Marissa blows up balloons for a party. She decides how big or small to make each balloon. Why does the air she blows into each balloon take up all the space inside the balloon

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:

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:

you can burn it.

Explanation:

that is the way to change paper

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:

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:

Letter D

Explanation:

Answer:C - 3.115 g

Explanation:Correct

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me with a wey some one made

Answer:ok

Explanation:

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:

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.

For more information about half-life, refer to the link:

https://brainly.com/question/24710827

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:

https://brainly.com/question/26905271

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