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2 years ago

Name an element in the fifth period (row) of the periodic table with a complete outer shell.

Chemistry

2 answers:

lorasvet [3.4K]2 years ago

4 0

Answer: The element in fifth period of the periodic table having complete outer shell is Xenon.

Explanation:

Elements are distributed in 7 periods and 18 groups. Groups are the vertical columns and periods are the horizontal rows.

Element having complete outer shells are present in Group 18 of the periodic table. These elements are known as noble gases.

Element present in Period 5 and Group 18 of the periodic table is Xenon.

The atomic number of Xenon is 54.

Hence, the element in fifth period of the periodic table having complete outer shell is Xenon.

katen-ka-za [31]2 years ago

3 0

I suppose it could be Xenon, but I'm not pretty sure about this.

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Using the balanced reaction below, drag and drop the terms into the correct location to solve the following problem:

k0ka [10]

Answer:

4.5 moles of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

C2H5OH + 3O2 —> 2CO2 + 3H2O

From the balanced equation above,

1 moles of C2H5OH reacted to produce 3 moles of H2O.

Finally, we shall determine the number of mole of H2O produced by the reaction of 1.5 moles of C2H5OH. This can be obtained as follow:

From the balanced equation above,

1 moles of C2H5OH reacted to produce 3 moles of H2O.

Therefore, 1.5 moles of C2H5OH will react to produce = 1.5 × 3/ 1 = 4.5 moles of H2O.

Thus, 4.5 moles of H2O will be produced.

6 0

3 years ago

You wish to prepare a tape-casting slip containing 50 vol% Al2O3 and 50 vol% polyvinyl butyral (PVB) binder. If the density of A

belka [17]

Answer: The mass of PVB required to produce 1000 grams of tape is 213.4 grams

Explanation:

To calculate the mass of aluminium oxide, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$ .......(1)

For $Al_2O_3$

We are given:

50% (v/v) of $Al_2O_3$

This means that 50 mL of aluminium oxide is present in 100 mL of tape

Calculating the mass of aluminium oxide by using equation 1:

Density of aluminium oxide = $3.98 g/cm^3$

Volume of aluminium oxide = $50mL=50cm^3$ (Conversion factor: $1mL=1cm^3$ )

Putting values in equation 1, we get:

$3.98g/cm^3=\frac{\text{Mass of aluminium oxide}}{50cm^3}\\\\\text{Mass of aluminium oxide}=(3.98g/cm^3\times 50cm^3)=199g$

Mass of aluminium oxide = 199 g

For PVB:

We are given:

50% (v/v) of PVB

This means that 50 mL of PVB is present in 100 mL of tape

Calculating the mass of PVB by using equation 1:

Density of PVB = $1.08 g/cm^3$

Volume of PVB = $50mL=50cm^3$

Putting values in equation 1, we get:

$1.08g/cm^3=\frac{\text{Mass of PVB}}{50cm^3}\\\\\text{Mass of PVB}=(1.08g/cm^3\times 50cm^3)=54g$

Mass of PVB = 54 g

Mass of tape = Mass of aluminium oxide + mass of PVB

Mass of tape = [199 + 54] g = 253 g

To calculate the mass of PVB required to produce 1000 g of tape, we use unitary method:

When 253 grams of tape is made, the mass of PVB required is 54 g

So, when 1000 grams of tape is made, the mass of PVB required will be = $\frac{54}{253}\times 1000=213.4g$

Hence, the mass of PVB required to produce 1000 grams of tape is 213.4 grams

4 0

2 years ago

Why does ice have a different density than liquid water?

gulaghasi [49]

Answer:

because it has lighter molecules.

4 0

3 years ago

PLEASE HELP!!

yanalaym [24]

Answer:

Remove a H2 molecule from the left side of the equation.

Explanation:

i took the quiz and got it right

8 0

2 years ago

Answer the following questions about the solubility of AgCl(s). The value of Ksp for AgCl(s) is 1.8 × 10−10.

Firlakuza [10]

Answer:

[Ag⁺] = 1.3 × 10⁻⁵M

s = 3.3 × 10⁻¹⁰ M

Because the common ion effect.

Explanation:

1. Value of [Ag⁺] in a saturated solution of AgCl in distilled water.

The value of [Ag⁺] in a saturated solution of AgCl in distilled water is calculated by the dissolution reaction:

AgCl(s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)

The ICE (initial, change, equilibrium) table is:

AgCl(s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)

I X 0 0

C -s +s +s

E X - s s s

Since s is very small, X - s is practically equal to X and is a constant, due to which the concentration of the solids do not appear in the Ksp equation.

Thus, the Ksp equation is:

Ksp = [Ag⁺] [Cl⁻]

Ksp = s × s

Ksp = s²

By substitution:

1.8 × 10⁻¹⁰ = s²

s = 1.34 × 10⁻⁵M

Rounding to two significant figures:

[Ag⁺] = 1.3 × 10⁻⁵M ← answer

2. Molar solubility of AgCl(s) in seawater

Since, the conentration of Cl⁻ in seawater is 0.54 M you must introduce this as the initial concentration in the ECE table.

The new ICE table will be:

AgCl(s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)

I X 0 0.54

C -s +s +s

E X - s s s + 0.54

The new equation for the Ksp equation will be:

Ksp = [Ag⁺] [Cl⁻]

Ksp = s × ( s + 0.54)

Ksp = s² + 0.54s

By substitution:

1.8 × 10⁻¹⁰ = s² + 0.54s

s² + 0.54s - 1.8 × 10⁻¹⁰ = 0

Now you must solve a quadratic equation.

Use the quadratic formula:

$s=\dfrac{-0.54\pm\sqrt{0.54^2-4(1)(-1.8\times 10^{-10})}}{2(1)}$

The positive and valid solution is s = 3.3×10⁻¹⁰ M ← answer

3. Why is AgCl(s) less soluble in seawater than in distilled water.

AgCl(s) is less soluble in seawater than in distilled water because there are some Cl⁻ ions is seawater which shift the equilibrium to the left.

This is known as the common ion effect.

By LeChatelier's principle, you know that an increase in the concentrations of one of the substances that participate in the equilibrium displaces the reaction to the direction that minimizes this efect.

In the case of solubility reactions, this is known as the common ion effect: when the solution contains one of the ions that is formed by the solid reactant, the reaction will proceed in less proportion, i.e. less reactant can be dissolved.

3 0

3 years ago