Answer:
155 °C
Explanation:
Step 1: Given data
Initial volume (V₁): 2.91 LInitial temperature (T₁): 23.0°CFinal volume (V₂): 4.20 LFinal temperature (T₂): ?Step 2: Convert the initial temperature to Kelvin
We will use the following expression.
K = °C + 273.15 = 23.0°C + 273.15 = 296.2 K
Step 3: Calculate the final temperature
Assuming an ideal gas behavior, we can calculate the final temperature using Charles' law.
V₁/T₁ = V₂/T₂
T₂ = V₂ × T₁/V₁
T₂ = 4.20 L × 296.2 K/2.91 L
T₂ = 428 K
Step 4: Convert the final temperature to Celsius
We will use the following expression.
°C = K - 273.15 = 428 - 273.15 = 155 °C
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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The rate constant for the decay of a radioactive element is 3.68 × 10⁻³ day⁻¹. What is the half-life of this element?
Answer:
half life=0.693/rate constant =188.3
The half-life of this element is 188.32 years
The formula for calculating the half-life of an element is expressed according to the equation:
[tex]t_{1/2}=\frac{ln 2}{\lambda}[/tex]
[tex]\lambda[/tex] is the decay constantt1/2 is the half-lifeGiven the following parameters:
The rate constant for the decay = 3.68 × 10⁻³ day⁻¹.
Substitute into the formula for calculating the half-life as shown:
[tex]3.68\times 10^{-3}=\frac{0.693}{\lambda}\\ 0.00368=\frac{0.693}{\lambda}\\\lambda=\frac{0.693}{0.00368}\\\lambda = 188.32 years[/tex]
Hence the half-life of this element is 188.32 years
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15. Calculate the critical angle of glass and water combination. Show your calculation. 16. What is the critical angle for the interface between Mystery A and glass
Answer:
15. Critical angle of glass and water combination, θ = 62.45°
16. Critical angle for the interface between Mystery A and glass, θ = 37.93°
Note; The question is incomplete. The complete question is as follows:
Medium Air Water Glass Mystery A Mystery B Table-2 Speed (m/s) 1.00 C 0.75 c 0.67 0.41 c 0.71 c n 1.00 1.33 1.50 Index of Refraction n of a given medium is defined as the ratio of speed of light in vacuum, c to the speed of light in a medium, v. n = c/v
Table-4: Incident Angle (degrees) Reflected Angle Refracted angle (degrees) (degrees) % Intensity of reflected ray 0 10 20 30 40 50 N/A N/A N/A 30 40 50 0 11.3 22.7 34.2 46.3 59.5 N/A N/A N/A 0.67 1.22 3.08 % Intensity of refracted ray 100 100 100 99.33 98.78 96.92
When rays travel from a denser medium to a less dense medium, we can define a critical angle of incidence θ such that refracted angle θ₂ = 90°. Applying Snell's law: Critical angle θ = sin-1(n₂/n₁).
When the angle of incidence is greater than the critical angle, 100% of the light intensity is reflected. This is called total internal reflection because all the light is reflected.
15. Calculate the critical angle of glass and water combination. Show your calculation.
16. What is the critical angle for the interface between Mystery A and glass?
Explanation:
15. Applying Snell's law; Critical angle θ = sin-1(n₂/n₁).
where n₂,refractive index of water = 1.33, n₁, refractive index of glass = 1.50 since glass is denser than water
θ = sin-1(1.33/1.50)
θ = 62.45°
Critical angle of glass and water combination, θ = 62.45°
16. Refractive index of mystery A , n = c/v
where v = 0.41 c
therefore, n = c / 0.41 c = 2.44
Critical angle for the interface between Mystery A and glass, θ = sin-1(n₂/n₁).
where n₂,refractive index of glass = 1.50, n₁, refractive index of mystery A = 2.44 since mystery A is denser than glass as seen from its refractive index
θ = sin-1(1.50/2.44)
θ = 37.93°
Critical angle for the interface between Mystery A and glass, θ = 37.93°
For the following reaction, 6.99 grams of oxygen gas are mixed with excess nitrogen gas . The reaction yields 10.5 grams of nitrogen monoxide . nitrogen ( g ) oxygen ( g ) nitrogen monoxide ( g ) What is the theoretical yield of nitrogen monoxide
Answer:
13.11 g.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below :
N2 + O2 —> 2NO
Next, we shall determine the mass of O2 that reacted and the mass of NO produced from the balanced equation. This is illustrated below:
Molar mass of O2 = 16x2 = 32 g/mol
Mass of O2 from the balanced equation = 1 x 32 = 32 g.
Molar mass of NO = 14 + 16 = 30 g/mol
Mass of NO from the balanced equation = 2 x 30 = 60 g.
From the balanced equation above,
32 g of O2 reacted to produce 60 g of NO.
Finally, we shall determine the theoretical yield of NO as follow:
From the balanced equation above,
32 g of O2 reacted to produce 60 g of NO.
Therefore, 6.99 g of O2 will react to produce = (6.99 x 60)/32 = 13.11 g of NO.
