Identify which properties are common to each of the following chemical families
(a) alkali metals
(b) alkaline earth metals
(c) halogens
(d) noble gases
Answers
The noble gases have a full outer shell of valence electrons, making them stable and unreactive. They are colorless, odorless gases at room temperature and have very low boiling points. Their lack of reactivity makes them useful in a variety of applications, including lighting and welding.
The properties that are common to each of the following chemical families include:
(a) Alkali metals The alkali metals have a single valence electron in their outermost shell, which is easily lost to form an ion with a charge of +1. They are the most reactive metals, reacting with water and air to produce hydrogen gas and an oxide layer, respectively. They are silvery-white and have a soft texture.
(b) Alkaline earth metals The alkaline earth metals have two valence electrons in their outermost shell, which they readily lose to form ions with a charge of +2. They are less reactive than the alkali metals, but they still react with oxygen to form an oxide layer on their surface. They are also silvery-white in color and have a harder texture than the alkali metals.
(c) Halogens The halogens have seven valence electrons in their outermost shell, making them highly reactive nonmetals. They readily form ions with a charge of -1 by gaining an electron. They are diatomic molecules at room temperature and can be found in a variety of colors and states of matter.
(d) Noble gases The noble gases have a full outer shell of valence electrons, making them stable and unreactive. They are colorless, odorless gases at room temperature and have very low boiling points. Their lack of reactivity makes them useful in a variety of applications, including lighting and welding. These properties are common to each of the following chemical families.
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Related Questions
which of the pairs of amino acids can form a salt bridge at physiological ph?
Answers
The pair of amino acids that can form a salt bridge at physiological pH is aspartic acid and histidine. Hence the correct option is (A).
A salt bridge is formed when the carboxylic acid group (-COOH) of one amino acid donates a proton (H+) to the amino group (-NH2) of another amino acid, which becomes positively charged. This creates an electrostatic attraction between the positively charged amino acid and the negatively charged carboxylate ion of the other amino acid. At physiological pH (around 7.4), the carboxylic acid group of aspartic acid (Asp) is deprotonated and has a negative charge, while the amino group of histidine (His) is protonated and has a positive charge. Therefore, Asp and His can form a salt bridge through electrostatic attraction between the negatively charged carboxylate group of Asp and the positively charged amino group of His. The other pairs of amino acids listed do not form salt bridges at physiological pH for the following reasons:
(B) Glutamic acid (Glu) and aspartic acid (Asp) both have negatively charged carboxylate groups at physiological pH, so they repel each other and do not form salt bridges.
(C) Two cysteines (Cys) can form a disulfide bond through the oxidation of their sulfur-containing side chains, but they do not form salt bridges.
(D) Arginine (Arg) has a positively charged guanidino group at physiological pH, while tyrosine (Tyr) has a neutral hydroxyl group. They do not have complementary charges to form a salt bridge.
(E) Lysine (Lys) has a positively charged amino group, while glutamic acid (Glu) has a negatively charged carboxylate group. They have complementary charges to form a salt bridge, but the pH is too low for Glu to be deprotonated and have a negative charge. Therefore, they do not form a salt bridge at physiological pH.
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Question - Which of the pairs of amino acids can form a salt bridge at physiological pH? Select from the following options:-
(A) aspartic acid and histidine
(B) glutamic acid and aspartic acid
(C) two cysteines
(D) arginine and tyrosine
(E) lysine and glutamic acid
3ch4o number of atoms
Answers
Which legislation covers various industrial safety issues? Factories & Industrial Undertakings Ordinance (Chapter 59) Factories & Industrial Undertakings Ordinance (Chapter 57) Factories & Industrial Undertakings Ordinance (Chapter 58) Factories & Industrial Undertakings Ordinance (Chapter 56)
Answers
The Factories and Industrial Undertakings Ordinance (Chapter 59) is the legislation that covers various industrial safety issues.
The Factories and Industrial Undertakings Ordinance is a piece of Hong Kong legislation. The Ordinance addresses a broad range of matters relating to the safety, health, and welfare of individuals employed in factories and other industrial undertakings. The ordinance was enacted in 1950.
