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Rasek [7]
3 years ago
7

Which of the following changes can alter the form of a substance but does not necessarily change it to an other substance?

Chemistry
1 answer:
Goshia [24]3 years ago
8 0

Answer:

  • The physical changes can alter the form of a substance but does not necessarily change it to an other substance

Explanation:

  • The physical properties of matter is one that does not depend on composition.It just change the form of matter into another form with changing into composition.
  • Physical properties include: appearance, texture, color, odor, melting point, boiling point, density, solubility, polarity, and many others.
  • The three states of matter are: solid, liquid, and gas. The melting point and boiling point are related to changes of the state of matter. All matter may exist in any of three physical states of matter.

Example:

  1. When water is boiled then it change into vapors. Now the form of water is change but its composition is same like water
  2. When water freezes into ice, now phase are different i.e water is liquid while ice is solid but both have same composition.
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onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency
mestny [16]

Answer : The excess reactant is, H_2O

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of CaCN_2 = 105.0 g

Mass of H_2O = 78.0 g

Molar mass of CaCN_2 = 80.11 g/mole

Molar mass of H_2O = 18 g/mole

Molar mass of CaCO_3 = 100.09 g/mole

First we have to calculate the moles of CaCN_2 and H_2O.

\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles

\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3

From the balanced reaction we conclude that

As, 1 mole of CaCN_2 react with 3 mole of H_2O

So, 1.31 moles of CaCN_2 react with 1.31\times 3=3.93 moles of H_2O

From this we conclude that, H_2O is an excess reagent because the given moles are greater than the required moles and CaCN_2 is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)

\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0
3 years ago
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IRISSAK [1]

Answer:

Explanation:according to question:

. Nacl (aq) + AgNO3 (aq) --> AgCl

(s) + NaNO3 (aq).balanced.

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