Therefore, the theoretical yield of nitrogen monoxide, NO is 13.11 g.
Write an equation to show how the base NaOH(s) behaves in water. Include states of matter in your answer. Click in the answer box to open the symbol palette.
Answer:
The reaction is given as:
[tex]NaOH(s)\rightarrow Na^+(aq)+OH^-(aq)[/tex]
Explanation:
Bases are defined as those chemical substances which give hydroxide ions in their aqueous solutions.
[tex]BOH(s)\rightarrow B^+(aq)+OH^-(aq)[/tex]
When sodium hydroxide is added to water it gets dissociated into two ions that are sodium ions and hydroxide ions. Along with this heat energy also releases during this reaction.
The reaction is given as:
[tex]NaOH(s)\rightarrow Na^+(aq)+OH^-(aq)[/tex]
The equation to show how NaOH behaves in water is NaOH → Na⁺ + (OH)⁻
The compound that produce negative hydroxide (OH−) ions when dissolved
in water are called bases .
This compounds NaOH (sodium hydroxide) is an example of a base.
When it dissolves in water it dissociate to form negative hydroxide (OH−)
ions and positive sodium (Na+) ions.
It can be represented by the following equation:
NaOH → Na⁺ + (OH)⁻
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Ozone (O 3) in the atmosphere can react with nitric oxide (NO): O 3(g) + NO(g) → NO 2(g) + O 2(g). Calculate the ΔG° for this reaction at 25°C. (ΔH° = –199 kJ/mol, ΔS° = –4.1 J/K·mol)
Answer:
ΔG° = 1022. 8 kJ
Explanation:
ΔH° = –199 kJ/mol
ΔS° = –4.1 J/K·mol
T = 25°C = 25 + 273 = 298K (Converting to kelvin temperature)
ΔG° = ?
The relationship between these varriables are;
ΔG° = ΔH° - TΔS°
ΔG° = –199 - 298 (–4.1)
ΔG° = -199 + 1221.8
ΔG° = 1022. 8 kJ
To calculate ΔG, the following equation should be used:
ΔG° = ΔH° - TΔS°
Given:
ΔH° = –199 kJ/mol,
ΔS° = –4.1 J/K·mol
T=25+273K=298K
Substitute the respective values:
ΔG° = ΔH° - TΔS°
=–199 kJ/mol-(298K×(-4.1 J/K·mol*(1KJ/1000J)))
=-197.78kJ/mol
Thus, we can conclude the value of ΔG°=-197.78kJ/mol
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Susan was investigating which glue would make the strongest craft stick tower. She
tested rubber cement, Elmer's glue-all, and Super Glue. She will test the strength of the
towers with books as the weight.
What is the independent variable?
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.".
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.
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.
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
what are the monomers of bakelite
Answer:
Bakelite is a polymer made up of the monomers phenol and formaldehyde. This phenol-formaldehyde resin is a thermosetting polymer.
Answer: The monomers of bakelite are formaldehyde and phenol
Explanation:
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.
03/08/2020
Question
1. (a) State the definition of a chemical formula.
(b) What does it tell about a compound?
(c) What information is conveyed by the formulation H2SO4?
Explanation:
According to your question,
no. a. ans would be like; chemical formula is defined as the an expression which determines no. and type of molecule of a compound.
b. no. ans; it tells that what type of compound is formed with the type and no. of atoms present in the atom.
c. no ans; the formulation of h2so4 states that it is acid named as hydrochloric acid which is formed by reacting of hydrogen (2 atoms ) ,sulpher (*1atom) and oxygen(4atoms).
Hope it helps...
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 pain reliever codeine is a weak base with a kb equal to 1.6 x 10^-6. what is the ph of a 0.05 m aqueous codeine solution
Answer:
[tex]pH=10.45[/tex]
Explanation:
Hello,
In this case, for the dissociation of the given base, we have:
[tex]base\rightleftharpoons OH^-+CA[/tex]
Whereas CA accounts for conjugated acid and OH⁻ for the conjugated base. In such a way, equilibrium expression is:
[tex]Kb=\frac{[OH^-][CA^+]}{[base]}[/tex]
And in terms of the reaction extent [tex]x[/tex] we can write:
[tex]1.6x10^{-6}=\frac{x*x}{0.05M-x}[/tex]
For which the roots are:
[tex]x_1=-0.000284M\\x_2=0.000282M[/tex]
For which clearly the result is the positive root which also equals the concentration of hydroxyl ions and we can compute the pOH:
[tex]pOH=-log([OH^-])=-log(0.000282)\\\\pOH=3.55[/tex]
And the pH:
[tex]pH=14-pOH=14-3.55\\\\pH=10.45[/tex]
Regards.
The pH of the solution is 10.45.
