Research Question: Does the type of gasoline put in a car effect how fast the car can drive?
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2KClO₃ → 2KCl + 3O₂
mole ratio of KClO₃ to O₂ is 2 : 3
∴ if moles of O₂ = 5 mol
then moles of KClO₃ =
= 3.33 mol
Mass of KClO₃ needed = mol of KClO₃ × molar mass of KClO₃
= 3.33 mol × ((39 × 1) + (35.5 × 1) + (16 × 3) g/mol
= 407.93 g
mole ratio of KClO₃ to O₂ is 2 : 3
∴ if moles of O₂ = 5 mol
then moles of KClO₃ =
= 3.33 mol
Mass of KClO₃ needed = mol of KClO₃ × molar mass of KClO₃
= 3.33 mol × ((39 × 1) + (35.5 × 1) + (16 × 3) g/mol
= 407.93 g
Answer:
Explanation:
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In this case, since the titration of acids like KHP with bases like NaOH are performed in a 1:1 mole ratio, it is possible for us to know that their moles are the same at the equivalence point, and the concentration, volume and moles are related as follows:
Thus, by solving for the volume of the base as NaOH, we obtain:
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Option (a) is correct.
A reducing agent is the one which loses electrons to other substance and an oxidizing agent is one which accepts electrons.
Here, In
, Cr has oxidation number 6+ in the L.H.S of the equation, but on R.H.S its oxidation number is 0 i.e. it Cr has gained electrons such that total charge is 0.
And the oxidation state of Al in the left-hand side of equation is 0 and in right-hand side, it is +6.i.e. it has donated its electrons to Cr.
Hence, Cr is the oxidizing agent and Al is the reducing agent.
A reducing agent is the one which loses electrons to other substance and an oxidizing agent is one which accepts electrons.
Here, In
And the oxidation state of Al in the left-hand side of equation is 0 and in right-hand side, it is +6.i.e. it has donated its electrons to Cr.
Hence, Cr is the oxidizing agent and Al is the reducing agent.