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2 years ago

What does X represent in the formula for the compound XCl4?

Chemistry

1 answer:

qaws [65]2 years ago

3 0

I believe the answer would be C

The the compound of interest i.e. XCl4, since there are 4 Cl atoms bonded to X. This signifies that the valency of X is 4.

There atomic number of C is 6. It's electronic configuration is giving by 1s2 2s2 2p2. Thus, there are 4 electrons in valence shell of C. This signifies that valency of C is 4. Hence the compound present in present case is CCl4.

Hope this helps :)

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Identify one disadvantage to each of the following models of electron configuration:

Murrr4er [49]

Answer:

The disadvantages of each of the given model of electron configuration have been mentioned below:

1). Dot Structures - They take up excess space as they do not display the electron distribution in orbitals.

2). Arrow and line diagrams make the counting of electrons and take up too much space.

3). Written Configurations do not display the electron distribution in orbitals and help in lose counting of electrons easily.

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2 years ago

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For this exercise, you can simulate the described conditions by changing the values in the run experiment tool of the simulation

Furkat [3]

1) ideal gas law: p·V = n·R·T.
p - pressure of gas.
V -volume of gas.
n - amount of substance.
R - universal gas constant.
T - temperature of gas.
n₁ = 0,04 mol, V₁ = 0,06 l.
n₂ = 0,07 mol, V₂ = 0,06 · 0,07 ÷ 0,04 = 0,105 l.
2) V₁ = 0,06 l, T₁ = 240,00 K.
T₂ = 340,00 K, V₂ = 340 · 0,06 ÷ 240 = 0,05 l.

3 0

2 years ago

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In the formula X2O5, the symbol X could represent an element in Group

topjm [15]

Answer: (3) 15

Explanation: We criss-cross down the oxidation numbers to get the subscripts for the correct formulas. That means the X has an oxidation number of 5. The element with the + oxidation number is always written first so it is +5. Of the groups names, only group 15 has +5 as an oxidation number.

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2 years ago

Express each aqueous concentration in the unit indicated.

MAXImum [283]

Answer:

a. ppb of trichloroethylene = 3 × 10⁶ ppb

b. ppm of Cl₂ = 3.8 ppm

c. Molarity = 0.0002 mol / L

d. Molarity = 0.0007 mol / L

e. For trace amount of concentrations

Explanation:

a. Given data

mass of trichloroethylene = 25 mg

Volume of water = 9.5 L

ppb of trichloroethylene = ?

Solution

As we know that

1 L = 1000 milliliters

9.5 L = 9.5 × 1000

9.5 L = 9500 millileters (ml)

we consider 25 mg = 25 millileters

ppb = (mass of solute / mass of solvent) × 1000,000,000 (1 billion)

ppb of trichloroethylene = (25 ÷ 9500) × 1000,000,000

ppb of trichloroethylene = 0.003 × 1000,000,000

ppb of trichloroethylene = 3 × 10⁶ ppb

B. Given data

Mass of Cl₂ = 38 g

volume of water = 1.00 × 10⁴ L ( 10000 L)

ppm of Cl₂ = ?

Solution

Volume of water in ml = 1 L = 1000 ml

Volume of water in ml = 10000 × 1000

Volume of water in ml = 10000000 ml

we take 38 g = 38 ml

Now we convert it to ppm

ppm = (mass of solute / mass of solvent) × 1000000 (1 million)

ppm of Cl₂ = ( 38 ÷ 10000000 ) × 1000000

ppm of Cl₂ = 0.0000038 × 1000000

ppm of Cl₂ = 3.8 ppm

C. Given data

Concentration of F⁻ ( Fluoride ion) = 2.4 ppm

Molarity = ?

Solution

As we know that 1 ppm = 0.001 g / L

2.4 ppm = 2.4 × 0.001 g/L

2.4 ppm = 0.0024 g/L

Mass of flouride ions = 0.0024 g

Now we find number of moles

moles = mass / molar mass

molar mass of F⁻ = 19 g/mol

moles of F⁻ = 0.0024 g / 19 g/mol

moles of F⁻ = 0.0002 mol

Molarity = mol of solute / liter of solution

Molarity = 0.0002 mol / 1 L

Molarity = 0.0002 mol / L

D. Given data

Concentration of NO₃⁻ ( nitrate ion) = 45 ppm

Molarity = ?

Solution

As we know that 1 ppm = 0.001 g / L

45 ppm = 45 × 0.001 g/L

45 ppm = 0.045 g/L

Mass of nitrate ions = 0.045 g

Now we find number of moles

moles = mass / molar mass

molar mass of NO₃⁻ = 62 g/mol

moles of NO₃⁻ = 0.045 g / 62 g/mol

moles of F⁻ = 0.0007 mol

Molarity = mol of solute / liter of solution

Molarity = 0.0007 mol / 1 L

Molarity = 0.0007 mol / L

E. Reason of expressing concentration in ppm and ppb

Scientist prefer ppm and ppb notations when the concentration difference of solute and solvent are very high.

As water contains contaminants is a very low amount we can say in trace amounts so scientist prefer ppm and ppb rather than molarity.

Example

Arcenic is an under ground water contaminant and its concentration of 10 μg/L is dangerous for health.

Lets change this in to molarity

mass = 10 μg

10 μg = 10 / 1000000

10 μg = 0.00001 g

now find out moles of Arcenic

moles = mass / molar mass

molar mass of arcenic = 75 g/mol

moles = mass / molar mass

moles of arcenic = 0.00001 g / 75 g/mol

moles of arcenic = 0.00000012 mol

Molarity = moles of solute / litres of solution

Molarity = 0.00000012 mol / 1 L

Molarity = 0.00000012 mol/ L

As we can see that in molarity it is a negligible amount so scientists express it in ppm and ppb

7 0

2 years ago

A concentrated binary solution containing mostly species 2 (but x2 ≠ 1) is in equilib- rium with a vapor phase containing both s

marusya05 [52]

Answer:

x1= 4.5 × 10^-3, y1= 0.9

Explanation:

A binary solution having two species is in equilibrium in a vapor phase comprising of species 1 and 2

Take the basis as the pressure of the 2 phase system is 1 bar. The assumption are as follows:

1. The vapor phase is ideal at pressure of 1 bar

2. Henry's law apply to dilute solution only.

3. Raoult's law apply to concentrated solution only.

Where,

Henry's constant for species 1 H= 200bar

Saturation vapor pressure of species 2, P2sat= 0.10bar

Temperature = 25°C= 298.15k

Apply Henry's law for species 1

y1P= H1x1...... equation 1

y1= mole fraction of species 1 in vapor phase.

P= Total pressure of the system

x1= mole fraction of species 1 in liquid phase.

Apply Raoult's law for species 2

y2P= P2satx2...... equation 2

From the 2 equations above

P=H1x1 + P2satx2

200bar= H1

0.10= P2sat

1 bar= P

Hence,

P=H1x1 + (1 - x1) P2sat

1bar= 200bar × x1 + (1 - x1) 0.10bar

x1= 4.5 × 10^-3

The mole fraction of species 1 in liquid phase is 4.5 × 10^-3

To get y, substitute x1=4.5 × 10^-3 in equation 1

y × 1 bar = 200bar × 4.5 × 10^-3

y1= 0.9

The mole fraction of species 1 in vapor phase is 0.9

4 0

2 years ago

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