Chapter 59 of the Factories and Industrial Undertakings Ordinance covers a range of topics related to industrial safety. It includes regulations for factories, safety management systems, mining installations, quarries, asbestos factories, and plants, noise in the workplace, and gas cylinders. These regulations aim to ensure the safety and health of workers in various industries by setting standards for machinery safety, ventilation, electrical safety, hazardous substance handling, noise control, and more. The ordinance provides guidelines for employers to create a safe working environment and imposes legal obligations to comply with these regulations. It plays a crucial role in preventing accidents, promoting worker well-being, and maintaining industrial safety standards.
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_Fe(CN)2+ _HI-> ___Fel2 + _HCN
Answers
Please help me with this, there is 2 screenshots
Answers
Answer: Celestial Bodies!
Explanation: Celestial Bodies are objects like stars, planets, ext.
If 2. 55 L of propane (C3H8) at 30 degree Celsius and 67. 2 kPa is completely burned in excess oxygen, what mass of carbon dioxide will be produced? R=0. 0821 L^ * atm/mol^ * K Write a balanced chemical equation: R=8. 314 L^ * kPa/mol^ * K
Answers
295 g of carbon dioxide will be produced.
The balanced chemical equation for the complete combustion of propane is:
\(C3H8 + 5O2 → 3CO2 + 4H2O\)
From the equation, we can see that 1 mole of propane produces 3 moles of carbon dioxide. We can use the ideal gas law to determine the number of moles of propane present in 2.55 L at 30°C and 67.2 kPa:
PV = nRT
n = PV/RT
n = (67.2 kPa)(2.55 L)/(0.0821 L·atm/mol·K)(303 K)
n = 2.24 mol
Therefore, the amount of carbon dioxide produced will be:
3 mol \(CO2\)/mol \(C3H8\) × 2.24 mol \(C3H8\) = 6.72 mol \(CO2\)
Finally, we can use the molar mass of carbon dioxide to convert moles to mass:
6.72 mol \(CO2\) × 44.01 g/mol \(CO2\) = 295 g \(CO2\)
Therefore, 295 g of carbon dioxide will be produced.
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In an experiment, 8.50 g of methane, CH4, was reacted with 15.9 g of oxygen gas to produce carbon dioxide and water. Determine the percentage yield if 9.77 g of carbon dioxide was obtained in the lab.
Answers
Answer:
89.3 %
Explanation:
M(CH4) = 12+ 4*1 = 16 g/mol
M(O2) = 2*16 = 32 g/mol
M(CO2) = 12 + 2*16 = 44 g/mol
8.50 g * 1 mol/16 g = 0.5313 mol CH4
15.9 g * 1 mol/32 g = 0.4969 mol O2
9.77 g * 1 mol/44 g = 0.2220 mol CO2
1) CH4 + 2O2 -----> CO2 + 2H2O
from reaction 1 mol 2 mol
given 0.5313 mol (0.4969 mol)
1 mol CH4 --- 2 mol O2
0.5313 mol CH4 --- x mol O2
x= 2*0.5313 = 1.0626 mol O2
We can see that for given amount of CH4 we do not have enough O2, so O2 is a limiting reactant.
2) CH4 + 2O2 -----> CO2 + 2H2O
from reaction 2 mol 1 mol
given 0.4969 mol x mol
x = 0.4969*1/2 = 0.2485 mol CO2 theoretical yield
3)
Practical yield CO2 = 0.2220 mol
Theoretical yield CO2 = 0.2485 mol
% yield = (0.2220/0.2485)*100% = 89.3 %
A geochemist examines a piece of metal that he found in the soil. He performs tests to identify the metal from its density, electrical conductivity, and melting point. Which statement best describes his investigation?
Answers
Answer:
what r the statements?
Explanation:
attach or write them
Answer:
A. He is determining physical properties that are sufficient to identify the metal.
Explanation:
Correct on Edge 2022!!!
Good luck everyone, you got this! Have a great day!
