2 ICl + H2 ----> I2 + 2 HCl
as given that rate is first order with respect to ICl and second order with respect to H2
The rate law will be
Rate = K [ICl] [ H2]^2
b) Given that K = 2.01 M^-2 s^-1
Concentrations are
[ICl] = 0.273 m and [H2] = 0.217 m
Therefore rate = 2.01 X (0.273)(0.217)^2 = 0.0258 M / s
Answer:
d. Two moles of carbon dioxide were produced from this reaction
Explanation:
The given chemical reaction can be written as follows;
2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
From the above chemical reaction, we have;
Two moles of C₂H₂ reacts with five moles of O₂ to produce four moles of CO₂ and two moles of H₂O
We have;
One mole of C₂H₂ will react with two and half moles of O₂ to produce <em>two moles of CO₂</em> and one mole of H₂O
Therefore, in the above reaction, when one mole of C₂H₂ is used, two moles of CO₂ will be produced.
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