.
i. 273.15 K and 22.414 L: in this case, v = 22.414 L / 1.00 mol = 22.414 L/mol

ii. 500 K and 100 cm³: in this case, v = 0.1 L / 1.00 mol = 0.100 L/mol

Whereas we can see a significant difference when the gas is at 500 K and occupy a volume of 0.100 L.
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