Answer:
glue that would make the strongest craft stick tower.
Explanation:
Independent variable: In statistics and research methods, the term "independent variable" is determined as a variable that is being changed, controlled, or altered in an experiment or research by the researcher or the experimenter to see its effect on DV or dependent variable. However, it is said that independent variable directly effect the dependent variable.
Which of the following best describes a limiting reactant? a. The reactant that limits the rate (or speed) of a chemical reaction. b. The reactant that limits the position of equilibrium in a reversible chemical change. c. The reactant that remains at the end of the reaction. d. The reactant that can produce the greatest amount of product. e. The reactant that is completely used up by a reaction.
Answer:
e is the suitable answer for that. I think it is correct.
Answer: The correct option is E ( the reactant that is completely used up by a reaction).
Explanation:
A LIMITING REACTANT can be defined as the reagent or the substance that is involved in a chemical reaction which determines when the reaction will stop. This is because it is COMPLETELY used up in the reaction. Reactants that are called limiting reactants is because the quantity of these reagents are capable of limiting the amount of products formed. And by doing so, the chemical reaction cannot proceed further with the absence of this reactant. Using an attached diagram below to illustrate further:
The reagents D and E reacts to form F as the product. In this reaction, reactant E is the limiting reagent because there is still some left over D in the products. Therefore, D was in excess when E was all USED UP.
Therefore the CORRECT option is E which states that the reactant that is completely used up by a reaction, best describes a limiting reactant.
Option A is WRONG because it's the concentration of both reactants in chemical equation can limit the speed of that reaction.
Option B is WRONG because it's when the concentration of a particular reactant is either increased or decreased can affect the position of equilibrium.
Option C and D are wrong because the reactant that remains in the end of a reaction and can produce the greatest amount of product is the one in EXCESS.
Which of the terms heat of vaporization and heat of fusion refers to condensation and which refers to melting?
Answer:
Heat of vaporization refers to condensation and heat of fusion refers to melting.
Explanation:
Heat of vaporization or heat of evaporation, is defined as the energy required to convert liquid substance into a gas which creates condensation. As a gas condenses to a liquid, heat is released.
Heat of fusion refers to melting because heat of fusion is defined as the energy required to change any amount of substance when it melts.
Hence, the correct answer is "Heat of vaporization refers to condensation and heat of fusion refers to melting.".
fishes can live inside a frozen pond why
Answer:
Underneath the frozen upper layer, the water remains in its liquid form and does not freeze. Also, oxygen is trapped beneath the layer of ice. As a result, fish and other aquatic animals find it possible to live comfortably in the frozen lakes and ponds. ... This irregular expansion of water is called anomalous expansion.
Explanation:
so pretty much its because there is water under the top frozen layer and air is trapped underneath the ice
Hydrazine, , emits a large quantity of energy when it reacts with oxygen, which has led to hydrazine used as a fuel for rockets: How many moles of each of the gaseous products are produced when 20.1 g of pure hydrazine is ignited in the presence of 20.1 g of pure oxygen
Answer:
[tex]1.25~mol~H_2O[/tex] and [tex]0.627~mol~N_2[/tex]
Explanation:
Our goal for this question is the calculation of the number of moles of the molecules produced by the reaction of hydrazine ([tex]N_2H_4[/tex]) and oxygen ([tex]O_2[/tex]). So, we can start with the reaction between these compounds:
[tex]N_2H_4~+~O_2~->~N_2~+~H_2O[/tex]
Now we can balance the reaction:
[tex]N_2H_4~+~O_2~->~N_2~+~2H_2O[/tex]
In the problem, we have the values for both reagents. Therefore we have to calculate the limiting reagent. Our first step, is to calculate the moles of each compound using the molar masses values (32.04 g/mol for [tex]N_2H_4[/tex] and 31.99 g/mol for [tex]O_2[/tex]):
[tex]20.1~g~N_2H_4\frac{1~mol~N_2H_4}{32.04~g~N_2H_4}=0.627~mol~N_2H_4[/tex]
[tex]20.1~g~O_2\frac{1~mol~O_2}{31.99~g~O_2}=0.628~mol~O_2[/tex]
In the balanced reaction we have 1 mol for each reagent (the numbers in front of [tex]O_2[/tex] and [tex]N_2H_4[/tex] are 1). Therefore the smallest value would be the limiting reagent, in this case, the limiting reagent is [tex]N_2H_4[/tex].
