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Lyrx [107]
2 years ago
9

Alex's teacher showed him a model of gas particles in a sealed container. How should Alex change the model to show the particles

of this gas at a higher temperature?​
Chemistry
1 answer:
Inessa05 [86]2 years ago
7 0

Answer:

  • Increased volume of particles in the container
  • greater vibration of particles

Explanation:

At higher temperature, the particles of the gas would be more active and vibrate more, or even have greater collisions. Alex can indicate this in the altered model to depict higher temperature.

Consequently, Charles law gives meaning to why there would be an increased volume of gas in the stable pressurized container, if the temperature were to be increased.

I hope this explanation was clear and concise?

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The following pairs of soluble solutions can be mixed. In some cases, this leads to the formation of an insoluble precipitate. D
Westkost [7]

Answer:

a) K_{2} S and NH_{4} Cl :

There are no insoluble precipitate forms.

b) Ca Cl_{2} and (NH_{4} )_{2} Co_{3} :

There are the insoluble precipitates of CaCo_{3}  forms.

c) Li_{2}S and MnBr_{2} :

There are the insoluble precipitates of MnS  forms.

d) Ba(No_{3} )_{2} and Ag_{2} So_{4} :                        

As Ag_{2} So_{4} is insoluble, therefore no precipitate forms.

e) Rb_{2}Co_{3} and NaCl:

There are no insoluble precipitates forms.

Explanation:

a)

Solubility rule suggests:- K_{2} S ⇒ soluble, NH_{4} Cl ⇒ soluble.

                                          KCl ⇒ soluble, (NH_{4})_{2} S  ⇒ soluble.

There are no insoluble precipitate forms.

b)

Solubility rule suggests:- Ca Cl_{2} ⇒ soluble, (NH_{4} )_{2} Co_{3} ⇒ soluble.

                                        CaCo_{3} ⇒ insoluble, NH_{4} Cl  ⇒ soluble.

There are the insoluble precipitates of CaCo_{3}  forms.

c)

Solubility rule suggests:- Li_{2}S ⇒ soluble, MnBr_{2} ⇒ soluble.

                                        LiBr ⇒ soluble, MnS  ⇒ insoluble.

There are the insoluble precipitates of MnS  forms.

d)

Solubility rule suggests:- Ba(No_{3} )_{2} ⇒ soluble, Ag_{2} So_{4} ⇒insoluble.

                                     

As Ag_{2} So_{4} is insoluble, therefore no precipitate forms.

e)

Solubility rule suggests:- Rb_{2}Co_{3} ⇒ soluble, NaCl ⇒ soluble.

                                        RbCl ⇒ soluble, Na_{2} Co_{3}  ⇒ soluble.

There are no insoluble precipitates forms.

6 0
2 years ago
Scientists discover two new elements, element A and element X, and are able to determine their atomic mass. Element A has an ato
olga nikolaevna [1]
17+76×3=228+17=245. D is right
5 0
2 years ago
Enter the appropriate symbol for an isotope of phosphorus-32 corresponding to the isotope notation AZX. Express your answer as a
Lelechka [254]

Answer:

^{32}_{15}P

Explanation:

Atomic number : It is defined as the number of electrons or number of protons present in a neutral atom.

However, when we talk about the atomic number of the ion, it is not equal to the number of electrons as electron can be gained or loosed.

This is why, more appropriately, the number of the protons which are present in the nucleus of the atom is called the atomic number.

Thus, atomic number of phosphorus = 15

Mass number is the number of the entities present in the nucleus which is the equal to the sum of the number of protons and electrons.

Given, Mass number = 32

Thus, the symbol of the isotope is:-

^{32}_{15}P

8 0
3 years ago
When 18.5 g of HgO(s) is decomposed to form Hg(l) and O2(g), 7.75 kJ of heat is absorbed at standard-state conditions. What is t
taurus [48]

Answer:

The standard enthalpy of formation of HgO is -90.7 kJ/mol.

Explanation:

The reaction between Hg and oxygen is as follows.

\text{Hg(l)}+\frac{1}{2}{O_{2}\rightarrow \text{HgO(s)}

From the given,

Molar mass of HgO = 216.59 g/mol

Mass of HgO decomposed = 18.5 g

Amount of heat absorbed = 7.75 kJ

From the reaction,

The standard  enthalpy of formation = +7.75\times\frac{kJ}{18.5 g}\frac{216.59}{1mol} \,\,= +90.7 kJ/mol

During the decomposition of 1 mol of HgO , 90.7 kJ of energy absorbed.

For the formation of 1 mol of HgO , 90.7 kJ of energy is release

Therefore, the enthalpy of formation of mercury(II)Oxide is -90.7 kJ/mol

5 0
3 years ago
SOMEONE PLEASE HELP ME ITS SCIENCE
andreev551 [17]

Answer:

2

Explanation:

The total number of atoms in silver sulfate Ag2SO4 is 2

6 0
3 years ago
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