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2 years ago

In which 1.0 gram sample are particles arranged in a rystal structure?

Chemistry

1 answer:

lilavasa [31]2 years ago

5 0

The Options are as follow,

(1) CaCl₂ (s) (3) CH₃OH (l)

(2) C₂H₆ (g) (4) Cal₂ (aq)

Answer:

Option-1 is the correct answer.

Explanation:

As we know crystal formation is the property of solids. Therefore, in given options we are given with four different states of matter.

Option A, CaCl₂ is in a solid state , so it can exist in crystal form.

Option 2, C₂H₆ (Ethane) is in gas form, so it cannot form crystals.

Option 3, CH₃OH (Methanol) is present in liquid form, so it fails to form crystals.

Option 4, CaI₂, it is dissolved in water, Hence, it is in aqueous state, Therefore it also lacks crystal structure.

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) Starting with 5.00 g barium chloride n hydrate yields 4.26 g of anhydrous barium chloride after heating. Determine the integer

RSB [31]

Answer:

Explanation:

Mass of water molecule = mass of hydrated salt - mass of anhydrous salt

Mass of water molecule = 5.00 - 4.26 = 0.74g of water molecule.

Number of moles = mass / molarmass

Molar mass of water = 18.015g/mol

No. of moles of water = 0.74 / 18.015 = 0.0411 moles.

Mass of BaCl2 present =?

1 mole of BaCl2 = 208.23 g

X mole of BaCl2 = 4.26 g

X = (4.26 * 1) / 208.23

X = 0.020

0.020 moles is present in 4.26g of BaCl2

Mole ratio between water and BaCl2 =

0.0411 / 0.020 = 2

Therefore 2 molecules of water is present the hydrated salt.

4 0

2 years ago

Two species, A and B, are separated on a 2.00 m long column which has 5.000 x 103 plates when the flow rate is 15.0 mL/min. A sp

sergiy2304 [10]

Explanation:

The given data is as follows.

$t_{m}$ = 30.0 sec, $t_{r1}$ = 5 min = $5 \times 60 sec$ = 300 sec

$t_{r2}$ = 12.0 min = $12 \times 60 sec$ = 720 sec

Formula for adjusted retention time is as follows.

$t'_{r} = t_{r} - t_{m}$

= 300 sec - 30.0 sec

= 270 sec

$t'_{r2}$ = 720 sec - 30 sec

= 690 sec

Formula for relative retention ( $\alpha$ ) is as follows.

$\alpha = \frac{t'_{r2}}{t'_{r1}}$

= $\frac{690 sec}{270 sec}$

= 2.56

Thus, we can conclude that the relative retention is 2.56.

4 0

2 years ago

Hydrochloric acid (75.0 mL of 0.250 M) is added to 225.0 mL of 0.0550 M Ba(OH)2 solution. What is the concentration of the exces

Shalnov [3]

Answer: The concentration of excess $[OH^-]$ in solution is 0.017 M.

Explanation:

1. $Molarity=\frac{moles}{\text {Volume in L}}$

moles of $HCl=Molarity\times {\text {Volume in L}}=0.250\times 0.075=0.019moles$

1 mole of $HCl$ give = 1 mole of $H^+$

Thus 0.019 moles of $HCl$ give = 0.019 mole of $H^+$

2. moles of $Ba(OH)_2=Molarity\times {\text {Volume in L}}=0.0550\times 0.225=0.012moles$

According to stoichiometry:

1 mole of $Ba(OH)_2$ gives = 2 moles of $OH^-$

Thus 0.012 moles of $Ba(OH)_2$ give = $2 \times 0.012=0.024$ moles of $OH^-$

$H^++OH^-\rightarrow H_2O$

As 1 mole of $H^+$ neutralize 1 mole of $OH^-$

0.019 mole of $H^+$ will neutralize 0.019 mole of $OH^-$

Thus (0.024-0.019)= 0.005 moles of $OH^-$ will be left.

$[OH^-]=\frac{\text {moles left}}{\text {Total volume in L}}=\frac{0.005}{0.3L}=0.017M$

Thus molarity of $[OH^-]$ in solution is 0.017 M.

4 0

2 years ago

Read 2 more answers

An aluminum ion has a charge of +3, and an oxide has a charge of -2. What would be the product of a reaction between these two e

kozerog [31]

The product of a reaction between these two elements is $Al_{2} O_{3}$ .

Explanation:

The oxidation state of an ion in a compound is equal to its charge.

The aluminum having a charge of +3 because oxidation state is +3

The oxide is having charge of -2

The product of these reactants will produce a chemical compound.

The compound formed is $Al_{2} O_{3}$ i.e Aluminium oxide. The compound while getting formed will share the charge and cation A+ will have the charge of anion and anion will have the charge of cation. This will result in a compound as there should be a neutral charge on the compound formed.

The +3 charge of the cation Al+ will go to anion oxide O2- and the charge of anion -2 will go with cation Al+. 

8 0

2 years ago

An atom of beryllium (m = 8.00 u) splits into two atoms of helium (m = 4.00 u) with the release of 92.2 kev of energy. if the or

galben [10]

The kinetic energy of the products is equal to the energy liberated which is 92.2 keV. But let's convert the unit keV to Joules. keV is kiloelectro volt. The conversion that we need is: 1.602×10⁻¹⁹ joule = 1 eV

Kinetic energy = 92.2 keV*(1,000 eV/1 keV)*(1.602×10⁻¹⁹ joule/1 eV) = 5.76×10²³ Joules

From kinetic energy, we can calculate the velocity of each He atom:
KE = 1/2*mv²
5.76×10²³ Joules = 1/2*(4)(v²)
v = 5.367×10¹¹ m/s

4 0

2 years ago

Read 2 more answers