With this in mind, we can calculate the number of moles for each product. In the case of [tex]N_2[/tex] we have a 1:1 molar ratio (1 mol of [tex]N_2[/tex] is produced by 1 mol of [tex]N_2H_4[/tex]), so:
[tex]0.627~mol~N_2H_4\frac{1~mol~N_2}{1~mol~N_2H_4}=~0.627~mol~N_2[/tex]
We can follow the same logic for the other compound. In the case of [tex]H_2O[/tex] we have a 1:2 molar ratio (2 mol of [tex]H_2O[/tex] is produced by 1 mol of [tex]N_2H_4[/tex]), so:
[tex]0.627~mol~N_2H_4\frac{2~mol~H_2O}{1~mol~N_2H_4}=~1.25~mol~H_2O[/tex]
I hope it helps!
What did J. J. Thomson's cathode ray experiment show about atoms?
Answer:
atoms contain tiny negatively charged subatomic particles or electrons
A variation of the acetamidomalonate synthesis can be used to synthesize threonine. The process involves the following steps: Ethoxide ion deprotonates diethyl acetamidomalonate, forming enolate anion 1; Enolate anion 1 makes a nucleophilic attack on acetaldehyde, forming tetrahedral intermediate 2; Protonation of the oxyanion forms alcohol 3; Acid hydrolysis yields dicarboxyamino alcohol 4; Decarboxylation leads to the final amino acid. Write out the mechanism on a separate sheet of paper, and then draw the structure of tetrahedral intermediate 2.
Answer:
See figure 1
Explanation:
For this reaction, we have the production of a carbanion as the first step. The base "ethoxide" can remove a hydrogen-producing a negative charge in the carbon (enolate anion 1). Then this negative charge can attack the carbon of the carbonyl group in the molecule acetaldehyde and the tetrahedral intermediate 2 is form. In the next step, we have the protonation of the oxygen to produce alcohol 3. A continuation we have the hydrolysis of the ester groups to produce the Dicarboxyamino alcohol and finally, we have a decarboxylation reaction we will produce the amino acid Threonine.
To further explanations see figure 1
I hope it helps!
The ph of the test tube can be calculated by knowing the concentration of hydroxide ions in solution. the ph = 14 + log 10 oh- for example the 0.1m stock of naoh has a ph = 14+ log 10 0.1= 13. using the dilution 5 ml 0.1 naoh 5ml h20 what is the ph of tube 1.
Answer:
Volume of the calcium hydroxide solution used is 0.0235 mL.
Explanation:
Moles of KHP =
According to reaction, 2 moles of KHP with 1 mole of calcium hydroxide , then 0.0330 moles of KHP will recat with ;
of calcium hydroxide
Molarity of the calcium hydroxide solution = 0.703 M
Volume of calcium hydroxide solution = V
Volume of the calcium hydroxide solution used is 0.0235 mL.
Does the amount of methanol increase, decrease, or remain the same when an equilibrium mixture of reactants and products is subjected to the following changes?
a. the catalyst is removed
b. the temp is increased
c. the volume is decreased
d. helium is added
e. CO is added
Answer:
a. Methanol remains the same
b. Methanol decreases
c. Methanol increases
d. Methanol remains the same
e. Methanol increases
Explanation:
Methanol is produced by the reaction of carbon monoxide and hydrogen in the presence of a catalyst as follows; 2H2+CO→CH3OH.
a) The presence or absence of a catalyst makes no difference on the equilibrium position of the system hence the methanol remains constant.
b) The amount of methanol decreases because the equilibrium position shifts towards the left and more reactants are formed since the reaction is exothermic.
c) If the volume is decreased, there will be more methanol in the system because the equilibrium position will shift towards the right hand side.
d) Addition of helium gas has no effect on the equilibrium position since it does not participate in the reaction system.
e) if more CO is added the amount of methanol increases since the equilibrium position will shift towards the right hand side.
Three structural isomers have the formula C5H12.C5H12. Draw and name the isomers using IUPAC names. Draw the isomer with five carbon atoms in main chain.
Answer:
Explanation:
Answer in attached file .
ΔH = +572 kJ for the decomposition of water by the reaction 2 H 2O( ) → 2 H 2(g) + O 2(g). How many grams of water can be decomposed by 5.00 × 10 3 kJ of energy? ΔH = +572 kJ for the decomposition of water by the reaction 2 H 2O( ) → 2 H 2(g) + O 2(g). How many grams of water can be decomposed by 5.00 × 10 3 kJ of energy?
