Answer:
Organic
Explanation:
Edge student
Which statement best describes the octet rule?
A. When an atom becomes an ion, it gains or loses electrons so that its valence shell holds eight electrons.
B. When an atom becomes an ion, it gains or loses protons so that its nucleus holds eight protons.
C. When an atom becomes an ion, it gains or loses eight electrons.
D. When an atom becomes an ion, it gains or loses eight neutrons.
Rupert had three substances. A brown substance was a liquid at room
temperature. He hit each of the other two with a hammer. A blue crystal
cracked but did not break. A silver substance flattened but did not crack.
Which two statements could be true?
A. The brown substance is ionic
B. The silver substance is ionic
C. The brown substance is molecular
D. The blue substance is ionic
Answer:
its C and D
C. The brown substance is molecular
D. The blue substance is ionic
Explanation:
did the test !
Two correct statements are B) The silver substance is ionic
C) The brown substance is molecular.
What kind of substance is silver?Silver is a chemical element with the symbol Ag and atomic wide variety 47. categorized as transition steel, Silver is stable at room temperature.
Which substance is molecular?It is a molecular substance, that's a substance with or more atoms, the smallest gadgets of remember joined together via a covalent bond. A covalent bond is a hyperlink created via the sharing of electrons that holds these atoms collectively.
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Which of the following is an example of a chemical change that occurs in your home?
A) Water boiling on the stove
B) Composting your own fertilizer
C) Smashing a window with a baseball
D) Adding food coloring to icing for a cake
Find ΔHrxn for the following reaction:
2PbS(s)+3O2(g)→2PbO(s)+2SO2(g)
Answer:
ΔH°rxn = -827.5 kJ
Explanation:
Let's consider the following balanced equation.
2 PbS(s) + 3 O₂(g) → 2 PbO(s) + 2 SO₂(g)
We can calculate the standard enthalpy of reaction (ΔH°rxn) from the standard enthalpies of formation (ΔH°f) using the following expression.
ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g) )] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g) )]
ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g) )] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g) )]
ΔH°rxn = [2 mol × (-217.32 kJ/mol) + 2 mol × (-296.83)] - [2 mol × (-100.4) + 3 mol × 0 kJ/mol]
ΔH°rxn = -827.5 kJ
The standard enthalpy of the reaction is -827.5 kJ/mol
The standard enthalpy of reaction [tex]\mathbf{\Delta H^0_{rxn}}[/tex] is the enthalpy change that happens in a system whenever one mole of the matter is converted through a chemical process under normal conditions.
The given reaction can be expressed as:
2PbS(s) + 3O₂(g) → 2PbO(s) + 2SO₂(g)
The standard enthalpy can be represented by the equation:
[tex]\mathbf{\Delta H^0_{rxn} = \sum \Delta _f ^0(products) - \sum \Delta _f^0(reactants)}[/tex]
At standard conditions, the standard enthalpies of formation of the given species are:
ΔH°f(PbO(s)) = 219 kJ/molΔH°f(SO₂(g)) = -296.83 kJ/molΔH°f(PbS(s)) = 100.4 kJ/molΔH°f(O₂(g)) = 0 kJ/mol
∴
[tex]\mathbf{ \Delta H^0_{rxn} = \Big[2 mol \times \Delta H^0_f(PbO(s)) + 2 mol \times \Delta H^0_ f(SO_2(g) )\Big] - \Big[2 mol \times \Delta H^0_f (PbS(s))} + \mathbf{ 3 mol \times \Delta H^0_f(O_2(g) )\Big] }}[/tex]
[tex]\mathbf{\Delta H^0rxn = [2 mol \times (-217.32 kJ/mol) + 2 mol \times (-296.83)] - [2 mol \times (-100.4)} \\ \mathbf{+ 3 mol \times 0 kJ/mol]}}[/tex]
[tex]\mathbf{\Delta H^0rxn = -827.5 \ kJ/mol}}[/tex]
Therefore, we can conclude that the standard enthalpy of the reaction is -827.5 kJ/mol
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You have a carbonate buffer with pH 10.3 and a concentration of 2.0 M. What is the buffer capacity of 100 mL of the buffer against 3.0 M CsOH?
