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timama [110]
3 years ago
12

For each of the following cases, identify the order with respect to the reactant, A. Case (A ----> product)

Chemistry
1 answer:
damaskus [11]3 years ago
5 0

Answer:

The answers to the question are

1. 2nd and above order order

2. 2nd order

3. 1/2 order

4.  1st order

5. 0 order

Explanation:

We have \frac{N}{N_{0} } = e^{-\lambda t}

1. For nth order reaction half life  t_{\frac{1}{2} } ∝ \frac{1}{[A_{0} ]^{n-1} }

Therefore for a 0 order reaction increasing concentration of the reactant there will increase   t_{\frac{1}{2} }

First order reaction    is independent  [A₀].

Second order reaction  [A₀] decrease,    increase.

Similarly for a  third order reaction

1. 2nd order

2. 2nd order reaction

3. Order of reaction is 1/2.

4. 1st order reaction.

5. Zero order reaction.

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Most marble is composed primarily of
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Answer:

It is composed primarily of the mineral calcite (CaCO3) and usually contains other minerals, such as clay minerals, micas, quartz, pyrite, iron oxides, and graphite.

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3 years ago
Read 2 more answers
1.31 times 10^-22 / 6.6262 times 10^-34
victus00 [196]
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4 0
3 years ago
Problem PageQuestion Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid w
quester [9]

Answer:

0.72g

Explanation:

Step 1:

We'll begin by writing a balanced equation for the reaction. This is illustrated below:

H2SO4 + 2NaOH —> Na2SO4 + 2H2O

Step 2:

Determination of the mass of sulphuric acid (H2SO4) and the mass of sodium hydroxide (NaOH) that reacted from the balanced equation. This is illustrated below:

Molar Mass of H2SO4 = (2x1) + 32 +(16x4) = 2 + 32 + 64 = 98g/mol

Molar Mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the balanced equation = 2 x 40 = 80g

Step 3

Determination of the limiting reactant. To do this, we need to know which of the reactant is excess.

Now let us consider using all of the mass of NaOH given to see if there will be left over for H2SO4. This is illustrated below:

From the balanced equation above,

98g of H2SO4 required 80g of NaOH.

Therefore, Xg of H2SO4 will require 1.6g of NaOH i.e

Xg of H2SO4 = (98x1.6)/80

Xg of H2SO4 = 1.96g

Now comparing the mass of H2SO4 that reacted ( i.e 1.96g) and the mass of H2SO4 given ( i.e 2.94g), we can see clearly that there are left over ( i.e 2.94 - 1.96 = 0.98g) of H2SO4. Therefore, H2SO4 is the excess reactant and NaOH is the limiting reactant.

Step 4:

Determination of the mass of water produced from the reaction. This is illustrated below:

The balanced equation for the reaction is given below:

H2SO4 + 2NaOH —> Na2SO4 + 2H2O

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g

From the balanced equation above,

80g of NaOH reacted to produced 36g of H2O.

Therefore, 1.6g of NaOH will react to produce = (1.6 x 36)/80 = 0.72g of H2O.

Therefore, the maximum mass of water (H2O) produced by the chemical reaction of aqueous sulfuric acid with solid sodium hydroxide is 0.72g

4 0
3 years ago
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