Answer:
315 g
Explanation:
Step 1: Write the thermochemical equation
2 H₂O(l) → 2 H₂(g) + O₂(g) ΔH = +572 kJ
Step 2: Calculate the molar of water decomposed by 5.00 × 10³ kJ of energy
According to the thermochemical equation, 572 kJ are required to decompose 2 moles of water.
5.00 × 10³ kJ × (2 mol/572 kJ) = 17.5 mol
Step 3: Calculate the mass corresponding to 17.5 moles of water
The molar mass of water is 18.02 g/mol.
17.5 mol × 18.02 g/mol = 315 g
A balloon has an initial volume of 2.954 L containing 5.50 moles of helium. More helium is added so that the balloon expands to 4.325 L. How much helium (moles) has been added if the temperature and pressure stay constant during this process.
Answer:
8.05 moles
Explanation:
5.50 / 2.954 = x / 4.325
x = 8.05
According to ideal gas equation, if the temperature and pressure stay constant during the process 0.520 moles have been added so that the balloon expands to 4.325 L.
What is ideal gas equation?The ideal gas equation is a equation which is applicable in a hypothetical state of an ideal gas.It is a combination of Boyle's law, Charle's law,Avogadro's law and Gay-Lussac's law . It is given as, PV=nRT where R= gas constant whose value is 8.314.The law has several limitations.The law was proposed by Benoit Paul Emile Clapeyron in 1834.
In the given example if pressure and temperature are constant then V=nR substituting V=4.325 l and R=8.314 so n=V/R=4.325/8.314=0.520 moles.
Thus, 0.520 moles of helium are added if the temperature and pressure stay constant during this process.
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Which are the chemical properties of water?
Answer:
See explanation.
Explanation:
Hello,
In this case, we can realize that water has a very simple atomic structure which consists of two hydrogen atoms bonded to one oxygen atom. The nature of the atomic structure of water causes its molecules to have unique electrochemical properties. The hydrogen side of the water molecule has a slight positive charge whereas at the other side of the molecule a negative charge exists. This molecular polarity causes water to be a powerful solvent and is responsible for its strong surface tension.
Moreover, water is involved in several both inorganic and organic chemical reactions leading to hydration, for example, the conversion of alkenes to alcohols, the hydrolysis of acyl halides, anhydrides, esters and amides to carboxylic acids and the hydration of a raft of inorganic salts that exist as hydrates only, such as copper (II) sulfate pentahydrate and so.
Best regards.
is the general formula of a certain hydrate. When 256.3 g of the compound is heated to drive off the water, 214.2 g of anhydrous compound is left. Further analysis shows that the percentage composition of the anhydrate is 21.90% Ca, 43.14% Se, and 34.97% O.. (Hint: Treat the anhydrous compound and water just as you have treated elements in calculating in the formula of the hydrate.) (Use an asterisk to enter the dot in the formula. If a subscript is 1, omit it.) Find the empirical formula of the anhydrous compound. Find the empirical formula of the hydrate.
Answer:
The general formula of the hydrate is Caa Seb Oc. nH2O. Based on the given information, the weight of the hydrated compound is 256.3 grams, the weight of the anhydrous compound is 214.2 grams.
Therefore, the weight of water evaporated is 256.3 g - 214.2 g = 42.1 grams
The molecular weight of water is 18 gram per mole. So, the number of moles of water will be,
Moles of water = weight of water/molecular weight
= 42.1 grams / 18 = 2.3
The given composition of calcium is 21.90 %. So, the concentration of calcium in anhydrous compound is,
= 214.2 * 0.2190 = 46.91 grams
The given composition of Se is 43.14 %. So, the concentration of selenium in anhydrous compound is,
= 214.2 * 0.4314 = 92.40 grams
The given composition of oxygen is 34.97%, So, the concentration of oxygen in anhydrous compound is,
= 214.2 * 0.3497 = 74.91 grams
The molecular weight of Ca is 40.078, the obtained concentration is 46.91 grams, stoichiometry will be, 46.91/40.078 = 1.17
The molecular weight of Se is 78.96, the obtained concentration is 92.40, stoichiometry will be,
92.40/78.96 = 1.17
The molecular weight of Oxygen is 15.999, the concentration obtained is 74.91, the stoichiometry will be,
74.91/15.999 = 4.68.
Thus, the formula becomes, Ca1.17. Se1.1e O4.68. 2.3H2O, the closest actual component is CaSeO4.2H2O
There are eight consitutional isomers with the molecular formula C4H11N.
name and draw a structural formulas for each amine.
Answer:
See figure 1
Explanation:
We have to remember that in the isomer structures we have to change the structure but we have to maintain the same formula, in this case [tex]C_4H_1_1N[/tex].
