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IRISSAK [1]
3 years ago
15

Give the characteristic of a first order reaction having only one reactant.a. The rate of the reaction is not proportional to th

e concentration of the reactantb. The rate of the reaction is proportional to the square of the concentration of the reactantc. The rate of the reaction is proportional to the square root of the concentration of the reactantd. The rate of the reaction is proportional to the natural logarithm of the concentration of the reactante. The rate of the reaction is directly proportional to the concentration of the reactant
Chemistry
1 answer:
kicyunya [14]3 years ago
3 0

Answer:

E) The rate of the reaction is directly proportional to the concentration of the reactant.

Explanation:

Give the characteristic of a first order reaction having only one reactant.

A) The rate of the reaction is not proportional to the concentration of the reactant.

B) The rate of the reaction is proportional to the square of the concentration of the reactant.

C) The rate of the reaction is proportional to the square root of the concentration of the reactant.

D) The rate of the reaction is proportional to the natural logarithm of the concentration of the reactant.

E) The rate of the reaction is directly proportional to the concentration of the reactant.

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How many grams of water are needed to dissolve 27.8 g of ammonium nitrate NH4NO3 in order to prepare a 0.452 m solution?
Vanyuwa [196]

Answer: 770 g water are needed to dissolve 27.8 g of ammonium nitrate NH_4NO_3 in order to prepare a 0.452 m solution

Explanation:

Molality : It is defined as the number of moles of solute present per kg of solvent

Formula used :

Molality=\frac{n\times 1000}{W_s}

where,

n= moles of solute

Moles of NH_4NO_3=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{27.8g}{80.0g/mol}=0.348moles  

W_s = weight of the solvent in g = ?

0.452=\frac{0.348\times 1000}{W_s}

W_s=770g

Thus 770 g water are needed to dissolve 27.8 g of ammonium nitrate NH_4NO_3 in order to prepare a 0.452 m solution

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Explanation:

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what is the balanced equation when copper metal is placed in a solution when platnium ii chloride is placed. what is the equatio
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Answer:

Cu~+~PtCl_2->Pt~+~CuCl_2

Explanation:

In this case, we can start with the <u>formula of Platinum (II) Chloride</u>. The cation is the atom at the left of the name (in this case Pt^+^2) and the anion is the atom at the right of the name (in this case Cl^-). With this in mind, the <u>formula would be</u> PtCl_2.

Now, if we used <u>metallic copper</u> we have to put in the reaction only the <u>copper atom symbol</u> Cu. So, we have as reagents:

Cu~+~PtCl_2->

The question now is: <u>What would be the products?</u> To answer this, we have to remember <u>"single displacement reactions"</u>. With a general reaction:

A~+~BC->AB~+~C

With this in mind, the reaction would be:

Cu~+~PtCl_2->Pt~+~CuCl_2

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