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1 year ago

How does the arrangement of atoms in a mineral relate to its properties?

Chemistry

2 answers:

Firdavs [7]1 year ago

7 0

Answer:

The arrangement of atoms in a mineral can make it a molecular, crystalline or amorphous material, interfering with its physical and chemical properties.

Explanation:

The arrangement of atoms determines how the atoms are distributed over the entire length of the mineral. According to this distribution, the material formed by these atoms may have a molecular, crystalline or amorphous constitution which will directly interfere in the properties of this material, as for example, crystalline materials have their atoms distributed in a very regular and strict way, which guarantees a rigidity in the material, since amorphous materials do not have a regular atomic arrangement, and do not have hardness and rigidity.

MakcuM [25]1 year ago

6 0

The arrangement of atoms in a mineral can change its physical and chemical properties.
Diamonds and coal are both made of carbon, however, their chemical and physical properties are very different.

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The crystalline hydrate cd(no3)2 ⋅ 4h2o(s) loses water when placed in a large, closed, dry vessel at room temperature: cd(no3)2⋅

Sever21 [200]

The given dehydration equation is,

$Cd(NO_{3})_{2}. 4H_{2}O (s) ---> Cd(NO_{3})_{2}(s) + 4 H_{2}O(g)$

Cadmiumnitrate tetrahydrate when heated dehydrates releasing the combined water as water vapor. The reaction produces 4 moles of gaseous product water vapor. So, the degree of disorder or randomness increases. Hence, the sign of change in entropy is positive.

This reaction is spontaneous at room temperature even if it is endothermic as the sign of change in entropy is positive.

8 0

2 years ago

Read 2 more answers

Calculate the theoretical density of iron, and then determine the number of vacancies per cm3 needed for a BCC iron crystal to h

Ann [662]

Answer:

Therefore the theoretical density of iron is 7.877 g/cm³ .

Therefore the number of vacancy per cm³ is $3.27 \times 10^{20}$

Explanation:

BCC structure contains 2 atoms per cell.

BCC : Body centered cubic.

Atomic mass of iron = 55.845 gram/mole.

Atomic mass of a chemical element is the mass of 1 mole of the chemical element.

Density: Density of a matter is the ratio of mass of the matter and volume of the matter.

Avogrdro Number is number of atoms per 1 mole.

Avogrdro Number= 6.023×10²³

Lattice parameter of iron = 2.866×10⁻⁸ cm

The theoretical density:

$\rho=\frac{\textrm{(x atom/cell)} \times \textrm{(atomic mass)}}{\textrm{avogadro's number}\times \textrm{(lattice parameter)}^3}$

$=\frac{2\times 55.845 }{(6.023\times 10^{23})\times (2.866\times 10^{-8})^3}$ g/cm³ [x= 2,Since it is BCC structure]

=7.877 g/cm³

Therefore the theoretical density of iron is 7.877 g/cm³ .

Now we have to find out the number of unit cell of iron crystal having density 7.874 g/cm³.

$\rho=\frac{\textrm{(x atom/cell)} \times \textrm{(atomic mass)}}{\textrm{avogadro's number}\times \textrm{(lattice parameter)}^3}$

$\Rightarrow 7.874 =\frac{x\times 55.845}{6.023\times 10^{23}\times (2.866\times 10^{-8})^3}$

$\Rightarrow x= \frac{7.874\times 6.023\times 10^{23}\times (2.866\times 10^{-8})^3}{55.845}$

⇒ x =1.9923 atom/cell

Therefore the vacancy of atom per cell = (2- 1.9923)=0.0077

$Vacancy\ per\ cm^3=\frac{\textrm{Vacancy per cm} ^3 }{\textrm{( lattice parameter)} ^3}$

$=\frac{0.0077}{(2.866\times 10^-8)^3}$

$=3.27 \times 10^{20}$

Therefore the number of vacancy per cm³ is $3.27 \times 10^{20}$

5 0

1 year ago

When a sample of a gas is heated in a sealed, rigid container from 200. K to 400. K, the pressure exerted by the gas is:

gavmur [86]

When a sample of a gas is heated in a sealed, rigid container from 200 degree Kelvin to 400 degree Kelvin, the pressure exerted by the gas is increased by a factor 2. Heating any gas actually increases the volume of the gas within a container. As the temperature of the gas rises, the molecules of the gas start moving faster and start striking the walls of the container in which it is kept with more force. The volume of the container tries to expand to accommodate the fast colliding molecules of the gas.