Answer:
Explanation:
pH = 10.3
[ H] = 10⁻¹⁰°³
= 5 x 10⁻¹¹ M
concentration of CsOH C = 3 M
pKa of carbonate = 6.35
Ka = 10⁻⁶°³⁵ = 4.46 x 10⁻⁷
Buffer capacity = 2.303 x C x Ka x [ H⁺] / ( Ka + [ H⁺]² )²
= 2.303 x 3 x 4.46 x 10⁻⁷ x 5 x 10⁻¹¹ / ( 4.46 x 10⁻⁷ + 25 x 10⁻²² )²
= 154 x 10⁻¹⁸ / 19.9 x 10⁻¹⁴
= 7.74 x 10⁻⁴ .
What is cellulose and what does it do?
Answer:
Cellulose is an organic compound with the formula , a polysaccharide consisting of a linear chain of several hundred to many thousands of β linked D-glucose units. Cellulose is an important structural component of the primary cell wall of green plants, many forms of algae and the oomycetes.
Explanation:
Answer:
Hope this helped.
Explanation:
Cellulose is a molecule, consisting of hundreds – and sometimes even thousands – of carbon, hydrogen and oxygen atoms. Cellulose is the main substance in the walls of plant cells, helping plants to remain stiff and upright. Humans cannot digest cellulose, but it is important in the diet as fibre.
Having the same number of valence electrons means having similar chemical properties.
True or False
Answer:
True
Explanation:
Because atoms with the same number of valence electrons react in similar ways with other elements.
How many liters of H2(g) at STP are produced per gram of Al(s) consumed in the following reaction? 2Al(s)+6HCl(aq)→2AlCl3(aq)+3H2(g)
Answer:
1.24 L of H₂ at STP .
Explanation:
2Al(s) +6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)
2 moles 3 x 22.4 L
2 x 27 g of Al reacts to give 3 x 22.4 L of H₂ at STP .
1 g of Al will react to give 3 x 22.4 / ( 2 x 27 ) L of H₂ at STP .
= 1.24 L of H₂ at STP .
The volume of hydrogen produced by 1 grams of Al has been 1.24 L.
The balanced chemical reaction has been given as:
[tex]\rm 2\;Al\;+\;6\;HCl\;\rightarrow\;2\;AlCl_3\;+\;3\;H_2[/tex]
From the equation, 2 moles of Aluminum gives 3 moles of Hydrogen
The mass of the compound from moles can be given as:
Mass = moles × molecular mass
Mass of 2 moles Al = 2 × 27 g
Mass of 2 moles Al = 54 g
Mass of 3 moles hydrogen = 3 × 2 g
Mass of 3 moles hydrogen = 6 g
From the equation,
54 g aluminum gives = 6 grams hydrogen
[tex]\rm 1\;gram\;aluminum\;=\;\dfrac{6}{54}\;\times\;1[/tex]
1 gram Aluminum = 0.11 grams hydrogen
The mass of hydrogen produced by 1 gram Al has been 0.11g. The moles equivalent to 0.11g hydrogen has been given as:
Mass = moles × molecular mass
0.11 g = moles × 2 g/mol
Mole of hydrogen = 0.055 mol
The moles of hydrogen produced by 1 gram of Al has been 0.055 mol.
According to the ideal gas equation, any gas at STP has 1 mole equivalent to 22.4 L. So,
1 mol = 22.4 L
0.055 mol = 0.055 × 22.4 L
0.055 mol = 1.244 L.
The volume of hydrogen produced by 1 grams of Al has been 1.24 L.
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What color is the acetic acid plus methyl orange solution and what does this tell you about where the equilibrium is
Answer:
Red
Explanation:
Acetic acid is an acidic medium. Recall that indicators are organic substances whose color changes in response to change in the pH of the solution. often times, the protonated and deprotonated forms of an indicator have different colors.
However, an equilibrium is set up when an indicator is in acid/ basic medium. Methyl orange is red in acid medium and yellow in basic medium.
Hence while in acetic acid, the equilibrium lies towards the protonated form of acetic acid, hence the solution appears red.
An unknown compound contains only carbon, hydrogen, and oxygen (CxHyOz). Combustion of 5.50 g of this compound produced 8.07 g of carbon dioxide and 3.30 g of water.