14 Copper(II) oxide reacts with hydrogen.
CuO + H2 → Cu + H2O
Which row is correct?
oxidising agent
reducing agent
A
H2
Cuo
B
Cuo
H2
с
H2O
Cu
D
Cu
H20
Answers
Answer:
B
CuO
H2
Explanation:
hydrogen will oxidized to form water
Copper(II) oxide reacts with hydrogen.
\(CuO + H_2\) → \(Cu + H_2O\)
Oxidising agent - \(Cuo\)
Reducing agent - \(H_2\)
What is a redox reaction?A chemical reaction that takes place between an oxidizing substance and a reducing substance.
The oxidizing substance loses electrons in the reaction, and the reducing substance gains electrons.
In the reaction \(CuO + H_2\) → \(Cu + H_2O\)
\(Cuo\) is the oxidising agent and \(H_2\) is the reducing agent because \(Cuo\) is oxidising \(H_2\) and \(H_2\) is the reducing agent because it is removing oxygen from \(Cuo\).
Hence, option B is correct.
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If the spheres represent an atom
and an anion of the same element, which sphere represents the atom?
Select one:
Answers
Answer:
a is the answer I know and if I'm right mark me brainlest
Perform the following
mathematical operation, and
report the answer to the
correct number of significant
figures.
3.96 * 0.1159 = [?]
Answers
The answer to the problem 3.96 * 0.1159 is : 0.459 (3 significant figures)
Meaning of Significant figures and the rule of engagementSignificant figures can be defined as numbers in a group of numbers that place number in the right value or worth.
Significant figures defined the value of measurement of a particular number.
The rules of engagement in this case is the set of rules or laid down principles that direct how we interact with this significant figures and how they are used.
For this case the significant figures of the product will be the same as the figure with the least significant figure in the multiplication.
In conclusion, The answer to the problem 3.96 * 0.1159 is : 0.459 (3 significant figures)
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How does the smell of food being cooked spread so fast ???
Answers
Answer:
When we cook food in the kitchen, that's the region of higher concentration of the smell. By diffusion, the smell spreads to the whole room and thereby whole house, so anyone standing at a distance, can smell it.
Answer:
smell of cooked food spread fast bc when something is heated up, the kinetic energy of the molecules/particles increases and some of them even mix with the air and moves around. the first answer is also correct (i just explained it more detaily)
so, this is the reason :)
mark as brainliest dagini ;)
which aromatic ring of phenyl benzoate would you expect to react faster for a nitration reaction? justify your answer
Answers
In phenyl benzoate, there are two aromatic rings: the phenyl ring and the benzoyl ring. The phenyl ring is expected to react faster for a nitration reaction than the benzoyl ring. This is because the phenyl ring is more electron-rich than the benzoyl ring, due to presence of the electron-donating substituent (the methoxy group) on the phenyl ring.
The reactivity of each ring towards nitration will depend on a number of factors, such as the electronic properties and steric hindrance of the substituents on each ring. In contrast, the benzoyl ring is relatively electron-poor, due to the presence of the carbonyl group, which withdraws electrons from the ring. This makes it less reactive towards electrophilic substitution reactions, such as nitration.
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URGENT! What occurs during a chemical reaction. :)
A.) Atoms of two or more elements are destroyed.
B.) Atoms of two or more elements oppose one another.
C.) Atoms of two or more elements bond together.
D.) Atoms of two or more elements trade protons.
Answers
Which phase do the chromosomes line up in the middle
Answers
Answer:
metaphase
Explanation:
the cells chromosomes line themselves in the middle of the cell through a type of cellular tug of war.
Answer:
metaphase
Explanation:
During metaphase, the cell's chromosomes align themselves in the middle of the cell through a type of cellular "tug of war." The chromosomes, which have been replicated and remain joined at a central point called the centromere, are called sister chromatids.
Of the compounds listed, which would be the most reactive toward alkylation?
A. NO2
B. OCH3
C. NH2
D. CH3
Answers
Among the given compounds, NH2 would be the most reactive toward alkylation. Option C is correct.
