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slega [8]
3 years ago
11

Which of the following statements is true for a synthesis reaction? A single product is formed. Multiple products are formed. A

single reactant breaks down. Multiple reactants break down.
Chemistry
2 answers:
pychu [463]3 years ago
8 0

Answer : The correct option is, A single product is formed.

Explanation :

  • Synthesis reaction : It is defined as the chemical reaction in which the multiple substances or the reactants combines then it forms a single product.

The synthesis reaction is represented as,

X+Y\rightarrow XY

For example : When the hydrogen gas react with oxygen gas then its forms water as a single product.

The balanced chemical reaction will be :

2H_2+O_2\rightarrow 2H_2O

  • Decomposition reaction : It is defined as a type of reaction in which a single larger compound decomposes to give two or more or multiple smaller molecules as a product.

The general representation of decomposition reaction is :

AB\rightarrow A+B

For example : When nitrogen dioxide decomposes then it gives oxygen gas and nitrogen gas as a products.

2NO_2\rightarrow 2O_2+N_2

Hence, the true statements for a synthesis reaction is, a single product is formed.

Papessa [141]3 years ago
4 0

Answer: A single product is formed.

Explanation: Synthesis reaction is one in which two or more substances react together to form a single product.

N_2+3H_2\rightarrow 2NH_3

A reaction in which a single reactant breaks down to form two or more than two products is called as decomposition.

CaCO_3\rightarrow CaO+CO_2


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Consider the insoluble compound nickel(II) hydroxide , Ni(OH)2 . The nickel ion also forms a complex with cyanide ions . Write a
natali 33 [55]

Answer: Equilibrium constant for this reaction is 2.8 \times 10^{15}.

Explanation:

Chemical reaction equation for the formation of nickel cyanide complex is as follows.

Ni(OH)_{2}(s) + 4CN^{-}(aq) \rightleftharpoons [Ni(CN)_{4}^{2-}](aq) + 2OH^{-}(aq)

We know that,

      K = K_{f} \times K_{sp}

We are given that, K_{f} = 1.0 \times 10^{31}

and,    K_{sp} = 2.8 \times 10^{-16}

Hence, we will calculate the value of K as follows.

     K = K_{f} \times K_{sp}

     K = (1.0 \times 10^{31}) \times (2.8 \times 10^{-16})

        = 2.8 \times 10^{15}

Thus, we can conclude that equilibrium constant for this reaction is 2.8 \times 10^{15}.

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3 years ago
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Explanation:

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6 0
3 years ago
A sample that contains only SrCO3 and BaCO3 weighs 0.846 g. When it is dissolved in excess acid, 0.234 g carbon dioxide is liber
In-s [12.5K]

Answer:28.605

Explanation:First, the molar mass of of SrCO3, BaCO3 and CO2 has to be calculated, (using the molar mass of each element Sr = 87.62, Ba = 137.327, C=12.011, O= 16.00)

The molar masses are;

SrCO3 = 87.62 + 12.011 + (3*16) = 147.631g/mol

BaCO3 = 79.904 + 12.011 + (3*16) = 197.34 g/mol

CO2 = 12.011 + (2*16) = 44.011 g/mol

To obtain one of the equations to solve the problem;

The sample is made of SrCO3 and BaCO3 and has a mass of 0.846 g. Representing the mass of SrCO3 as ma and that of BaCO3 as mb. The first equation can be written as:

ma + mb = 0.846g                 (1)

To obtain another equation in order to be able to determine the different percentages of the compounds (SrCO3 and BaCO3) that make of the sample, a relationship can be obtained by determining the relationship between the number of moles of CO2 formed as the mass of the SrCO3 and BaCO3;

The number of moles of CO2 formed = (mass of CO2)/(molar mass) =0.234/44.011 =0.00532moles

CO2 contains 1 mole of carbon (C) so therefore 0.00532 moles of CO2 contains 0.00532 moles of C

The sample produced 0.00532 moles of CO2, therefore the number of moles SrCO3 and BaCO3 that produced this amount can be calculated using the formula;

= (mass )/(molar mass)

No of moles of SrCO3 and BaCO3 will be ma/147.631 and mb/197.34 moles respectively

The total amount of C molecules produced by SrCO3 and BaCO3 will be 0.00532 moles of C

The second equation can be written as

ma/147.631 + mb/197.34= 0.00532          (2)

Solving Equation (1) and (2) simultaneously;

ma = 0.604g; mb = 0.242g

Therefore the percentage of BaCO3   = (mass of BaCO3 )/(mass of sample )*100

                                                         = 0.242/(0.846 )*100

                                                         = 28.605%

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