In the formula, we have 1 nitrogen atom. Therefore we will have as a main functional group the amine group.
In the amines, we have different types of amines. Depending on the number of carbons bonded to the "N" atom. In the primary amines, we have only 1 C-H. In the secondary amines, we have two C-N bonds and in the tertiary amines, we have three C-N bonds.
With this in mind, we can have:
-) Primary amines:
1) n-butyl amine
2) sec-butyl amine including 2 optical isomers
3) isobutyl amine
4) tert-butyl amine
-) Secondary amines:
5) N-methyl n-propyl amine
6) N-methyl isopropyl amine
7) N, N-diethyl amine
-) Tertiary amines:
8) N-ethyl N, N-dimethyl amine
See figure 1
I hope it helps!
A 3.140 molal solution of NaCl is prepared. How many grams of NaCl are present in a sample containing 2.692 kg of water
Answer:
494.49 g of NaCl.
Explanation:
Data obtained from the question include the following:
Molality of NaCl = 3.140 m
Mass of water = 2.692 kg
Mass of NaCl =.?
Next, we shall determine the number of mole of NaCl in the solution.
Molality is simply defined as the mole of solute per unit kilogram of solvent. Mathematically, it is expressed as
Molality = mole of solute /Kg of solvent
With the above formula, we can obtain the number of mole NaCl in the solution as follow:
Molality of NaCl = 3.140 m
Mass of water = 2.692 kg
Mole of NaCl =..?
Molality = mole of solute /Kg of solvent
3.140 = mole of NaCl /2.692
Cross multiply
Mole of NaCl = 3.140 x 2.692
Mole of NaCl = 8.45288 moles
Finally, we shall covert 8.45288 moles of NaCl to grams. This can be obtained as follow:
Mole of NaCl = 8.45288 moles
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl =.?
Mole = mass /Molar mass
8.45288 = mass of NaCl /58.5
Cross multiply
Mass of NaCl = 8.45288 × 58.5
Mass of NaCl = 494.49 g.
Therefore, 494.49 g of NaCl are present in the solution.
The mass of NaCl in 3.140 molal NaCl solution has been 494.493 grams.
Molality can be defined as the mass of solute present in 1000 grams of solvent.
Molality = [tex]\rm \dfrac{moles\;of\;solute}{mass\;of\;solvent\;(kg)}[/tex]
The moles of NaCl has been calculated as;
3.140 = [tex]\rm \dfrac{moles\;of\;NaCl}{mass\;of\;water\;(kg)}[/tex]
3.140 = [tex]\rm \dfrac{moles\;of\;NaCl}{2.692\;kg}[/tex]
Moles of NaCl = 3.140 [tex]\times[/tex] 2.692
Moles of NaCl = 8.45288 mol
The moles can be expressed as;
Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
Molecular weight of NaCl = 58.5 g/mol
The mass of NaCl can be calculated as:
8.45288 mol = [tex]\rm \dfrac{mass\;of\;NaCl}{58.5\;g/mol}[/tex]
Mass of NaCl = 58.5 g/mol [tex]\times[/tex] 8.45288 mol
Mass of NaCl = 494.493 grams.
The mass of NaCl in 3.140 molal NaCl solution has been 494.493 grams.
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A gas mixture contains 3.50 moles of helium, 5.00 moles of krypton and 7.60 moles of neon. A) What is the mole fraction for each gas
Answer:
.217, .311, and .472, respectively.
Explanation:
The total number of moles of gas is 3.50 + 5.00 + 7.60 = 16.10 (to preserve significant digits).
X of helium=3.50/16.10 = .217
X of krypton=5.00/16.10 = .311
X of neon=7.60/16.10 = .472
For a spontaneous process, which of the following MUST be true?
A. TΔS>ΔH
B. ΔG>0
C. ΔSuniv>0
D. ΔSsys>ΔSsurr
Answer:
C; ΔSuniv>0
Explanation:
In this question, we want to select which of the options must be true.
What we should understand is that for a process to be spontaneous, the change in entropy must be greater than 0 i.e the change in entropy must be positive.
Looking at the options we have; option C is the correct answer.
Option B looks correct but it is wrong. This is because if change in universal entropy is greater than zero, then change in Gibbs free energy must be less than zero for spontaneity to occur
It can be deduced that for a spontaneous process, B. ΔG>0.
What is a spontaneous process?It should be noted that a spontaneous process simply means a process that occurs without input of matter or electrical energy.