The addition of an alkyl group to a molecule is referred to as alkylation. The donor molecule is usually an alkyl halide, and the acceptor molecule can be an alkene, an amine, or a phenol in many situations.
Alkylation is widely utilized in the chemical industry to produce specific chemicals. The given compounds are NO2, OCH3, NH2, and CH3. Let's examine each of the options:
Option A: NO2Nitro groups (-NO2) are electron-withdrawing, and as a result, they do not participate in alkylation. As a result, NO2 is not the most reactive option towards alkylation. Therefore, the answer cannot be option A.
Option B: OCH3Methoxy (-OCH3) groups, like nitro groups, are electron-withdrawing, which makes them less reactive to alkylation. As a result, OCH3 cannot be the most reactive option towards alkylation. Therefore, the answer cannot be option B.
Option C: NH2Amines (-NH2) are electron donating, and therefore they are the most reactive towards alkylation. As a result, NH2 is the most reactive option towards alkylation. Therefore, the answer is option C.
Option D: CH3Methyl (-CH3) groups, like amines, are electron donating, but not as powerful as amines. As a result, CH3 can be a reactive option towards alkylation, but not the most reactive option. Therefore, the answer cannot be option D.
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if i have 3m h2 and 3m i2, and i let it reach equilibrium and measure 6m hi, how much h2 is there in molarity?
Answers
Let us assume the balanced chemical equation of H2 and I2 forming HI is as follows: H2(g) + I2(g) 2HI(g) in the chemical reaction at equilibrium is said to be constant.
Hence, we can write the expression for the equilibrium constant (Kc) of the above chemical reaction as follows:
Kc = [HI]2 / [H2][I2] We know the concentration of HI is 6M, and we have to find out the molarity of H2 present. Now that we know the concentration of I2, we can use the expression above to solve for the molarity of H2. However, we don't have a concentration of I2.
Therefore, we must use the balanced chemical reaction to find the molarity of H2 as follows:
3H2(g) + 3I2(g) → 6HI(g) 2H2(g) + I2(g) → 2HI(g)
Since 2 mol of HI are formed from 1 mol of I2, if we use up all the I2, we will produce 4 mol of HI using 2 mol of H2. Therefore, we have:
2H2(g) + I2(g) → 4HI(g)At equilibrium, we have a total of 6 M of HI and 0.5 M of I2. Since I2 is a limiting reagent, we must use 0.5 M to solve for the molarity of H2 as follows:
2H2(g) + I2(g) → 4HI(g) 2x 0.5 4x x = 0.5 / 2 = 0.25 M
Therefore, the molarity of H2 is 0.25 M.
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1. The equilibrium constant, Kc, for the following reaction is 1.89×10-6 at 506 K.
NH4Cl(s) NH3(g) + HCl(g)
Calculate Kc at this temperature for the following reaction:
NH3(g) + HCl(g) NH4Cl(s)
Kc =
Answers
If the equilibrium constant, Kc, for the following reaction is 1.89 × 10⁻⁶at 506 K, Kc for the reaction: NH3(g) + HCl(g) → NH4Cl(s)is 5.291 × 10⁵.
The given reaction is: NH4Cl(s) → NH3(g) + HCl(g)
To calculate Kc for the reaction: NH3(g) + HCl(g) → NH4Cl(s) we must reverse the given reaction. We must use the relationship: If aA + bB ⇌ cC + dD then Kc = [C]c[D]d/[A]a[B]b where [A], [B], [C], [D] are molar concentrations at equilibrium, while a, b, c, d are the coefficients of the balanced chemical equation. To reverse the given reaction, we must invert the value of the Kc. The equation is: NH4Cl(s) ⇌ NH3(g) + HCl(g)Kc = 1.89 × 10⁻⁶at 506 K
Now, we will invert this equation. Kc(NH3)(HCl) = 1/Kc(NH4Cl)
Now, substituting the values: Kc(NH3)(HCl) = 1/(1.89 × 10⁻⁶)Kc(NH3)(HCl) = 5.291 × 10⁵
Therefore, Kc for the reaction:NH3(g) + HCl(g) → NH4Cl(s)is 5.291 × 10⁵.
