Answer:
In an experiment, a student transferred 4.50 mL of a liquid into a pre-weighed beaker (the weight of which was determined to be 35.986 g ).
Explanation:
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<em>Answer:</em>
- Conc. of K+ ions = 0.90 M
- Coc. of SO4∧-2 = 0.45 M
<em>Explanation:</em>
<em>Data Given:</em>
Conc. of H2SO4 = 0.450
As sulphoric acid is a strong electrolyte, it completely dissociate into ions.
H2SO4 ⇆ 2K+ + SO4∧-2
.450 M K2SO4 means that there is .450 mols of K2SO4 in every liter of solution.
K2SO4 : K+ K2SO4 : SO4∧-2
1 = 2 1 = 1
0.450 = 2× 0.450 = 0.90 0.450 = 0.450×1 = 0.450
<em> Result:</em>
Conc. of potassium ion will be 0.90M
Coc. of sulphate ions will be 0.45 M
Answer:
Neon is an inner gas because it’s outer <u>__shell__</u> is full of electrons.
Answer:
Explanation:
Homogeneous mixture is a mixture in which the components of the mixture are in the same proportion throughout any sample extracted from the mixture while an heterogeneous mixture is a mixture in which the components of the mixture differ in term of proportion when different samples of the mixture are extracted and compared.
For example, a sandy water will have some parts (usually the bottom) of the mixture with more sand than other parts of the mixture, hence, it (sandy water) is a heterogeneous mixture. While salty and ocean water has it's salt dissolved in the same proportion throughout the water in the mixture, hence salty and/or ocean water is a homogeneous mixture.
Sandy water can be separated by filtration (i.e using a filter paper to separate the sand from the water when the mixture is poured over a filter paper) while salty and ocean water can be separated by distillation (i.e boiling of the mixture so the water molecules can boil and move through a tube as gas or steam into another container where they are cooled and converted back to liquid or water while leaving the solid salt component of the mixture in the boiling tube).
1) At tne same temperature and with the same volume, initially the chamber 1 has the dobule of moles of gas than the chamber 2, so the pressure in the chamber 1 ( call it p1) is the double of the pressure of chamber 2 (p2)
=> p1 = 2 p2
Which is easy to demonstrate using ideal gas equation:
p1 = nRT/V = 2.0 mol * RT / 1 liter
p2 = nRT/V = 1.0 mol * RT / 1 liter
=> p1 / p2 = 2.0 / 1.0 = 2 => p1 = 2 * p2
2) Assuming that when the valve is opened there is not change in temperature, there will be 1.00 + 2.00 moles of gas in a volumen of 2 liters.
So, the pressure in both chambers (which form one same vessel) is:
p = nRT/V = 3.0 mol * RT / 2liter
which compared to the initial pressure in chamber 1, p1, is:
p / p1 = (3/2) / 2 = 3/4 => p = (3/4)p1
So, the answer is that the pressure in the chamber 1 decreases to 3/4 its original pressure.
You can also see how the pressure in chamber 2 changes:
p / p2 = (3/2) / 1 = 3/2, which means that the pressure in the chamber 2 decreases to 3/2 of its original pressure.