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ch4aika [34]
3 years ago
11

BALANCE THE FOLLOWING EQUATION

Chemistry
1 answer:
Musya8 [376]3 years ago
4 0

Answer:

1) Sc(C2H3O2)3 + Na3BO3 = 3 NaC2H3O2 + ScBO3

2) 2 Ba3(PO4)2 + 3 Ca2SiO4 = 3 Ba2SiO4 + 2 Ca3(PO4)2

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When the concentrations of CH 3 Br and NaOH are both 0.100 M, the rate of the reaction is 0.0030 M/s. What is the rate of the re
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Answer : The rate of the reaction if the concentration of CH_3Br is doubled is, 0.006 M/s

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Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The balanced equations will be:

CH_3Br+NaOH\rightarrow CH_3OH+NaBr

In this reaction, CH_3Br and NaOH are the reactants.

The rate law expression for the reaction is:

\text{Rate}=k[CH_3Br][NaOH]

As we are given that:

[CH_3Br] = concentration of CH_3Br = 0.100 M

[NaOH] = concentration of NaOH = 0.100 M

Rate = 0.0030 M/s

Now put all the given values in the above expression, we get:

0.0030M/s=k\times (0.100M)\times (0.100M)

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\text{Rate}=k[CH_3Br][NaOH]

\text{Rate}=(0.3M^{-1}s^{-1})\times (2\times 0.100M)\times (0.100M)

\text{Rate}=0.006M/s

Thus, the rate of the reaction if the concentration of CH_3Br is doubled is, 0.006 M/s

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4 years ago
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