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How many base units are there in the code for one amino acid?
Answers
There are three base units in the code for one amino acid.
What are the 3 bases that code for an amino acid?
Codons are made up of any triplet combination of the four nitrogenous bases adenine (A), guanine (G), cytosine (C), or uracil (U). Of the 64 possible codon sequences, 61 specify the 20 amino acids that makeup proteins and three are stop signals.
What is the 3 base code?
A three-base sequence in DNA or RNA is known as a codon. The codes in the coding strand of DNA and in messenger RNA aren't, of course, identical, because in RNA the base uracil (U) is used instead of thymine (T).
Why do codons have 3 bases?
The more bases there are per codon the more information you can code for. There are only 22 different amino acids, in consequence, we need a minimum of 3 bases per codon.
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What happens to a liquid when you keep on cooling it until it changes?
Answers
Answer:
it turns into a solid
Would the h2co3/hco3- system work effectively as a buffer under the conditions found in the patient 10 hours after aspirin ingestion? explain
Answers
In the conditions found in a patient 10 hours after aspirin ingestion, where acidosis may be present, the H2CO3/HCO3- system may not work effectively as a buffer.
The effectiveness of the H2CO3/HCO3- system as a buffer depends on the pKa of the weak acid and the pH of the solution. In the case of a patient 10 hours after aspirin ingestion, the conditions may not be suitable for the H2CO3/HCO3- system to work effectively as a buffer.
Aspirin, or acetylsalicylic acid, is metabolized in the body to salicylic acid, which can cause acidosis and decrease the pH of the blood. This decrease in pH can shift the equilibrium of the H2CO3/HCO3- system, affecting its buffering capacity.
Explanation:
The H2CO3/HCO3- system acts as a buffer in the blood by maintaining a stable pH. Carbonic acid (H2CO3) is a weak acid that can dissociate into bicarbonate ions (HCO3-) and hydrogen ions (H+). The bicarbonate ions can act as a reservoir for H+ ions, preventing large changes in pH. The effectiveness of this buffer system is determined by the pKa of carbonic acid, which is around 6.1.
However, when aspirin is ingested, it undergoes metabolism in the body to salicylic acid. Salicylic acid can cause an increase in H+ ions and lead to acidosis, a condition characterized by a decrease in blood pH. Acidosis shifts the equilibrium of the H2CO3/HCO3- system towards the formation of more carbonic acid (H2CO3) and decreases the concentration of bicarbonate ions (HCO3-). As a result, the buffering capacity of the H2CO3/HCO3- system may be compromised, and it may not effectively maintain the pH within the desired range.
Therefore, in the conditions found in a patient 10 hours after aspirin ingestion, where acidosis may be present, the H2CO3/HCO3- system may not work effectively as a buffer. Additional medical intervention may be required to restore and maintain the appropriate pH balance in the body.
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the henry's law constant for h2 is 8.1×10−4 matm at 25∘c. what pressure of hydrogen is needed to maintain a h2 concentration of 0.42 m?\
Answers
518.5 atm pressure of hydrogen is needed to maintain a \(H_2\) concentration of 0.42 m
The given Henry's law constant for \(H_2\) is 8.1 × 10^-4 M atm^-1 at 25°C. To find the pressure of hydrogen needed to maintain a \(H_2\) concentration of 0.42 M, we can use Henry's law.
The equation for Henry's law is:
C = kH*P
where C is the concentration of gas in moles per liter, P is the partial pressure of the gas in atmospheres, and kH is Henry's law constant in M/atm.
Plugging the values in Henry's law equation, we get:
0.42 = (8.1 × 10^-4)P
Dividing both sides by (8.1 × 10^-4), we get:
P = (0.42)/(8.1 × 10^-4)
P = 518.5 atm
Hence, the pressure of hydrogen needed to maintain a \(H_2\) concentration of 0.42 M is 518.5 atm.
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Help me balance this equation please.