Required:
a. How many moles of carbon, C, were in the original sample?
b. How many moles of hydrogen, H, were in the original sample?
Answer:
a. 0.183 mol C
b. 0.366 mol H
Explanation:
Assuming total combustion, all of the carbon in the unknown compound turned into carbon dioxide, CO₂.
So first we calculate the CO₂ moles produced, using its molecular weight:
8.07 g CO₂ ÷ 44 g/mol = 0.183 mol CO₂This means in the unknown compound there were 0.183 moles of carbon, C.
Conversely, all of the hydrogen in the unknown compound turned into water, H₂O.
Calculating the H₂O moles:
3.30 g ÷ 18 g/mol = 0.183 mol H₂OWe multiply the water moles by two, as there are 2 H moles per H₂O mol:
0.183 * 2 = 0.366 mol H.Help, 8th grade Science
Find the density of a liquid that has a mass of 24 grams and a volume of 6 ml.
Answer:
4
Explanation:
24/6 is 4.
Answer:
Mathematically: D = m/v. If you know what liquid you have, you can look up its density in a table. Once you know that, all you have to do to find the mass of the liquid is to measure its volume. Once you know density and volume, calculate mass using this relationship: mass = density • volume.
Explanation:
The atomic notation for a particular atom of boron is ' B. The
atomic number is while the mass number is
Answer:
5;11
Explanation:
Determine each type of reaction. 2 C 2 H 2 ( g ) + 5 O 2 ( g ) ⟶ 4 C O 2 ( g ) + 2 H 2 O ( l ) 2CX2HX2(g)+5OX2(g)⟶4COX2(g)+2HX2O(l) Choose... N H 4 N O 3 ( s ) ⟶ N 2 O ( g ) + 2 H 2 O ( l ) NHX4NOX3(s)⟶NX2O(g)+2HX2O(l) Choose... C O ( g ) + 2 H 2 ( g ) ⟶ C H 3 O H ( l ) CO(g)+2HX2(g)⟶CHX3OH(l) Choose... 2 F e ( s ) + 6 H C l ( a q ) ⟶ 2 F e C l 3 ( a q ) + 3 H 2 ( g ) 2Fe(s)+6HCl(aq)⟶2FeClX3(aq)+3HX2(g) Choose... C a C l 2 ( a q ) + N a 2 C O 3 ( a q ) ⟶ 2 N a C l ( a q ) + C a C O 3 ( s ) CaClX2(aq)+NaX2COX3(aq)⟶2NaCl(aq)+CaCOX3(s) Choose...
Answer:
2 C 2 H 2 ( g ) + 5 O 2 ( g ) ⟶ 4 C O 2 ( g ) + 2 H 2 O ( l )- combustion reaction
N H 4 N O 3 ( s ) ⟶ N 2 O ( g ) + 2 H 2 O ( l )- decomposition reaction
C O ( g ) + 2 H 2 ( g ) ⟶ C H 3 O H ( l ) - combination reaction
2 F e ( s ) + 6 H C l ( a q ) ⟶ 2 F e C l 3 ( a q ) + 3 H 2 ( g )- Redox reaction
C a C l 2 ( a q ) + N a 2 C O 3 ( a q ) ⟶ 2 N a C l ( a q ) + C a C O 3 ( s )- double displacement reaction
Explanation:
We can determine the type of reaction by considering the reactants and products.
Combustion is a reaction between a substance and oxygen which produces heat and light. The first reaction is the equation for the combustion of ethyne.
A decomposition reaction is one in which a single reactant breaks down to form products. The second reaction is the decomposition of ammonium nitrate.
A combination reaction is said to occur when two elements or compounds react to form a single product. The third reaction is the combination of carbon dioxide and methane to form methanol.
An oxidation-reduction reaction is a reaction in which there is a change in oxidation number of species from left to right of the chemical reaction equation. The fourth reaction is the oxidation of iron (0 to +3 state) and reduction of hydrogen (+1 to 0 state).
A double displacement reaction is a reaction in which ions exchange partners from left to right in the reaction equation. The fifth reaction is a double displacement reaction. Both Na^+ and Ca^2+ exchanged partners from left to right of the reaction equation.
