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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Is buoyancy a physical or chemical property?
Do all irreversible reactions go to completion? If so why?
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.
what are all the ways you can change a piece of binder paper?
Answer:
you can burn it.
Explanation:
that is the way to change paper
54. Copper(I) sulfide is formed when copper and sulfur are
heated together. In this reaction, 127 g of copper reacts
with 41 g of sulfur. After the reaction is complete, 9 g of
sulfur remains unreacted. What is the mass of copper
sulfide formed?
SEL
Ma
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.
What is a limiting reagent?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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Please give formulas of how you solved I only need help on a but if you can do all that I would appreciate that.
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
which of the following would form an ionic bond?
a.strontium and oxygen
b. sulfur and oxygen
Group 17 is called the halogen family, and the group to its right is called the noble gases. How are these elements alike and how are they different?
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;
Halogens have 7 electrons in their outermost shell whereas noble gases have 8 electrons in theirs. Halogens are highly reactive elements, noble gases are non-reactive. Halogens are made up of electronegative elements where as noble gases are neither electropositive nor electronegative.How can the shape of a molecule determine its polarity.
Answer:
Letter D
Explanation:
N2 + 3H2 Right arrow. 2NH3 What is the percent yield of NH3 if the reaction of 26.3 g of H2 produces 79.0 g of NH3? Use Percent yield equals StartFraction actual yield over theoretical yield EndFraction times 100..
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:
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Explanation:
Strontium-90 is one of the products of the fission of uranium-235. This strontium isotope is radioactive, with a half-life of 28.1 yr. Calculate how long it will take for 0.900 g of the isotope to be reduced to 0.170 g by decay.
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 =
67.6 years is the time the isotope take to decay from 0.900g to 0.170gThe 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:
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