Which of the following best illustrates the law of corkervation of energy?
A
The work done stretching a spring is transformed into stored energy in the spring
B
Leak-proof batteries are able to store chemical energy for long periods of time.
С
Ajogger conserves her energy by reducing the rate at which she jogs.
D
Ice water changes into liquid water when it absorbs heat energy.

Answer:

beryllium has a higher ionization energy because its radius is smaller. boron has a higher ionization energy because its radius is smaller.

Answer:

The answer is that light is faster than sound.

Explanation:

A good example is of lightning and thunder during a thunderstorm.

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Answer:

f = 0.6× 10¹⁶ Hz

Explanation:

Given data:

Wavelength of radiation = 5.00 × 10⁻⁸ m

Frequency of radiation = ?

Solution:

Formula:

Speed of light = wavelength × frequency

speed of light = 3 × 10⁸m/s

Now we will put he values in formula.

3 × 10⁸m/s = 5.00 × 10⁻⁸ m ×  f

f = 3 × 10⁸m/s / 5.00 × 10⁻⁸ m

f = 0.6× 10¹⁶ s⁻¹

s⁻¹ = Hz

f = 0.6× 10¹⁶ Hz

Lead + Iron(II) Sulfate = Lead Sulfate + Iron

Lead - Pb

Plumbum Element 82 Bulk Lead 鉛 Lead (native) Lead-210 Pb-210 210Pb

Iron(II) Sulfate - FeSO4

Iron Sulfate (1:1) Iron(II) Sulfate (1:1) Ferrous Sulfate (anhydrous) Ferrous Sulfate Ferrous Sulfate (anh.) Iron(2+) Sulfate (anh.) FeSO4 Fe(II)SO4

Lead Sulfate - PbSO4

Anglesite

iron - Fe

Element 26 Fer 26Fe Eisen Ferrum Hierro Ferrum Metallicum 鐵

Answer:

3.1 moles of ammonia

18.67× 10²³ molecules

Mass = 52.7 g

Explanation:

Given data:

Number of atoms of hydrogen = 5.68×10²⁴ atoms

A) Number of molecules of ammonia = ?

Solution:

First of all we will calculate the number of moles of hydrogen.

1 mole = 6.022× 10²³ atoms

5.68×10²⁴ atoms × 1 mol / 6.022× 10²³ atoms

0.94×10¹ mol

9.4 moles of hydrogen

Moles of ammonia:

3 moles of hydrogen are present in one mole of ammonia.

9.4 moles of hydrogen = 1/3×9.4 =

3.1 moles of ammonia

Number of molecules of ammonia:

1 mole contain 6.022× 10²³ molecules.

3.1 mol × 6.022× 10²³ molecules / 1 mol

18.67× 10²³ molecules

c) Mass of sample = ?

Mass = number of moles × molar mass

Mass = 3.1 moles × 17 g/mol

Mass = 52.7 g

Answer:

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Explanation:

Answer:

159g

Explanation:

Copper (I) sulfide is formed by the chemical reaction between copper and sulfur. Copper is a limiting reagent that results in the formation of 190.26 grams of copper(I) sulfide.

A limiting reagent is defined as the reactant that gets completely utilized in the reaction and affects the amount of the product formation. On the contrary, the excess reagent is present in more amounts.

Given,

Mass of copper = 127 gm

Mass of sulfur = 41 gm

Moles of copper:

n = mass ÷ molar mass

n = 127 ÷ 63.546

n = 1.99

Moles of sulfur:

n = 41 ÷ 32.065

n = 1.27

The reaction is shown as:

Cu + S → CuS

Here, 9 g of sulfur remains unreacted and hence is an excess reagent, and copper being consumed is a limiting reagent.

Here, 1 mole of copper reacts to form 1 mole of copper sulfide hence, 1.99 moles of Cu produces, 1.99 moles of copper sulfide.