NaCl + H2O —→ Cl2+ NaOH + H2
Answers
Answer:
2NaCl + 2H2O --> Cl2 + 2NaOH + H2
Explanation:
1. Determine the molecular formula of an oxide of iron in which the mass of iron and oxygen are 69.9% and 30% respectively given that the molar mass of the oxide 159.898/mol, find the empirical and molecular formula.
2. a crystalline salt when heated becomes anhydrous and loses 51.2% of its weight the anhydrous salt analysis gave the percent composition as magnesium is equal to 20.0% and sulphur is equal to 26.66% and oxygen is equal to 53.33%.
3. In three moles of Ethane calculate the following
1. calculate number of carbon atoms.
2. number of moles of hydrogen atoms
3. number of molecules of Ethane.
Answers
1a. The empirical formula of the compound is Fe₂O₃
1b. The molecular formula of the compound is Fe₂O₃
2a. The molecular formula of the anhydrous salt is MgSO₄
2b. The formula of the crystalline salt is MgSO₄.7H₂O
3i. The number of mole of carbon atoms in the compound is 6 moles
3ii. The number of mole of hydrogen atoms in the compound is 18 moles
3iii. The number of molecules in 3 moles of ethane is 1.806×10²⁴ molecules
1a. How to determine the empirical formulaFe = 69.9%O = 30%Empirical formula =?Divide by their molar mass
Fe = 69.9 / 56 = 1.248
O = 30 / 16 = 1.875
Divide by the smallest
Fe = 1.248 / 1.248 = 1
O = 1.875 / 1.248 = 3/2
Multiply by 2 to express in whole number
Fe = 1 × 2 = 2
O = 3/2 × 2 = 3
Thus, the empirical formula of the compound is Fe₂O₃
1b. How to determine the molecular formulaEmpirical formula = Fe₂O₃Molar mass of compound = 159.89 g/molMolecular formula = ?Molecular formula = empirical × n = molar mass
[Fe₂O₃]n = 159.89
[(56×2) + (16×3)]n = 159.89
160n = 159.89
n = 159.89 / 160
n = 1
Molecular formula = [Fe₂O₃]n
Molecular formula = [Fe₂O₃] × 1
Molecular formula = Fe₂O₃
2a. How to determine the molecual formula of the anhydrous saltWe'll begin by calculating the empirical formula
Mg = 20.0% S = 26.66% O = 53.33%Empirical formula =?Divide by their molar mass
Mg = 20.0 / 24 = 0.83
S = 26.66 / 32 = 0.83
O = 53.33 / 16 = 3.33
Divide by the smallest
Mg = 0.83 / 0.83 = 1
S = 0.83 / 0.83 = 1
O = 3.33 / 0.83 = 4
Thus, the empirical formula of the anhydrous salt is MgSO₄
The molecular formula of the anhydrous salt can be obtained as follow:
Empirical formula = MgSO₄Molar mass of compound = 120 g/molMolecular formula = ?Molecular formula = empirical × n = molar mass
[MgSO₄]n = 120
[24 + 32 + (16×4)]n = 159.89
120n = 120
n = 120 / 120
n = 1
Molecular formula = [MgSO₄]n
Molecular formula = [MgSO₄] × 1
Molecular formula = MgSO₄
2b. How to determine the formula of the crystalline saltWater (H₂O) = 51.2%Anhydrous salt (MgSO₄) = 100 - 51.2 = 48.8%Formula of crystalline salt =?Divide by their molar mass
MgSO₄ = 48.8 / 120 = 0.4
H₂O = 51.2 / 18 = 2.8
Divide by the smallest
MgSO₄ = 0.4 / 0.4 = 1
H₂O = 2.8 / 0.4 = 7
Thus, the formula of the crystalline salt is MgSO₄.7H₂O
3i. How to determine the mole of carbon atoms in 3 moles of C₂H₆1 mole of C₂H₆ contains 2 moles of carbon atoms.
Therefore,
3 moles of C₂H₆ will contain = 3 × 2 = 6 moles of carbon atoms
3ii. How to determine the mole of hydrogen atoms in 3 moles of C₂H₆1 mole of C₂H₆ contains 6 moles of hydrogen atoms.