In this case, for a spontaneous process, it's true that ΔG>0, it should be noted that a spontaneous process related to the second law in thermodynamics. This is characterized by an increase in entropy.
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1A. A strontium hydroxide solution is prepared by dissolving 10.45 g of Sr(OH)2 in water to make 41.00 mL of solution. What is the molarity of this solution?
1B. Next the strontium hydroxide solution prepared in part (a) is used to titrate a nitric acid solution of unknown concentration. Write a balanced chemical equation to represent the reaction between strontium hydroxide and nitric acid solutions.
1C. If 23.9 mL of the strontium hydroxide solution was needed to neutralize a 31.5 mL aliquot of the nitric acid solution, what is the concentration (molarity) of the acid?
Answer:
1. 0.00352 M
2. 2HNO3(aq) + Sr(OH)2(aq) -----> Sr(NO3)2(aq) + 2H2O(l)
3. 0.00534 M
Explanation:
1.
Mass of strontium hydroxide= 10.45 g
Volume of solution = 41.00 ml
Number of moles = mass of Sr(OH)2/molar mass of Sr(OH)2 = 10.45g/121.63 g/mol= 0.0859 moles
Molarity= number of moles × volume = 0.0859 ×41/1000 = 0.00352 M
2.
2HNO3(aq) + Sr(OH)2(aq) -----> Sr(NO3)2(aq) + 2H2O(l)
3.
Concentration of acid CA= the unknown
Volume of acid VA= 31.5 ml
Concentration of base CB= 0.00352 M
Volume of base VB= 23.9 ml
Number of moles of acid NA= 2
Number of moles of base NB= 1
From;
CAVA/CBVB = NA/NB
CAVANB= CBVBNA
CA= CBVBNA/VANB
CA= 0.00352 × 23.9 ×2/31.5 ×1
CA= 0.00534 M
A. The molarity of the Sr(OH)₂ solution is 2.09 M
B. The balanced equation for the reaction is
2HNO₃ + Sr(OH)₂ —> Sr(NO₃)₂ + 2H₂O
C. The molarity of the acid, HNO₃ is 3.17 M
A. Determination of the molarity of the Sr(OH)₂ solution
We'll begin by calculating the number of mole in 10.45 g of Sr(OH)₂Mass of Sr(OH)₂ = 10.45 g
Molar mass of Sr(OH)₂ = 88 + 2(16 + 1) = 122 g/mol
Mole of Sr(OH)₂ =?Mole = mass / molar mass
Mole of Sr(OH)₂ = 10.45 / 122
Mole of Sr(OH)₂ = 0.0857 mole Finally, we shall determine the molarity of Sr(OH)₂Mole of Sr(OH)₂ = 0.0857 mole
Volume = 41 mL = 41 / 1000 = 0.041 L
Molarity of Sr(OH)₂ =?Molarity = mole / Volume
Molarity of Sr(OH)₂ = 0.0857 / 0.041
Molarity of Sr(OH)₂ = 2.09 MB. The balanced equation for the reaction.
2HNO₃ + Sr(OH)₂ —> Sr(NO₃)₂ + 2H₂OC. Determination of the molarity of the acid, HNO₃.
From the balanced equation above,
The mole ratio of the acid, HNO₃ (nA) = 2
The mole ratio of the base, Sr(OH)₂ (nB) = 1
From the question given above,
Volume of base, Sr(OH)₂ (Vb) = 23.9 mL
Molarity of base, Sr(OH)₂ (Mb) = 2.09 M
Volume of acid, HNO₃ (Va) = 31.5 mL
Molarity of acid, HNO₃ (Ma) =?MaVa / MbVb = nA/nB
(Ma × 31.5) / (2.09 × 23.9) = 2
(Ma × 31.5) / 49.951 = 2
Cross multiply
Ma × 31.5 = 49.951 × 2
Ma × 31.5 = 99.902
Divide both side by 31.5
Ma = 99.902 / 31.5
Ma = 3.17 MThus, molarity of the acid, HNO₃ is 3.17 M
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Which solution has the greatest buffer capacity? Select the correct answer below: 1 mole of acid and 1 mole of base in a 1.0 L solution
Answer:
The answer is
Explanation:
1 mole of acid.
Hope this helps....
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A buffer that is 1 M in acid and base will have the greatest capacity of buffer, and therefore the greatest buffer capacity.
What do you mean by the buffer solution ?A weak acid and the conjugate base of the weak acid, or a weak base and the conjugate acid of the weak base, are combined to form the buffer solution, a water-based solvent solution.