Reactions are the formation of the products from the reactant. The types of reactions are combustion, decomposition, combination, Redox and double displacement.
What are the types of reactions?The reaction is a chemical change in the properties of the reactant that forms the products. It can be of various types based on the formation of the product.
The first reaction is combustion as the reactants react and use oxygen to form heat, carbon dioxide and water. The combustion reaction of ethyne can be shown as,
[tex]\rm 2 C _{2} H _{2} ( g ) + 5 O _{2} ( g ) \rightarrow 4 C O _{2} ( g ) + 2 H _{2} O ( l )[/tex]
The second reaction is decomposition in which a single reactant decomposes to form two or more products. The decomposition of ammonium nitrate can be shown as,
[tex]\rm N H _{4} N O _{3} ( s ) \rightarrow N _{2} O ( g ) + 2 H _{2} O ( l )[/tex]
The third reaction is a combination reaction in which two compound or elements combines to form one product. The combination reaction between carbon monoxide and hydrogen to form methanol can be shown as,
[tex]\rm C O ( g ) + 2 H _{2} ( g ) \rightarrow C H _{3} O H ( l )[/tex]
The fourth reaction is redox and includes the oxidation and the reduction of the species of the reaction. In the reaction, iron undergoes oxidation and hydrogen reduction. The redox reaction can be shown as,
[tex]\rm 2 F e ( s ) + 6 H C l ( a q ) \rightarrow 2 F e C l _{3} ( a q ) + 3 H _{2} ( g )[/tex]
The fifth reaction is a double displacement reaction in which the calcium and sodium interchange their position in the product formation. The reaction can be shown as,
[tex]\rm C a C l _{2} ( a q ) + N a _{2} C O _{3} ( a q ) \rightarrow 2 N a C l ( a q ) + C a C O _{3} ( s )[/tex]
Therefore, the type of reactions is 1. combustion, 2. decomposition, 3. combination, 4. redox and 5. double displacement.
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Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory to standardize strong base solutions. It has the unwieldy formula of KHC8H4O4. This is often written in shorthand notation as KHP.
Required:
How many grams of KHP are needed to exactly neutralize 25.6 mL of a 0.527 M potassium hydroxide solution?
Answer:
2.75 g
Explanation:
The number of moles of KOH is obtained from the equation;
n=CV
C= concentration
V= volume of solution
n= 0.527 × 25.6/1000
n= 0.0135
Reaction equation;
KOH(aq) + KHC8H4O4(aq) --->(KC8H4O4)K(aq) + H2O(l)
1 mole of KHP reacts with 1 mole of KOH
Hence; 0.0135 moles of KHP reacts with 0.0135 moles of KOH
From;
Number of moles = reacting mass/molar mass
Reacting mass = 0.0135 × 204
= 2.75 g of KHP
A certain chemical reaction releases of heat for each gram of reactant consumed. How can you calculate the heat produced by the consumption of of reactant? Set the math up. But don't do any of it. Just leave your answer as a math expression. Also, be sure your answer includes all the correct unit symbols.
Complete Question
The complete question is shown in the first uploaded image
Answer:
So the math expression is
[tex]heat = \frac{ 35. 7 KJ * 1900 \ gram }{ 1 \ gram }[/tex]
Explanation:
From the question we are told that
The heat released for 1 gram of reactant consumed is [tex]H = 37.5 \ KJ/g [/tex]
The mass of reactant considered is [tex]m = 1.9 \ kg = 1900 \ g[/tex]
So if
[tex]37.5 \ KJ [/tex] is produced for 1 gram
Then
x kJ is produced for 1900 g
=> [tex]x = \frac{ 35. 7 KJ * 1900 \ gram }{ 1 \ gram }[/tex]
So the heat released is
[tex]heat = \frac{ 35. 7 KJ * 1900 \ gram }{ 1 \ gram }[/tex]
Josh heated a certain amount of blue copper sulfate crystals to get 2.1 g of white copper sulfate powder and 1.4 g of water. What is most likely the mass of the blue copper sulfate that he heated and why?
Answer: The mass of blue copper sulfate is 3.5 g
Explanation:
Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.
This also means that total mass on the reactant side must be equal to the total mass on the product side.
