Boyle's law gives the relationship between pressure of a gas and volume.
It states that the pressure of a gas is inversely proportional to volume of gas at a constant temperature.
PV = k
where P - pressure , V - volume and k - constant
P1V1 = P2V2
where parameters for the first instance are on the left side of the equation and parameters for the second instance are on the right side of the equation.
1.50 atm x 500.0 mL = P x 150.0 mL
P = 5 atm
applied pressure should be 5 atm to compress the gas.
Answer:
See Explanation
Explanation:
It is a common observation that a strip of aluminium metal in aqueous copper(II)Sulfate does not show any visible reaction. Aluminium is normally expected to displace copper in solution since it is higher than copper in the electrochemical series.
The reason for this is that aluminium forms an oxide film around its surface which prevents reaction with aqueous copper(II)Sulfate. This oxides film protects the aluminium surface such that it is now unable to react with the aqueous copper(II)Sulfate
Answer:
Sry but i hope u will get the right answer and have a great christmas only 3 more days!
Explanation:
Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
Therefore, since the masses of both of the reactants are given, one computes the available moles of sulfuric acid and those moles of it consumed by the sodium hydroxide as shown below:
In such a way, since there is more available sulfuric acid than it that is consumed, the sodium hydroxide is the limiting reagent, consequently, the maximum mass of sodium sulfate turns out:
Best regards.
The formula of compound is LiClO4.3H2O
<em><u>calculation</u></em>
- <em><u> </u></em>find the mole of each element
that is moles for Li,Cl,O and that of H2O
- moles = % composition/ molar mass
For Li = 4.330/ 6.94 g/mol= 0.624 moles
Cl=22.10/35.5=0.623 moles
39.89/16 g/mol =2.493 moles
H20= 33.69/18 g/mol= 1.872 moles
- find the mole ratio by dividing each moles by smallest number of mole ( 0.624 moles)
that is for Li= 0.624/0.623= 1
Cl= 0.623/0.623=1
O = 2.493/0.623 =4
H2O= 1.872/0.623=3
<h3>Therefore the formula=LiClO4.3H2O</h3><h3 />