In a biological system, a buffer's keep intracellular and extracellular pH levels within a relatively small range and to withstand pH fluctuations brought on by both internal and external factors.
A buffer is a substance that can withstand a pH shift when acidic or basic substances are added. It may balance out little quantities of additional acid or base, keeping the pH stable.
Thus, 1 M in acid and base solution has the greatest buffer capacity.
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A laboratory technician drops a 0.0850 kg sample of unknown solid material, at a temperature of 100 oC, into a calorimeter. The calorimeter can, initially at 19.0 oC, is made of 0.150 kg of copper and contains 0.20 kg of water. The final temperature of the calorimeter can, and contents is 26.1 oC. Compute the specific heat of the sample.
Answer:
The specific heat of the sample [tex]\mathbf{c_3 = 1011.056 \ J/kg.K}[/tex]
Explanation:
Given that:
mass of an unknown sample [tex]m_3[/tex] = 0.0850
temperature of the unknown sample [tex]t_{unknown}[/tex] = 100° C
initial temperature of the calorimeter can = 19° C
mass of copper [tex]m_1[/tex] = 0.150 kg
mass of water [tex]m_2[/tex]= 0.20 kg
the final temperature of the calorimeter can = 26.1° C
The objective is to compute the specific heat of the sample.
By applying the principle of conservation of energy
[tex]Q = mc \Delta T[/tex]
where;
[tex]Q_1 +Q_2 +Q_3 = 0[/tex]
i.e
[tex]m_1 c_1 \Delta T_1 +m_2 c_2 \Delta T_2+m_3 c_3 \Delta T_3 =0[/tex]
the specific heat capacities of water and copper are 4.18 × 10³ J/kg.K and 0.39 × 10³ J/kg.K respectively
the specific heat of the sample [tex]c_3[/tex] can be computed by making [tex]c_3[/tex] the subject of the above formula:
i.e
[tex]c_3 = \dfrac{m_1 c_1 \Delta T_1 +m_2 c_2 \Delta T_2}{m_3 c_3 \Delta T_3}[/tex]
[tex]c_3 = \dfrac{ 0.150 \times 0.39 \times 10^3 \times (26.1 -19) + 0.20 \times 4.18 \times 10^3 \times (26.1 -19) }{0.0850 \times (100-26.1 )}[/tex]
[tex]c_3 = \dfrac{ 0.150 \times 0.39 \times 10^3 \times (7.1) + 0.20 \times 4.18 \times 10^3 \times (7.1) }{0.0850 \times (73.9)}[/tex]
[tex]c_3 = \dfrac{415.35 + 5935.6 }{6.2815}[/tex]
[tex]c_3 = \dfrac{415.35 + 5935.6 }{6.2815}[/tex]
[tex]c_3 = \dfrac{6350.95}{6.2815}[/tex]
[tex]\mathbf{c_3 = 1011.056 \ J/kg.K}[/tex]
The specific heat of the sample [tex]\mathbf{c_3 = 1011.056 \ J/kg.K}[/tex]
Without doing any calculations, determine the sign of ΔSsys for each of the following chemical reactions. Drag the appropriate items to their respective bins.
1. 2H30' (aq) + CO23- (aq) - CO2(g) +3H2O(1)
2. CH4(g) + 202,(g) - CO2(g) + 2H2O(l)
3. Mg (s) + Cl2(g) - MgCǐ2(s)
4. SO3(g) + H2O(I) - H2SO4(I)
A. ΔSsys greater than
B. ΔSsys smaller than
Answer:
Answers are in the explanation.
Explanation:
In a chemical reaction we can determine the sign of ΔSsys based on the states of products and reactants knowing that:
Entropy of gases >>> entropy of liquid > entropy of solids.
The entropy of solids is lower than entropy of liquids that is lower than entropy of gases.
In the reactions:
1. 2H₃O⁺(aq) + CO₃²⁻(aq) → CO₂(g) +3H₂O(l)
As 1 gas is produced, entropy of products is higher than entropy of reactants. That means ΔSsys > 0 (That because ΔSsys is ΔSProducts - ΔSReactants)
2. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
3 moles of gas are converted in 1 mole of gas in products. Entropy of reactants is higher than entropy of products, ΔSsys < 0.
3. Mg(s) + Cl₂(g) → MgCǐ₂(s)
You have 1 mole of gas in reactants and 1 mole of solid in products. ΔSProducts <<< ΔSReactants. ΔSsys < 0.
4. SO₃(g) + H₂O(I) → H₂SO₄(I)
1 mole of gas in reactants, a liquid in products. ΔSProducts <<< ΔSReactants. ΔSsys < 0.
