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0.50 moles of oxygen gas will react.
Balanced chemical eqation for reaction of calcium with oxygen:
2Ca + O₂ → 2CaO
n(Ca) = 1.0 mol; amount of calcium
From balanced chemical reaction, we can see that 2 moles of calcium reacts with one mole of oxygen gas. Ratio of moles is 2 : 1.
n(Ca) : n(O₂) = 2 : 1; ratio of the amount of substance of calcium and oxygen gas
n(O₂) = n(Ca) / 2; two times less amount of oxygen than calcium
n(O₂) = 1.0 mol / 2
n(O₂) = 0.5 mol; amount of oxygen gas needed for this chemical reaction.
More about stoichiometry: brainly.com/question/16060223
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The melting point of a particular compound is fixed and it is an important identification of an unknown compound. The practice to determine the melting point of an unknown material In different ratio with a known material is important to get the exact melting point of the unknown material. In different ratio the melting point of the unknown material will be fixed as the melting point of a pure material doesn't depend on the ratio in which they are mixed with other material. To get the exact melting point it is always good to get the melting point twice in different ratio.
Answer:
104.352°C
Explanation:
Data Given:
Boiling point of water = 100.0°C
Kb (boiling point constant = 0.512°C/m
Concentration of the Mg₃(PO₄)₂ = 8.5 m
Solution:
Formula Used to find out boiling point
ΔTb = m.Kb . . . . . . (1)
where
ΔTb = boiling point of solution - boiling point of water
So,
we can write equation 1 as under
ΔTb = Tb (Solution) -Tb (water)
As we have to find out boiling point so rearrange the above equation
Tb (Solution) = m.Kb + Tb (water) . . . . . . . (2)
Put values in Equation 2
Tb (Solution) = (8.5 m x 0.512°C/m ) + 100.0°C
Tb (Solution) = 4.352 + 100.0°C
Tb (Solution) = 104.352°C
so the boiling point of Mg₃(PO₄)₂ 8.5 m solution = 104.352°C
Molarity = moles of solute/volume of solution in liters.
There are 2.66 moles of KOH, the solute, and the volume of the solution is 0.750 L.
The molarity of this solution would thus be (2.66 moles KOH)/(0.750 L) = 3.55 M KOH.