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omeli [17]
3 years ago
5

Write the rate law for the following elementary reaction:

Chemistry
1 answer:
Harrizon [31]3 years ago
6 0

Answer:

Rate law = k1 [N2O5]^2

Explanation:

Rate law only cares about reactants and rate law can only be determined experimentally or by using the coefficients of reactants in elementary reactions

If 2A---->B is an elementary reaction... rate law for this is rate=k[A]^2

So if we look at your question...2N2O5---->2N2O4 + 02 (I think you are missing O2 in question because otherwise equation is unbalanced)

Rate law = k1 [N2O5]^2

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PLEASE HELP!!
wariber [46]

Answer:

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A hydrate of CoCl2 has a mass of 17.40 g before heating. After heating, the mass of the anhydrous compound is found to be 10.27
melisa1 [442]

Mass CoCl2 = 10.27 g
moles CoCl2 = 10.27 g/ 129.839 g/mol=0.07910

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moles CaF2 = 85.8 g/ 78.0748 g/mol=1.10
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5 0
3 years ago
g Enter your answer in the provided box. If 30.8 mL of lead(II) nitrate solution reacts completely with excess sodium iodide sol
Ronch [10]

Answer:

M=0.0637M

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

Pb(NO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaNO_3(aq)

Thus, for 0.904 g of precipitate, that is lead (II) iodide, we can compute the initial moles of lead (II) ions in lead (II) nitrate:

n_{Pb^{2+}}=0.904gPbI_2*\frac{1molPbI_2}{461gPbI_2}*\frac{1molPb(NO_3)_2}{1molPbI_2}  *\frac{1molPb^{2+}}{1molPb(NO_3)_2} =1.96x10^{-3}molPb^{2+}

Finally, the resulting molarity in 30.8 mL (0.0308 L):

M=\frac{1.96x10^{-3}molPb^{2+}}{0.0308L}\\ \\M=0.0637M

Regards.

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3 years ago
5. I have a mixture of salt, water,
dolphi86 [110]

Iron is left in the filter and salt solution (salt and water) passes into the cup.

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