Mass of CuS is calculated as:

Mass = moles × molar mass

Mass = 1.99 mol × 95.611 g/mol

= 190.26 grams

Therefore, 127 gm of copper and 41 gm of sulfur produces 190.26 grams of copper sulfide.

Learn more about limiting reagents, here:

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Explanation:

The halogen family and noble gases are similar in just one particular way, they are groups of non-metals. All members of these two groups are categorized as non-metals.

Here are some of the differences between them;

Answer:

you can burn it.

Explanation:

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Answer:

2.1 moles of water formed.

Explanation:

Given data:

Moles of water formed = ?

Moles of Ni(OH) = 4.20 mol

Solution:

Chemical equation:

2Ni(OH)    →  Ni₂O + H₂O

Now we will compare the moles of Ni(OH)  with water.

             Ni(OH)           :            H₂O

                2                 :              1

              4.20             :            1/2×4.20 = 2.1 mol

2.1 moles of water formed.

Answer:

Letter D

Explanation:

Answer:

53.4%

Explanation:

N₂ + 3H₂ → 2NH₃

Given that:

26.3 g of H₂ react with N₂ to produce 79.0 g of NH₃

Then:

The number of moles of H₂ = 26.3g of H₂ * (1 mol of H₂/ 2.02g of H₂)

= 13.05 mol of H₂

The number of moles of NH₃ = 13.05 mol of H₂ * ( 2mol of NH₃/ 3 mol of H₂)

= 8.697 mol of NH₃

The mass of NH₃ = 8.697 mol of NH₃ *( 17.04g of NH₃/ 1 mol of NH₃) = 148.1 g of NH₃

The percent yield = actual yield/ theoretical yield * 100%

The percent yield = ( 79.0 g/ 148.1 g )* 100%

The percent yield ≅ 53.4 %

Answer:

53.4

Explanation:

Answer:C - 3.115 g

Explanation:Correct

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Answer:ok

Explanation:

Answer:

True

Explanation:

All irreversible reactions go to completion because the process cannot proceed in the backward direction.

A reversible reaction is one that does not go into completion in one direction.

An irreversible direction completely uses up the reactants to form the products and goes into completion.

These reactions can only proceed in one direction and not both.

Answer:

67.6 years is the time the isotope take to decay from 0.900g to 0.170g

Explanation:

The radioactive decay follows first order law:

Ln [A] = -kt + ln[A]₀

Where [A] is concentration after time t,

k is decay constant:

k = ln 2 / t(1/2)

k = ln2 / 28.1 years

k = 0.02467 years⁻¹

[A]₀ = Initial concentration.

We can replace concentration and use the mass of the isotope:

Ln [A] = -kt + ln[A]₀

Ln [0.170g] = -0.02467 years⁻¹t + ln[0.900g]

-1.667 = -0.02467 years⁻¹t

t =

The time required for the sample to reduce to 0.170 grams has been 67.75 years.

Half-life can be defined as the times required by the substance to reduce to half of its initial concentration.

The half-life can be expressed as:

Amount left = Initial amount [tex]\times[/tex] [tex]\rm \dfrac{1}{2}^\dfrac{time}{Half-life}[/tex]

From the given:

0.17 g = 0.9 [tex]\rm \times\;\dfrac{1}{2}^\dfrac{t}{28.1}[/tex]

[tex]\rm \dfrac{0.17}{0.9}[/tex] = [tex]\rm \dfrac{1}{2}^\dfrac{t}{28.1}[/tex]

0.188 = [tex]\rm (0.5)^\dfrac{t}{28.1}[/tex]

Taking log on both the sides,

log 0.188 = [tex]\rm \dfrac{t}{28.1}\;\times[/tex] log 0.5

2.411 = [tex]\rm\dfrac{t}{28.1}[/tex]

t = 2.411 [tex]\times[/tex] 28.1 years

Time = 67.75 years.

The time required for the sample to reduce to 0.170 grams has been 67.75 years.

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