5 0

1 year ago

Question 1

Nezavi [6.7K]

Answer:

The heat of reaction is $Q = 170.92 \ J$

The enthalpy is $\Delta H = -170.91 \ J$

The concentration of HCl is $M__{HCl}} =2.468 M$

Explanation:

From the question we are told that

The volume of $NaOH$ is $V_{NaOH} = 45 \ cm^3 = \frac{45} {1000} = 0.045 L$

The number of concentration of $NaOH$ is $M__{NaOH}} = 2M$

The volume of HCl is $V_{HCl} = 20 cm^3$

The number of concentration of $HCl$ is $M__{HCl}} = 1 M$

The temperature difference is $\Delta T = 9.4 ^o C$

Now the heat of reaction is mathematically represented as

$Q = m * c_p * \Delta T$

Where

$c_p$ is the specific heat of water with value $c_p = 4200 J /kg ^o C$

$m = m__{NaOH}} + m__{HCl}}$

Now $m__{NaOH}} = V_{NaOH} * M_{NaOH} * Z_{NaOH}$

where $Z_{NaOH}$ is the molar mass of NaOH with the value of 0.04 kg/mol

So $m__{NaOH}} = 0.045 * 2 * 0.040$

$m__{NaOH}} =0.0036\ kg$

While

$m__{HCl}} = V_{HCl} * M_{HCl} + Z_{HCl}$

Where $Z_{HCl}$ is the molar mass of $HCl$ with the value of 0.03646 kg/mol

$m__{NaOH}} = 0.020 * 1 * 0.03646$

$m__{NaOH}} = 0.000729 kg$

$m = 0.0036 + 0.000729$

$m = 0.00433$

=> $Q = 0.00433 * 4200 * 9.4$

$Q = 170.92 \ J$

The enthalpy is mathematically represented as

$\Delta H = - Q$

=> $\Delta H = -170.91 \ J$

From the second question we are told that

The volume of HCl is $V_{HCl_1} = 10cm^3 = \frac{10}{1000} = 0.010 L$

The volume of NaOH is $V_{NaOH_1 } = 30 cm^3 = \frac{30}{1000} = 0.03 L$

The concentration of NaOH is $M_{NaOH} = 2 M$

The first temperature change is $\Delta T = 4.5 ^oC$

The second volume of $V_{HCl_2} = 20 cm^3 = \frac{20}{1000 } = 0.020 m^3$

The mass of NaOH is

$m__{NaOH}} = V_{NaOH} * M_{NaOH} * Z_{NaOH}$

substituting values

$m__{NaOH}} = 0.03 * 2 * 40$

$m__{NaOH}} = 3.6 \ g$

The mass of the product formed is

$m = \frac{Q}{c_p * \Delta T}$

substituting values

$m = \frac{170.91}{4200 * 9} * 1000$

The multiplication by 1000 is to convert it from kg to grams

$m = 4.5 g$

Now the mass of HCl is

$m__{HCl}} = m - m__{NaOh}}$

substituting values

$m__{HCl}} = 4.5 -3.6$

$m__{HCl}} = 0.9 \ g$

Now the concentration of HCl is

$M__{HCl}} = \frac{m_{HCl}}{(Z_{HCl} * *1000) * V_{HCl_1}}}$

The multiplication of $Z_{HCl}$ is to convert it from kg/mol to g/mol

$M__{HCl}} = \frac{0.9}{36.46 * 0.01}}$

$M__{HCl}} =2.468 M$

3 0

1 year ago

At 700 K, Kp for the following equilibrium is (5.6 x 10-3) 2HgO(s)--> 2Hg(l) + O2(g) Suppose 51.2 g of mercury(II) oxide is p

Vesnalui [34]

Answer: Option (B) is the correct answer.

Explanation:

According to the given reaction equation, formula to calculate $\Delta n$ is as follows.

$\Delta n$ = coefficients of gaseous products - gaseous reactants

= 1 - 0

= 1

Also we know that,

$K_{p} = K_{c} \times (RT)^{\Delta n}$

$5.6 \times 10^{-3} = K_{c} \times (0.0821 \times 700)^{1}$

$K_{c} = 0.097 \times 10^{-3}$

For the equation, $2HgO(s) \rightarrow 2Hg(l) + O_{2}(g)$

Activity of solid and liquid = 1

As, $K_{p} = \frac{P^{2}_{Hg} \times P_{O_{2}}}{P^{2}_{HgO}}$

$5.6 \times 10^{-3} = P_{O_{2}}$

Hence, $P_{O_{2}}$ = 0.0056 atm

Thus, we can conclude that partial pressure of oxygen gas at equilibrium is 0.0056 atm.

5 0

1 year ago