Therefore,
3 moles of C₂H₆ will contain = 3 × 6 = 18 moles of hydrogen atoms
3iii. How to determine the number of moleculesFrom Avogadro's hypothesis,
1 mole of ethane = 6.02×10²³ molecules
Therefore,
3 moles of ethane = 3 × 6.02×10²³ molecules
3 moles of ethane = 1.806×10²⁴ molecules
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Complete question
2. A crystalline salt when heated becomes anhydrous and loses 51.2% of its weight the anhydrous salt analysis gave the percent composition as magnesium is equal to 20.0% and sulphur is equal to 26.66% and oxygen is equal to 53.33%. Ccalculate the molecular formula of the anhydrous and the crystalline salt. The molecular weight of the anhydrous salt is 120
The value of ΔfH⊖ for NH3 is -91.8kJ mol−1. Calculate enthalpy change for the following reaction 2NH3(g)→N2(g)+3H2(g).
Answers
Calculated by reaction using the formula,ΔH = ∑(ΔfH(products)) - ∑(ΔfH(reactants))Here,ΔfH⊖ for NH3 = -91.8 kJ/mol.
The balanced chemical equation for the given reaction is 2 NH3(g) → N2(g) + 3 H2(g)So, the enthalpy change for the given reaction is,ΔH = ∑(ΔfH(products)) - ∑(ΔfH(reactants))ΔH = [ΔfH⊖ (N2) + 3ΔfH⊖ (H2)] - [2ΔfH⊖ (NH3)]Substituting the respective values,ΔH = [(0 + 3 × 0) kJ/mol] - [2 × (-91.8 kJ/mol)]ΔH = 183.6 kJ/mol.
Enthalpy change can be calculated by using the formula,ΔH = ∑(ΔfH(products)) - ∑(ΔfH(reactants))Where,ΔH = enthalpy change for the reactionΔfH⊖ = standard enthalpy of formationThe balanced chemical equation for the given reaction is 2 NH3(g) → N2(g) + 3 H2(g)So, the enthalpy change for the given reaction is,ΔH = ∑(ΔfH(products)) - ∑(ΔfH(reactants))ΔH = [ΔfH⊖ (N2) + 3ΔfH⊖ (H2)] - [2ΔfH⊖ (NH3)]Substituting the respective values,ΔH = [(0 + 3 × 0) kJ/mol] - [2 × (-91.8 kJ/mol)]ΔH = 183.6 kJ/mol.
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A small amount of liquid bromine is added to a container which is then sealed.
Br2 (l) " Br2 (g)
Answers
Answer:
This answer is....
Br2 (l)
What do red foxes need to live
Answers
Answer:
food, water ,shelter
Explanation:
hope this helps you out!!!!!
Answer:
species of mammals to feed on
Explanation:
they need this to live
do compounds that can hydrogen bond to the silica gel generally have a lower or higher rf value than compounds that cannot?
Answers
Yes, compounds that can hydrogen bond to the silica gel have a lower Rf value than compounds that cannot hydrogen bond to the silica gel.
The Rf value (retention factor) is the distance traveled by the compound from the starting line to the position of the compound's center divided by the distance traveled by the solvent from the starting line to the position of the solvent's center.
In paper chromatography, the Rf value aids in the identification of the components of a mixture. In paper chromatography, the Rf value ranges from 0 to 1.
The Rf value of a compound is affected by the solvent, temperature, pressure, and the nature of the stationary and mobile phases.
When a compound can hydrogen bond with the stationary phase, it will stay on the silica gel, and its Rf value will be lower than the Rf value of the compound that cannot hydrogen bond with the stationary phase, and hence the answer is compounds that can hydrogen bond to the silica gel have a lower Rf value than compounds that cannot.
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How many chiral carbon atoms are present in a molecule of 2,3,4-trichloropentane?
Answers
Answer:
The answer is 2.
Explanation:
compare and contrast a Covalent and an ionic bond