M(Cs)=133 g/mol
M(O)=16 g/mol
M(CsxOy)=298 g/mol
w(Cs)=0.89
w(O)=0.11
CsxOy
x=M(CsxOy)w(Cs)/M(Cs)
x=298*0.89/133=2
y=M(CsxOy)w(O)/M(O)
y=298*0.11/16=2
Cs₂O₂ cesium peroxide
Answer:
Explanation:
The elements in Group I of the periodic table are called alkali metals. They are called alkali metals because they react with water to form alkali solutions. These metals are very reactive; hence they have to be stored under oil to protect them from corrosion by air and waterwaterwater
Partial pressure of gas A is 1.31 atm and that of gas B is 0.44 atm.
The partial pressure of a gas in a mixture can be calculated as
Pi = Xi x P
Where Pi is the partial pressure; Xi is mole fraction and P is the total pressure of the mixture.
Therefore we have Pa = Xa x P and Pb = Xb x P
Let us find Xa and Xb
Χa = mol a/ total moles = 2.50/(2.50+0.85) = 2.50/3.35 = 0.746
Xb = mol b/total moles = 0.85/(2.50+0.85) = 0.85/3.35 = 0.254
Total pressure P is given as 1.75 atm
Pa = Xa x P = 0.746 x 1.75 = 1.31atm
Partial pressure of gas A is 1.31 atm
Pb = Xb x P = 0.254 x 1.75 = 0.44atm
Partial pressure of gas B is 0.44 atm.
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Answer: It would be Sp2, because H3O+ has planar structure.
<h2>Answer </h2>
Option C - 320J
<u>Explanation </u>
Since ethanol solid at −120 °C and is only cooling down (it won’t change states)
. The amount of Thermodynamic properties values c is given in form of solid, liquid and gas. Amount of energy released is calculated below.
Formula,
= change in temperature x specific heat capacity for solid ethanol x 40
=> 0.5 x 16x 40 = 320J
Therefore, the 320J of heat is released when 40.0g of ethanol cools.