Let us represent codeine with the generic formula BH. We can set up the ICE table as follows;
:B(aq) + H2O(l) ⇄ BH(aq) + OH^-(aq)
I 0.05 0 0
C -x +x +x
E 0.05 - x x x
We know that the Kb of codeine is 1.6 x 10^-6, Hence;
1.6 x 10^-6 = x^2/0.05 - x
1.6 x 10^-6 (0.05 - x ) = x^2
8 x 10^-8 - 1.6 x 10^-6x = x^2
x^2 + 1.6 x 10^-6x - 8 x 10^-8 = 0
x = 0.00028 M
The concentration of hydroxide ions = 0.00028 M
Given that pOH = - log[0.00028 M]
pOH = 3.55
pH + pOH = 14
pH = 14 - 3.55
pH = 10.45
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Calculate the equilibrium constant K c for the following overall reaction: AgCl(s) + 2CN –(aq) Ag(CN) 2 –(aq) + Cl –(aq) For AgCl, K sp = 1.6 × 10 –10; for Ag(CN) 2 –, K f = 1.0 × 10 21.
Answer:
1.6x10¹¹ = Kc
Explanation:
For the reaction:
AgCl(s) + 2CN⁻(aq) ⇄ Ag(CN)₂⁻(aq) + Cl⁻(aq)
Kc is defined as:
Kc = [Ag(CN)₂⁻] [Cl⁻] / [CN⁻]²
Ksp of AgCl is:
AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)
Where Ksp is:
Ksp = [Ag⁺] [Cl⁻] = 1.6x10⁻¹⁰
In the same way, Kf of Ag(CN)₂⁻ is:
Ag⁺(aq) + 2CN⁻ ⇄ Ag(CN)₂⁻
Kf = [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺] = 1.0x10²¹
The multiplication of Kf with Ksp gives:
[Ag⁺] [Cl⁻] * [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺] = Ksp*Kf
[Ag(CN)₂⁻] [Cl⁻] / [CN⁻]² = Ksp*Kf
Obtaining the same expression of the first reaction
That means Ksp*Kf = Kc
1.6x10⁻¹⁰*1.0x10²¹ = Kc
1.6x10¹¹ = Kc1A. 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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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
What did J. J. Thomson's cathode ray experiment show about atoms?
Answer:
atoms contain tiny negatively charged subatomic particles or electrons
If 37.5 mL of 0.100 M calcium chloride reacts completely with aqueous silver nitrate, what is the mass of AgCl (MM
Answer:
OPTION C is correct
C) 1.07 g
Explanation:
CaCl2(aq) + 2 AgNO3(aq) → 2 AgCl(s) + Ca(NO3)3(aq)
But we know molarity molarity= number of moles of solute/ volume of the solution
M= n/V
From the equation above
number of moles of Cacl2 = (37.5 ×0 .100 × 10^-3) = 0.00375 moles.
Then
1 mole of Cacl2 yields 2 moles of Agcl2
0.00375 moles of Cacl2 will produce let say Y.
Y= (0.00375 ×2)/1
= 0.0075 moles.
Number of moles of Agcl2 = mass /molar mass of Agcl
Molar mass of Agcl = 178
Then mass = 178 ×0.0075 = 1.047
Therefore, the mass of agcl precipitate is
1.07 g
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.
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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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 qualitative researcher may select a ______ case that is unusually rich in information pertaining to the research question.
A) critical
B) typical
C) deviant
D) rare
Answer:
D. Rare
Explanation:
Qualitative research has to do with non-numerical data and is used to understand the beliefs of a group of people which can be gotten from surveys, questionnaires, interviews, etc.
A qualitative researcher may select a RARE case that is unusually rich in information pertaining to the research question.
This is because he wants to get an insight into the why, how, where and when of that particular ase as it's not a usual occurrence.
which of the following compounds are polar: CH2Cl2, HBr?
Answer:
CH2Cl2 is polar
Explanation:
Compound has a molar mass of and the following composition: elementmass % carbon47.09% hydrogen6.59% chlorine46.33% Write the molecular formula of .
The given question is incomplete. The complete question is:
Compound X has a molar mass of 153.05 g/mol and the following composition:
element mass %
carbon 47.09%
hydrogen 6.59%
chlorine 46.33%
Write the molecular formula of X.
Answer: The molecular formula of X is [tex]C_6H_{10}Cl_2[/tex]
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of C= 47.09 g
Mass of H = 6.59 g
Mass of Cl = 46.33 g
Step 1 : convert given masses into moles.
Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{47.09g}{12g/mole}=3.92moles[/tex]
Moles of H =[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.59g}{1g/mole}=6.59moles[/tex]
Moles of Cl =[tex]\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{46.33g}{35.5g/mole}=1.30moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{3.92}{1.30}=3[/tex]
For H = [tex]\frac{6.59}{1.30}=5[/tex]
For Cl =[tex]\frac{1.30}{1.30}=1[/tex]
The ratio of C : H: Cl= 3: 5 :1
Hence the empirical formula is [tex]C_3H_5Cl[/tex]
The empirical weight of [tex]C_3H_5Cl[/tex] = 3(12)+5(1)+1(35.5)= 76.5g.
The molecular weight = 153.05 g/mole
Now we have to calculate the molecular formula.
[tex]n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{153.05}{76.5}=2[/tex]
The molecular formula will be=[tex]2\times C_3H_5Cl=C_6H_{10}Cl_2[/tex]
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.
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!
