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3 years ago

Breaking an object such as a mirror is an example of physical change, chemical change,physical reaction, chemical reaction

Chemistry

1 answer:

dangina [55]3 years ago

5 0

Answer:

option A= Physical change

Explanation:

physical change:

" It is a change in which no new substance is formed"

Breaking of object like glass is the example of physical change because it is not change into another object. It effect the form of object but can not change the chemical composition.

Chemical changes:

" it is a change in which one substance is converted into new product through chemical reaction".

During the chemical changes the types and the number of atom remain same but their arrangement changed.

for example: burning of wood, baking of cake, digesting food, resting of iron etc.

Physical reaction:

" physical reaction occur during the molecular rearrangement. There is no chemical change occur"

In this type of changes no bonds are break to form new bonds, for example boiling point.

Chemical reaction:

" chemical reaction occur when molecules are chemically react with each others and bonds formation and breaking is also occur"

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Tcecarenko [31]

Answer:

$1.25~mol~H_2O$ and $0.627~mol~N_2$

Explanation:

Our goal for this question is the calculation of the number of moles of the molecules produced by the reaction of hydrazine ( $N_2H_4$ ) and oxygen ( $O_2$ ). So, we can start with the reaction between these compounds:

$N_2H_4~+~O_2~->~N_2~+~H_2O$

Now we can balance the reaction:

$N_2H_4~+~O_2~->~N_2~+~2H_2O$

In the problem, we have the values for both reagents. Therefore we have to calculate the limiting reagent. Our first step, is to calculate the moles of each compound using the molar masses values (32.04 g/mol for $N_2H_4$ and 31.99 g/mol for $O_2$ ):

$20.1~g~N_2H_4\frac{1~mol~N_2H_4}{32.04~g~N_2H_4}=0.627~mol~N_2H_4$

$20.1~g~O_2\frac{1~mol~O_2}{31.99~g~O_2}=0.628~mol~O_2$

In the balanced reaction we have 1 mol for each reagent (the numbers in front of $O_2$ and $N_2H_4$ are 1). Therefore the smallest value would be the limiting reagent, in this case, the limiting reagent is $N_2H_4$ .

With this in mind, we can calculate the number of moles for each product. In the case of $N_2$ we have a 1:1 molar ratio (1 mol of $N_2$ is produced by 1 mol of $N_2H_4$ ), so:

$0.627~mol~N_2H_4\frac{1~mol~N_2}{1~mol~N_2H_4}=~0.627~mol~N_2$

We can follow the same logic for the other compound. In the case of $H_2O$ we have a 1:2 molar ratio (2 mol of $H_2O$ is produced by 1 mol of $N_2H_4$ ), so:

$0.627~mol~N_2H_4\frac{2~mol~H_2O}{1~mol~N_2H_4}=~1.25~mol~H_2O$

I hope it helps!

4 0

4 years ago

ammonia (NH3(g) Hf=-45.9 kJ/mol) reacts. with oxygen to produce nitrogen and water (H2O(g) Hf = -241.8 kJ/mol according to the e

kherson [118]

Answer:

ΔH°_rxn = -195.9 kJ·mol⁻¹

Explanation:

4NH₃(g) + 3O₂(g) ⟶ 2N₂(g) +6H₂O(g)

ΔH°_f/(kJ·mol⁻¹): -45.9 0 0 -241.8

The formula relating ΔH°_rxn and enthalpies of formation (ΔH°_f) is

ΔH°_rxn = ΣΔH°_f(products) – ΣΔH°_f(reactants)

ΣΔH°_f(products) = -6(241.8) = -1450.8 kJ

ΣΔH°_f(reactants) = -4(45.9) = -183.6 kJ

ΔH°_rxn = (-1450.8 + 183.6) kJ = -1267.2 kJ

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