The chemical equation for the heating of copper sulfate crystals is:
Let the mass of blue copper sulfate be 'x' grams
We are given:
Mass of copper sulfate powder = 2.1 grams
Mass of water = 1.4 grams
Total mass on reactant side = x
Total mass on product side = (2.1 + 1.4) g
So, by applying law of conservation of mass, we get:
Hence, the mass of blue copper sulfate is 3.5 grams
Cathode rays are deflected toward a negatively charged plate in an electric field.
True
False
Answer:
true. cathode rays are deflected
Problem:
[Ar]4s2
Identify the period (p) , group (g) and valence electrons block of the element
Answer:
it is Calcium (Ca)
4th period, 2nd group, 2 valence electrons
Carbonic acid, H2CO3, has two acidic hydrogens. A solution containing an unknown concentration of carbonic acid is titrated with potassium hydroxide. It requires 22.9 mL of 1.430 M KOH solution to titrate both acidic protons in 54.2 mL of the carbonic acid solution.
Required:
a. Write a balanced net ionic equation for the neutralization reaction. Include physical states.
b. Calculate the molarity of the carbonic acid solution.
Answer:
a. H₂CO₃(aq) + KOH(aq) ⇄ K₂CO₃(aq) + H₂O(l)
b. 0.603 M
Explanation:
Step 1: Write the neutralization reaction
H₂CO₃(aq) + KOH(aq) ⇄ K₂CO₃(aq) + H₂O(l)
Step 2: Calculate the reacting moles of KOH
22.9 mL of 1.430 M KOH react.
0.0229 L × (1.430 mol/L) = 0.0327 mol
Step 3: Calculate the reacting moles of H₂CO₃
The molar ratio of H₂CO₃ to KOH is 1:1. The reacting moles of H₂CO₃ are 1/1 × 0.0327 mol = 0.0327 mol.
Step 4: Calculate the molarity of H₂CO₃
0.0327 moles of H₂CO₃ are in a volume of 54.2 mL. The molarity of H₂CO₃ is:
M = 0.0327 mol/0.0542 L = 0.603 M
Which 2 main body systems work alongside the digestive system?
Which pair of elements will form a covalent bond?
Aluminum metal and bromine liquid (red) react violently to make aluminum bromide (white powder). One way to represent this equilibrium is:
Al(s) + 3/2 Br2(l)AlBr3(s)
We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above.
1) 2 AlBr3(s) 2 Al(s) + 3 Br2(l)
2) 2 Al(s) + 3 Br2(l) 2 AlBr3(s)
3) AlBr3(s) Al(s) + 3/2 Br2(l)
Answer:
Explanation:
Al(s) + 3/2 Br₂(l) = AlBr₃(s)
K = [ AlBr₃] / [ Al] [ Br₂]³/²
K² = [ AlBr₃]² / [ Al ] ² [ Br₂]³
2 AlBr₃ = 2 Al(s) + 3 Br₂(l) =
K₁ = [ Al ] ² [ Br₂]³ / [ AlBr₃]²
K₁ = ( 1 / K² ) = K⁻²
2 ) 2 Al(s) + 3 Br₂(l) = 2 AlBr₃(s)
K₂ = [ AlBr₃ ]² / [ Al ]² [ Br₂ ]³
K₂ = K²
3 )
AlBr₃(s) = Al(s) + 3/2 Br₂(l)
K₃ = [ Al ] [ Br₂ ] ³/² / [ AlBr₃ ]
= ( 1 / K ) = K⁻¹
Which profile best shows the topography alone line AD
A gas has a volume of 300 mL and a pressure of 2 atm. What volume will the gas occupy when the pressure is
increased to 7 atm (total)?
Answer:
The answer is 85.71 mLExplanation:
The new volume can be found by using the formula for Boyle's law which is
[tex]P_1V_1 = P_2V_2[/tex]
where
P1 is the initial pressure
P2 is the final pressure
V1 is the initial volume
V2 is the final volume
Since we are finding the new volume
[tex]V_2 = \frac{P_1V_1}{P_2} \\[/tex]
We have
[tex]V_2 = \frac{300 \times 2}{7} = \frac{600}{7} \\ = 85.714285...[/tex]
We have the final answer as
85.71 mLHope this helps you
The molar mass of gallium (Ga) is 69.72 g/mol.
