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3 years ago

In the periodic table which elements typically have similar properties

1 answer:

7 0

Elements which appear in the same column have similar properties (periodicity). For example, all of the elements in group XVII (17), the Halogens, all react in a similar fashion; they all like to attract one additional electron and form a -1 anion.

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A 232-lb fullback runs the 40-yd dash at a speed of 19.8 ± 0.1 mi/h.

Neporo4naja [7]

Answer:

(a) 7.11 x 10⁻³⁷ m

(b) 1.11 x 10⁻³⁵ m

Explanation:

(a) The de Broglie wavelength is given by the expression:

λ = h/p = h/mv

where h is plancks constant, p is momentum which is equal to mass times velocity.

We have all the data required to calculate the wavelength, but first we will have to convert the velocity to m/s, and the mass to kilograms to work in metric system.

v = 19.8 mi/h x ( 1609.34 m/s ) x ( 1 h / 3600 s ) = 8.85 m/s

m = 232 lb x ( 0.454 kg/ lb ) = 105.33 kg

λ = h/ mv = 6.626 x 10⁻³⁴ J·s / ( 105.33 kg x 8.85 m/s ) = 7.11 x 10⁻³⁷ m

(b) For this part we have to use the uncertainty principle associated with wave-matter:

ΔpΔx > = h/4π

mΔvΔx > = h/4π

Δx = h/ (4π m Δv )

Again to utilize this equation we will have to convert the uncertainty in velocity to m/s for unit consistency.

Δv = 0.1 mi/h x ( 1609.34 m/mi ) x ( 1 h/ 3600 s )

= 0.045 m/s

Δx = h/ (4π m Δv ) = 6.626 x 10⁻³⁴ J·s / (4π x 105.33 kg x 0.045 m/s )

= 1.11 x 10⁻³⁵ m

This calculation shows us why we should not be talking of wavelengths associatiated with everyday macroscopic objects for we are obtaining an uncertainty of 1.11 x 10⁻³⁵ m for the position of the fullback.

5 0

3 years ago

Calculate the mass of water produced when 42 g of propane, c3h8, is burned with 115 g of oxygen

gavmur [86]

The balanced combustion reaction of propane, C₃H₈, is

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

Molar mass of propane: 44 g/mol
Moles of propane = 42 g * (1 mol/44g) = 0.9545 mol propane

Molar mass of oxygen: 32 g/mol
Moles of oxygen = 115 g * (1 mol/32 g) = 3.594 mol oxygen

Moles of oxygen needed to completely react propane:
0.9545 mol propane * (5 mol O₂/1 mol propane) = 4.7725 mol oxygen

Since the available oxygen is only 3.594 moles and propane needs 4.7725 moles, that means oxygen is our limiting reactant. We base the amount of water produced here.

Molar mass of water: 18 g/mol
Mass of water produced = 3.594 mol O₂ * (4 mol H₂O/5 mol O₂) * (18 g/mol)
Mass of water produced = 258.768 grams

6 0

3 years ago

The daily production of carbon dioxide from an 780.0 mw coal-fired power plant is estimated to be 3.3480 x 104 tons (not metric)

valkas [14]

The production of $CO_{2}$ is $3.3480\times 10^{4} tons/day$ . Converting mass into kg,

1 ton=907.185 kg, thus,

$3.3480\times 10^{4} tons=3.037\times 10^{7} kg$

Thus, production of $CO_{2}$ will be $3.037\times 10^{7} kg/ day$ .

The specific volume of $CO_{2}$ is $0.0120 m^{3}/kg$ .

Volume of $CO_{2}$ produced per day can be calculated as:

$V=Specific volume\times mass$

Putting the values,

$V=0.0120 m^{3}/kg\times 3.037\times 10^{7} kg=364440 m^{3}/day$

Thus, volume of $CO_{2}$ produced per year will be:

$V=\frac{365 days}{1 year}(364440 m^{3}/day)=1.33\times 10^{8}m^{3}/year$

Thus, in 4 year volume of $CO_{2}$ produced will be:

$V=1.33\times 10^{8}m^{3}/year\times 4 years=5.32\times 10^{8}m^{3}$

8 0

3 years ago

Hi! I need help with this question, if it is possible, please answer immediately.

Dennis_Churaev [7]

Answer:

Explanation:

The paper clip allows elictricity to pass, unlike the eraser or paper

6 0

3 years ago

Arrange the elements in decreasing order of first ionization energy.

just olya [345]

Answer:

The decreasing order of first ionization energy: Se > Ge > In > Cs

The decreasing order of first ionization energy: x > y > z

Explanation:

Ionization energy refers to the energy needed to completely pull out an electron from the valence shell of a neutral gaseous atom.

First ionization energy is the energy involved in the removal of first valence electron.

In the periodic table, down the group, as atomic radius of elements increases, the ionization energy decreases

Whereas, across a period, as atomic radius of elements decreases, the ionization energy increases.

PART (A):

Position of the given elements in the periodic table:

Indium (In): Group 13, period 5

Germanium (Ge): Group 14, period 4

Selenium (Se): Group 16, period 4

Caesium (Cs): Group 1, period 6

Thus, the increasing order of atomic radius: Se < Ge < In < Cs

Therefore, the decreasing order of first ionization energy: Se > Ge > In > Cs

PART (B):

Given elements:

element x: radius = 110 pm

element y: radius = 199 pm

element z: radius = 257 pm

Thus, the increasing order of atomic radius: x < y < z

Therefore, the decreasing order of first ionization energy: x > y > z

5 0

3 years ago