All categories

2 years ago

Which material has a crystalline structure at room temperature ( 20 degrees Celsius )

Chemistry

2 answers:

Sauron [17]2 years ago

8 0

It's 2, glass. Water, nitrogen, and sucrose don;t have a crystalline structure.

Musya8 [376]2 years ago

8 0

Answer:

Sucrose. Option 4 is correct.

Explanation:

Crystallinity can be defined as the degree of structural order in a solid.

Let us take a look at each of the material mentioned in the question..

The first one is WATER. Water has the molecular formula of H2O, it is bent at an angle approximately 104.5 degrees. Because of its molecular form as liquid, WATER is amorphous and NOT crystalline.

The second material is GLASS: GLASS do not crystallized when they become hardened.

The third material is Nitrogen, NITROGEN is a non metallic element of group 15. Although solid nitrogen have crystalline modifications, they still exist as amorphous.

The fourth material is sucrose. SUCROSE is a disaccharide. It is a crystalline white solid.

You might be interested in

Select all that reasons that the reaction contents does not conduct at this low point.

Alexxx [7]

It's all three of the answers

5 0

2 years ago

Which neutral atom is isoelectronic with o+?

sukhopar [10]

Isoelectronic means equal number of electrons.

O+ is formed when the atom of O loses 1 electron.

The number of electrons of neutral O atom equals its number of protons.

Number of protons identifies the atomic number and position of the element in the periodic table.

The positon of O in the periodic table is A = 8, so it has 8 electrons and O+ has 8 - 1 = 7 electrons.

The neutral atom with one electron less than O is of the element to the left of O in the periodic table (A = 7). That element is N.

Therefore, the neutral atom isoelectronic with O+ is N (both have 7 electrons).

5 0

2 years ago

Calculate the molality of 2.0 M MgCl2 solution. The density of the solution is 1.127 g/mL. (The molar mass of MgCl2 = 95.211 g/m

Elan Coil [88]

The answer is 2.135 mol/Kg

Given that molarity is 2M, that is, 2 moles in 1 liter of solution.

Density of solution is 1.127 g/ml

Volume of solution is 1L or 1000 ml

mass of solution (m) = density × volume

m₁ = density × volume = 1.127 × 1000 = 1127 g

mass of solute, m₂ = number of moles × molar mass

m₂ = 2 × 95.211

m₂ = 190.422 g

mass of solvent = m₁ - m₂

= 1127 - 190.422

= 936.578 g

= 0.9366 Kg

molality = number of moles of solute / mass of solvent (in kg)

= 2 / 0.9366

= 2.135 mol/Kg

8 0

2 years ago

Read 2 more answers

A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re

Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of $\gamma$ .

$\gamma=\frac{C_p}{C_v}$

As, $C_p=R+C_v$

So, $\gamma=\frac{R+C_v}{C_v}$

Given :

$C_v=20.8J/K\\\\R=8.314J/K$

$\gamma=\frac{8.314+20.8}{20.8}=1.4$

Now we have to calculate the initial volume of gas.

Formula used :

$P_1V_1=nRT_1$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$V_1$ = initial volume of gas = ?

$T_1$ = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

$(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)$

$V_1=5.80L$

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

$P_1V_1^{\gamma}=P_2V_2^{\gamma}$

where,

$P_1$ = initial pressure of gas = 4.25 atm

$P_2$ = final pressure of gas = 2.50 atm

$V_1$ = initial volume of gas = 5.80 L

$V_2$ = final volume of gas = ?

$\gamma$ = 1.4

Now put all the given values in above formula, we get:

$(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}$

$V_2=8.47L$

Now we have to calculate the final temperature of gas.

Formula used :

$P_2V_2=nRT_2$

where,

$P_2$ = final pressure of gas = 2.50 atm

$V_2$ = final volume of gas = 8.47 L

$T_2$ = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in above formula, we get:

$(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2$

$T_2=257.9K\approx 258K$

Now we have to calculate the work done.

$w=nC_v(T_2-T_1)$

where,

w = work done = ?

n = number of moles of gas =1.0 mol

$T_1$ = initial temperature of gas = 300 K

$T_2$ = final temperature of gas = 258 K

$C_v=20.8J/K$

Now put all the given values in above formula, we get:

$w=(1.0mol)\times (20.8J/K)\times (258-300)K$

$w=-873.6J$

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0

2 years ago

The accepted value is 1.43. Which correctly describes this student’s experimental data?

Natalija [7]

Answer:

Neither accurate nor precise

Explanation:

The values were not near or even the same as the accepted value thus making it neither accurate nor precise.

4 0

2 years ago

Read 2 more answers