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3 years ago

2NaOH+CO2----->Na2CO3+H2O

Chemistry

1 answer:

kondaur [170]3 years ago

8 0

Answer:

13.2 g Na2CO3

Explanation:

Convert 10.0 g NaOH to mol.

10.0 g x 1 mol/39.997 g = 0.250 mol

Use mol ration given by the equation: 2 mol NaOH to 1 mol Na2CO3

0.250 mol NaOH x 1 mol Na2CO3/2 mol NaOH = 0.125 mol Na2CO3

Finally, convert the moles of Na2CO3 to grams.

0.125 mol Na2CO3 x 105.99 g/1 mol = 13.2 g

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Explanation:

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Many double-displacement reactions are enzyme-catalyzed via the "ping pong" mechanism, so called because the reactants appear to

zhenek [66]

Answer:

D. It will decrease by a factor of 4

Explanation:

According to the question , the equation follows :

$A+B\rightarrow C+D$

Rate law : This states the rate of reaction is directly proportional to concentration of reactants with each reactant raised to some power which may or may not be equal to the stoichiometeric coefficient.

$Rate\ \alpha [A]^{a}[B]^{b}$

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STEP": First, find out the power "a" and "b"

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According to question, doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,

r' = 2r if [A'] = 2[A]

Here [B] is uneffected means [B']=[B]

hence new rate =

$r'=[A']^{a}[B']^{b}$

Put the value of [A'] , [B'] and r' in the above equation:

$2r=[2A]^{a}[B]^{b}$ ...........(2)

Divide equation (1) by (2) we , get

$\frac{2r}{r}=\frac{[2A]^{2}[B]^{b}}{[A]^{a}[B]^{b}}$

$2= 2(\frac{A}{A})^{a}\times (\frac{B}{B})^{b}$

Here A and A cancel each other

B and B cancel each other

We get,

$2= 2^{a}\times 1^{b}$

1^b = 1 ( power of 1 = 1)

$2= 2^{a}$

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1 + b = 3

b =3 -1 = 2

b = 2

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$r=[A]^{a}[B]^{b}$

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Look in the question now, it is asked to calculate the concentration of [B],if cut in half

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[B']=1/2[B]

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$r'=[A]^{1}[B']^{2}$

$r'=[A]^{1}(\frac{1}{2}[B])^{2}$

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But

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Compare equation (a) and (b) , we get

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2 years ago

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