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3 years ago

Which characteristic best identifies a decomposition reaction?

Chemistry

1 answer:

gtnhenbr [62]3 years ago

6 0

Answer:

Decomposition reaction = Those reactions in which single compound decompose down to give two or more products are called decomposition reaction .

Explanation:

Those reactions in which single compound decompose down to give two or more products are called decomposition reaction .

For the Decomposition reaction to take place , a large amount of energy ias needed. This is because for decomposition bonds are required to breaks. Hence energy is needed to break the bonds as they have strong force of attraction.

They occur in presence of HEAT , LIGHT , ELECTRICITY.

examples of decomposition reactions are:

$CaCO_{3}\rightarrow CaO+CO_{2}$

Here only one CaCO3 is producing two substances which are : CaO and CO2

This is also known as decompositon of Calcium carbonate and it require a lot of heat energy.

$2AgBr\rightarrow Ag + Br_{2}$

This is also a decomposition reaction , This occur in the presence of sunlight

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4Fe + 30₂ ⇒ Fe₂0₃

DaniilM [7]

Answer:

The nuber of each one should be same

5 0

2 years ago

A sample of 28 Mg decays initially at a rate of 53500 disintegrations per minute, but the decay rate falls to 10980 disintegrati

prisoha [69]

Answer : The half life of 28-Mg in hours is, 6.94

Explanation :

First we have to calculate the rate constant.

Expression for rate law for first order kinetics is given by:

$k=\frac{2.303}{t}\log\frac{a}{a-x}$

where,

k = rate constant

t = time passed by the sample = 48.0 hr

a = initial amount of the reactant disintegrate = 53500

a - x = amount left after decay process disintegrate = 53500 - 10980 = 42520

Now put all the given values in above equation, we get

$k=\frac{2.303}{48.0}\log\frac{53500}{42520}$

$k=9.98\times 10^{-2}hr^{-1}$

Now we have to calculate the half-life.

$k=\frac{0.693}{t_{1/2}}$

$9.98\times 10^{-2}=\frac{0.693}{t_{1/2}}$

$t_{1/2}=6.94hr$

Therefore, the half life of 28-Mg in hours is, 6.94

8 0

3 years ago

19. I accidentally mixed bleach and ammonia when cleaning one day. It released 55 liters of deadly chlorine (CI) gas. How many g

user100 [1]

55,000 grams
Hope this helps you out :)

3 0

3 years ago

Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this li

meriva

Answer:

The standard enthalpy change for the reaction at $25^{0}\textrm{C}$ is -2043.999kJ

Explanation:

Standard enthalpy change ( $\Delta H_{rxn}^{0}$ ) for the given reaction is expressed as:

$\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]$

Where $\Delta H_{f}^{0}$ refers standard enthalpy of formation

Plug in all the given values from literature in the above equation:

$\Delta H_{rxn}^{0}=[3mol\times (-393.509kJ/mol)]+[4mol\times (-241.818kJ/mol)]-[1mol\times (-103.8kJ/mol)]-[5mol\times (0kJ/mol)]=-2043.999kJ$

4 0

3 years ago

Two solids, A and B, are located in the same family on the periodic table. A sample of each is placed in a beaker of HCL. Substa

Vlad [161]

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4 0

3 years ago