Determining the sign of ΔSsys for each given chemical reaction, the following can be obtained without calculations:
1. ΔSsys > 0.
2. ΔSsys < 0
3. ΔSsys < 0
4. ΔSsys < 0
Recall:
The states of the reactants and the products in a chemical reaction determines the sign of ΔSsys of the reaction.The entropy of gasses is greater than the entropy of liquid and solids.The entropy of solids is less than the entropy of liquid and gasses.Gasses have the highest entropy, while solids have the least.Thus:
In the first chemical reaction, 1 mole of gas is produced, therefore: ΔSsys > 0.
In the second chemical reaction, 3 moles of gasses gives a products of 1 mole of gas, therefore: ΔSsys < 0.
In the third chemical reaction, 1 mole of gas gives 1 mole of solid as product, therefore: ΔSsys < 0.
In the fourth chemical reaction, 1 mole of gas gives 1 mole of liquid as product, therefore: ΔSsys < 0.
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Data is collected for the gas phase reaction 2 A + B + 3 C → Products at 470 K.What is the order of the reaction with respect to A?
Explanation:
The equation is given as;
2 A + B + 3 C → Products
The order of the reaction refers to the extent at which the rate depends n the concentration of the reactant.
The order of reaction is experimentally obtained. It can also be obtained from the rate law of the reaction.
If the rate law is given as;
rate law = k [A]²[B][C]³
Then the order is second order with respect to A.
The order is second order with respect to A.
Reaction series;Given that;
2A + B + 3C → Products at 470 K
Find:
Order of reaction with respect to A
Computation:
The reaction that takes place refers to how much the rate is influenced by the reactant concentration.
The reaction order is determined empirically. This can also be derived from the reaction's rate law.
Rate law = k[A]²[B][C]³
So, The order is second order with respect to A.
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Calculate the energy required to heat of 1.50 kg silver from -7.8 C to 15.0 C . Assume the specific heat capacity of silver under these conditions is .0235 J*g^-1*K^-1 . Be sure your answer has the correct number of significant digits.
Answer:
804 J
Explanation:
Step 1: Given data
Mass of silver (m): 1.50 kgInitial temperature: -7.8 °CFinal temperature: 15.0 °CSpecific heat capacity of silver (c): 0.0235J·g⁻¹K⁻¹Step 2: Calculate the energy required (Q)
We will use the following expression.
Q = c × m × ΔT
Q = 0.0235J·g⁻¹K⁻¹ × (1.50 × 10³g) × [15.0°C-(-7.8°C)]
Q = 804 J
the following glassware was found in a lab drawer: 12 beakers, 10 flasks and 60 test tubes. what percent of the glassware are test tubes?
Answer:
73.1707317073%
=> Approximately 73.2%
Explanation:
Total = 60 + 10 + 12
=> 82
Test tubes are 60/82
=> 30/41
=> 73.1707317073%
=> Approximately 73.2%
Answer:
73.17%
Explanation:
To find the percentage of test tubes to the overall glassware, we need to get the number of test tubes divided by the total number of glassware.
12 beakers + 10 flasks + 60 test tubes = 82 glassware
% test tube = 60 / 82 = .7317 ==> 73.17 %
So 73.17 % of the glassware was test tubes.
Cheers.
which of the following compounds are polar: CH2Cl2, HBr?
Answer:
CH2Cl2 is polar
Explanation:
When NaOH is added to water, the (OH) = 0.04 M. What is the [H30*]?
What is the PH of the solution?
Answer:
[H₃O⁺] = 2.5 × 10⁻¹³ M
pH = 12.6
Explanation:
Step 1: Given data
Concentration of OH⁻: 0.04 M
Step 2: Calculate the concentration of H₃O⁺
Let's consider the self-ionization of water reaction.
2 H₂O(l) ⇄ OH⁻(aq) + H₃O⁺(aq)
The ionic product of water is:
Kw = [OH⁻] × [H₃O⁺] = 10⁻¹⁴
[H₃O⁺] = 10⁻¹⁴ / [OH⁻]
[H₃O⁺] = 10⁻¹⁴ / 0.04
[H₃O⁺] = 2.5 × 10⁻¹³ M
Step 3: Calculate the pH
The pH is:
pH = -log [H₃O⁺] = -log 2.5 × 10⁻¹³ = 12.6
Can solid FeBr2 react with Cl2 gas to produce solid FeCl2 and Br2 gas? Why or why not?