Calculate the number of atoms in a 27.2 mg sample of Ga.
Write your answer in scientific notation using three significant figures.
atoms Ga
Answer:
2.35 x 10²⁰ atoms Ga
Explanation:
After converting from mg to g, use the molar mass as the unit converter to convert to moles. Then using Avogadro's number, 6.022 x 10²³ convert from moles to atoms of Ga.
[tex]27.2mgGa*\frac{1g}{1000mg} *\frac{1 mol Ga}{69.72gGa} *\frac{6.022*10^2^3 atoms Ga}{1 molGa} = 2.349 * 10^2^0 atoms Ga[/tex]
Then round to 3 significant figures = 2.35 x 10²⁰ atoms Ga.
The number of atoms in 27.2 mg sample of Ga is 2.35 × 10²⁰ atoms
StoichiometryFrom the question, we are to calculate the number of atoms in a 27.2 mg sample of Ga.
First, we will determine the number of moles of Ga present
Using the formula,
[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass} [/tex]
Mass = 27.2 mg = 0.0272 g
Molar mass = 69.72 g/mol
Then,
[tex]Number\ of\ moles \ of\ Ga = \frac{0.0272}{69.72} [/tex]
[tex]Number\ of\ moles \ of\ Ga = [/tex] 0.000390132 moles
Now, for the number of atoms present
From the formula
Number of atoms = Number of moles × Avogadro's constant
Then,
Number of Ga atoms = 0.000390132 × 6.022×10²³
Number of Ga atoms = 2.35 × 10²⁰ atoms
Hence, the number of atoms in 27.2 mg sample of Ga is 2.35 × 10²⁰ atoms
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Which is the best molecule to build collegen, muscle repair, and tendon repair?
a
carohydrates
b
lipids
c
nucleic acids
d
Proteins
Answer:
d
Explanation:
collagen is made of proteins
Which of the following best describes the structure of a nucleic acid?
a
Carbon ring(s)
b
Globular or fibrous
c
Single or double helix
d
Hydrocarbon(s)
A student ran the following reaction in the laboratory at 751 K: N2(g) + 3H2(g) 2NH3(g) When she introduced 3.47×10-2 moles of N2(g) and 6.38×10-2 moles of H2(g) into a 1.00 liter container, she found the equilibrium concentration of H2(g) to be 6.25×10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction.
Answer:
Kc = 4.86×10⁻⁶
Explanation:
We begin from the equation:
N₂ + 2H₂ ⇄ 2NH₃
We start from 3.47×10⁻² moles of N₂(g) and 6.38×10⁻² moles of H₂(g), so when we reach the equilibrium, we get 6.25×10⁻² moles of H₂.
This data means, that in the reaction we made react:
6.38×10⁻² - x = 6.25×10⁻²
x = 1.3×10⁻³ moles of H₂
As stoichiometry is 1:3, we will know that the moles of N₂ that have been reacted were:
1.3×10⁻³ moles / 3 = 4.33×10⁻⁴ moles of N₂
So, in the equilibrium we would have:
3.47×10⁻² moles of N₂ - 4.33×10⁻⁴ moles of N₂ = 0.0343 moles of N₂
How many ammonia, would we have in the equilibrium?
4.33×10⁻⁴ mol . 2 = 8.66×10⁻⁴ moles (from stoichiometry with N₂, 1:2)
(1.3×10⁻³ mol . 2) / 3 = 8.66×10⁻⁴ moles (from stoichiometry with H₂, 2:3)
Let's make the expression for Kc
Kc = [NH₃]³ / [N₂] . [H₂]²
(8.66×10⁻⁴ )³ / (0.0343 . (6.25×10⁻²)² = 4.86×10⁻⁶
How many moles of H2 are needed to produce 24 moles of NH3?
Answer:
36 mol of H2
Explanation:
The balanced equation of the reaction is given as;
3H2 + N2 --> 2NH3
From the reaction;
It takes 3 mol of H2 reacting with 1 mol of N2 to form 2 mol of NH3
3 mol of H2 = 2 mol of NH3
x mol of H2 = 24 mol of NH3
x = (24 * 3) / 2 = 36 mol of H2