All categories

2 years ago

Table salt, NaCl, is an example of an amorphous solid. TRUE FALSE

Chemistry

2 answers:

SVETLANKA909090 [29]2 years ago

8 0

Answer:

False

Explanation:

Before answering the question, it is important to understand;

An amorphous solid is any noncrystalline solid in which the atoms and molecules are not organized in a definite lattice pattern. Such solids include glass, plastic, and gel. Emphasis on the word noncrystalline.

Table salt on the other hand is an ionic solid made up of Ions of opposite charges (Sodium ion and Chlorine ion) strongly attract each other; those of like charges repel. As a result ions in an ionic compound are arranged in a particular manner.

The ions in NaCl are ordered hence are referred to as crystalline solids and not amorphous.

prohojiy [21]2 years ago

3 0

False, it is an example of an Ionic solid

You might be interested in

Estimate the increase in the molar entropy of O2(g) when the temperature is increased at constant pressure from 298 K to 348 K,

Tamiku [17]

Explanation:

It is known that relation between entropy, heat energy and temperature is as follows.

dS = $\frac{Q}{T}$

$\int dS = \int \frac{Q}{T}$

Also we know that at constant pressure, Q = $\Delta H = C_{p} - dT$

$\int_{S_{1}}^{S_{2}} dS = \int_{T_{1}}^{T_{2}} C_{p} \frac{dT}{T}$

$\Delta S = C_{p} \int_{T_{1}}^{T_{2}} \frac{dT}{T}$

As the given data is as follows.

$T_{1}$ = 298 K, $T_{2}$ = 348 K

$C_{p}$ = 29.355 J/K mol

Now, putting the given values into the above formula as follows.

$\Delta S = C_{p} ln {T_{1}}^{T_{2}} \frac{dT}{T}$

= $29.355 [ln (348) - ln (298)]$

= $29.355 [5.85 - 5.69]$

= 4.48 J/k mol

Thus, we can conclude that the increase in the molar entropy of given oxygen gas is 4.48 J/k mol.

6 0

2 years ago

What mass of lif is contained in 18.66 g of a 20.0% by mass solution of lif in water?

aev [14]

Answer is: mass of lithium fluoride is 3,732 grams.
m(solution) = 18,66 g.
ω(solution) = 20% = 20% ÷ 100% = 0,2.
m(LiF) = ?
ω(solution) = m(LiF) ÷ m(solution).
m(LiF) = ω(solution) · m(solution).
m(LiF) = 0,2 · 18,66 g.
m(LiF) = 3,732 g.

7 0

2 years ago

if 5.50 mol of calcium carbide (CaC 2 ) reacts with an excess of water, how many moles of acetylene (C 2 H 2 ) , a gas used in w

Leona [35]

Answer:

5.5moles

Explanation:

CaC2 + 2H2O —> Ca(OH)2 + C2H2

From the equation, the following were observed:

1mole of CaC2 reacted to produced 1mol of C2H2.

Therefore, 5.5moles of CaC2 will also produce 5.5moles of C2H2

5 0

2 years ago

50 kg of N2 gas and 10kg of H2 gas are mixed to produce NH3 gas calculate the NH3gas formed. Identify the limiting reagent in th

statuscvo [17]

Answer:

1. H2 is the limiting reactant.

2. 56666.67g ( i.e 56.67kg) of NH3 is produced.

Explanation:

Step 1:

The equation for the reaction. This is given below:

N2 + H2 —> NH3

Step 2:

Balancing the equation.

N2 + H2 —> NH3

The above equation can be balanced as follow :

There are 2 atoms of N on the left side and 1 atom on the right side. It can be balance by putting 2 in front of NH3 as shown below:

N2 + H2 —> 2NH3

There are 6 atoms of H on the right side and 2 atoms on the left side. It can be balance by putting 3 in front of H2 as shown below

N2 + 3H2 —> 2NH3

Now the equation is balanced.

Step 3:

Determination of the masses of N2 and H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:

N2 + 3H2 —> 2NH3

Molar Mass of N2 = 2x14 = 28g/mol

Molar Mass of H2 = 2x1 = 2g/mol

Mass of H2 from the balanced equation = 3 x 2 = 6g

Molar Mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol

Mass of NH3 from the balanced equation = 2 x 17 = 34g

From the balanced equation above,

28g of N2 reacted with 6g of H2 to produce 34g of NH3

Step 4:

Determination of the limiting reactant. This is illustrated below:

N2 + 3H2 —> 2NH3

Let us consider using all the 10kg (i.e 10000g) of H2 to see if there will be any left of for N2.

From the balanced equation above,

28g of N2 reacted with 6g of H2.

Therefore, Xg of N2 will react with 10000g of H2 i.e

Xg of N2 = (28 x 10000)/6

Xg of N2 = 46666.67g

We can see from the calculations above that there are leftover for N2 as only 46666.67g reacted out of 50kg ( i.e 50000g) that was given. Therefore, H2 is the limiting reactant.

Step 5:

Determination of the mass of NH3 produced during the reaction. This is illustrated below:

N2 + 3H2 —> 2NH3

From the balanced equation above,

6g of H2 reacted to produce 34g of NH3.

Therefore, 10000g of H2 will react to produce = ( 10000 x 34)/6 = 6g of 56666.67g of NH3.

Therefore, 56666.67g ( i.e 56.67kg) of NH3 is produced.

3 0

2 years ago

Read 2 more answers

A mixture of three gases has a pressure at 298 K of 1380 mm Hg. The mixture is analysed and is found to contain 1.27 mol CO2, 3.

enot [183]

Answer:

The partial pressure of Ar is 356.04 mm Hg (= 0.4685 atm)

Explanation:

Step 1: Data given

A mixture of three gases has a total pressure of 1380 mm Hg (=1.81579 atm) at 298 K

Moles of CO2 = 1.27 moles

Moles of CO = 3.04 moles

Moles of Ar = 1.50 moles

Step 2: Calculate total number of moles

Total number of moles = n(CO2)+ n(CO)+ n(Ar) = 1.27 mol+ 3.04 mol+ 1.50 mol = 5.81 moles

Step 3: Calculate mol fraction Ar

Mol fraction Ar = 1.50 mol/5.81 mol = 0.258

Step 4: Calculate partial pressure

1380 mm Hg * 0.258 moles Ar = 356.04 mm Hg = 0.4685 atm

The partial pressure of Ar is 356.04 mm Hg (= 0.4685 atm)

8 0

2 years ago