Answer:
Yes
Explanation:
The balanced equation of the reaction is;
FeBr2 (aq) + Cl2 (g) → FeCl2 (aq) + Br2 (aq)
This reaction is possible because chlorine is more electronegative than bromine and can displace it from its salt.
In group seventeen, electro negativity decreases down the group. Hence as we move down the group, elements become less electronegative and can be displaced from their salt by more electronegative elements found earlier in the group.
Hence chlorine can displace bromine in FeBr2 to form FeCl2.
Answer:
Yes, because Cl2 has higher activity than Br2
Explanation:
a reaction mixture initially contains 10.0 atm N2 and 10.0 atm H2. If the equilibrium pressure of NH3 is measured to be 6.0 atm, find the equilibrium constant (Kp) for the reaction. g
Answer:
[tex]Kp=5.14[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]N_2+3H_2\rightarrow 2NH_3[/tex]
Thus, the equilibrium expression is written as:
[tex]Kp=\frac{p_{NH_3}^2}{p_{N_2}p_{H_2}^3}[/tex]
And in terms of the reaction extent:
[tex]Kp=\frac{(2x)^2}{(10-x)(10-3x)^3}[/tex]
Thus, from the equilibrium pressure of ammonia we can compute the reaction extent:
[tex]p_{NH_3}=2x=6.0 atm\\\\x=3.0atm[/tex]
Therefore, the equilibrium constant turns out:
[tex]Kp=\frac{(2*3.0)^2}{(10.0-3.0)(10.0-3*3.0)^3}\\\\Kp=5.14[/tex]
Regards.
3. Strontium-90 is produced during the nuclear fission of uranium-235 and is part of nuclear fallout created by weapons testing. If the half-life of Sr-90 is 28 days, how long will it take for grass contaminated with Sr-90 to be safe (<2 percent of the starting radioactivity) for cattle to eat?
A. 158 days
B. 28 days
C. 1 year
D. 158 years
Answer:158 days (D)
Explanation:
It will take 158 days for grass contaminated with Sr-90 to be safe for cattle to eat. Therefore, option (A) is correct.
What is the half-life period?The half-life of a radioactive material is defined as the time that is needed to reduce the initial quantity of a radioactive element to half after disintegration.
The half-life of a radioactive element can be described as the characteristic of the element and does not influence by the initial amount of the radioactive substance.
Given, the half-life of the Strontium-90 = 28 days
The rate constant of the decay can be determined as:
[tex]t_{\frac{1}{2} } =\frac{0.693}{k}[/tex]
[tex]k=\frac{0.693}{t_{\frac{1}{2} } }[/tex]
k = 0.693/28
k = 0.025 day⁻¹
The concentration of Strontium-90 reduced to less than 2% is safe. Therefore final concentration [A] = 2 % = 0.02
[tex]t = \frac{2.303}{k} log \frac{[A_o]}{[A]}[/tex]
[tex]t = \frac{2.303}{0.02475} log \frac{1}{[0.02]}[/tex]
t = 158 days
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When balancing redox reactions under basic conditions in aqueous solution, the first step is to:________.
a. balance oxygen
b. balance hydrogen
c. balance the reaction as though under acidic conditions
d. none of the above
Answer:
When balancing redox reactions under basic conditions in aqueous solution, the first step is to balance oxygen.
Explanation:
Oxidation-reduction reactions or redox reactions are those in which an electron transfer occurs between the reagents. An electron transfer implies that there is a change in the number of oxidation between the reagents and the products.
The gain of electrons is called reduction and the loss of electrons oxidation. That is to say, there is oxidation whenever an atom or group of atoms loses electrons (or increases its positive charges) and in the reduction an atom or group of atoms gains electrons, increasing its negative charges or decreasing the positive ones.
The oxidation and reduction half-reactions, in a basic medium, adjust the oxygens and hydrogens as follows:
In the member of the half-reaction that presents excess oxygen, you add as many water molecules as there are too many oxygen. Then, in the opposite member, the necessary hydroxyl ions are added to fully adjust the half-reaction. Normally, twice as many hydroxyl ions, OH-, are required as water molecules have previously been added.
In short, you first adjust the oxygens with OH-, then you adjust the H with H₂O, and finally you adjust the charge with e-
So, when balancing redox reactions under basic conditions in aqueous solution, the first step is to balance oxygen.
Answer:
c. balance the reaction as though under acidic conditions
Explanation:
When balancing redox reactions under basic conditions, a good technique is to first balance the reaction as though under acidic conditions. We then adjust the result to reflect the basic conditions.
