Answer:
Part A is just T2 = 58.3 K
Part B ∆U = 10967.6 x C[tex]_{V}[/tex] You can work out C[tex]_{V}[/tex]
Part C
Part D
Part E
Part F
Explanation:
P = n (RT/V)
V = (nR/P) T
P1V1 = P2V2
P1/T1 = P2/T2
V1/T1 = V2/T2
P = Pressure(atm)
n = Moles
T = Temperature(K)
V = Volume(L)
R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.
bar = 0.986923 atm
N = 14g/mol
N2 Molar Mass 28g
n = 3.5 mol N2
T1 = 350K
P1 = 1.5 bar = 1.4803845 atm
P2 = 0.25 bar = 0.24673075 atm
Heat Capacity at Constant Volume
Q = nCVΔT
Polyatomic gas: CV = 3R
P = n (RT/V)
0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))
V = (nR/P) T
V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K
V = (0.28721/1.4803845) x 350
V = 0.194 x 350
V = 67.9036 L
So V1 = 67.9036 L
P1V1 = P2V2
1.4803845 atm x 67.9036 L = 0.24673075 x V2
100.52343693 = 0.24673075 x V2
V2 = P1V1/P2
V2 = 100.52343693/0.24673075
V2 = 407.4216 L
P1/T1 = P2/T2
1.4803845 atm / 350 K = 0.24673075 atm / T2
0.00422967 = 0.24673075 /T2
T2 = 0.24673075/0.00422967
T2 = 58.3 K
∆U= nC[tex]_{V}[/tex] ∆T
Polyatomic gas: C[tex]_{V}[/tex] = 3R
∆U= nC[tex]_{V}[/tex] ∆T
∆U= 28g x C[tex]_{V}[/tex] x (350K - 58.3K)
∆U = 28C[tex]_{V}[/tex] x 291.7
∆U = 10967.6 x C[tex]_{V}[/tex]
Can the pH scale be utilized for all acids (Arrhenius, Bronsted-Lowery, and Lewis)? Give examples of substances from each definition category that can/cannot use the pH scale and explain your reasoning.
Explanation:
Before proceeding to answering the questions, let us go over some definitions.
pH scale: The pH scale measures how acidic or basic a substance is. The pH scale ranges from 0 to 14. A pH of 7 is neutral. A pH less than 7 is acidic.
pH stands for Potential of Hydrogen. It refers to the hydrogen ion concentration in a solution.
The keywords being; Hydrogen Ion concentration.
A Bronsted-Lowry acid is a chemical species that donates one or more hydrogen ions in a reaction.
A Lewis acid is any substance that can accept a pair of nonbonding electrons. It is an electron pair acceptor.
An Arrhenius acid is a substance that dissociates in water to form hydrogen ions or protons.
Based on the definitions of given above, it is obseved that both Bronsted lowry and arrhenius acids deals with hydrogen ions. Hence both of this acids can be measured using the pH scale. The lewis acid on the other hand do not necessarily contain hydrogen ions, hence the pH scale cannot be utilized for it.
Examples includes;
Arrhenius acid; Nitric Acid – HNO3 etc
Lewis acid; boron trifluoride (BF3) and aluminum fluoride (AlF3) etc
Bronsted lowry acid; HCl etc
What is the color of litmus solution in sodium hydroxide?
Answer:
Sodium Hydroxide turns blue litmus red .
4. Alex is a soil specialist and he sells manure best suited for wheat. He goes to a village where farmers are
planning to sow wheat which grows best in clayey and loamy soil. He tests the soil sample and finds
that the soil is more suitable for the growth of cotton rather than wheat. He tells the villagers about
this despite knowing that they will not buy his manure.
(a) Why is clayey-loamy soil preferred for growing wheat?
(b) Which value is shown by Alex?
Please answer the question fast
Answer:
Explanation:
1) Clayey-loamy soil tends to be excellent at retaining water which is critical for proper growth of wheat thus this kind of soil is preferred.
2) Alex is showing honesty as he informing the villagers about the actual truth and reality (he is not basing his answers on judgements rather he is basing it on scientific knowledge and being honest about it)
Determine the number of moles of air present in 1.35 L at 750 torr and 17.0°C.
Which equation should you use?
Answer:
0.0560 mol
Explanation:
You need to use the ideal gas law equation:
[tex]PV = nRT[/tex]
P = 750 torr
V = 1.35 L
n = moles
R = 62.364 L torr [tex]mol^{-1}[/tex] [tex]K^{-1}[/tex]
T = 17.0 ˚C + 273.15 = 290.15 K
Rearranging the equation for n:
[tex]n= \frac{PV}{RT}[/tex]
[tex]n= \frac{(750)(1.35)}{(62.364)(290.15)}[/tex]
n = 0.0560 mol
Answer:
n=PV/RT
Explanation:
on edge 2020
which two compounds are structural isomers of each other
Answer:
Explanation:
Structural isomer is a type of isomer in which molecules with the same molecular formula have different bonding patterns and atomic organization.
For example Pentan-1-ol, pentan-2-ol, and pentan-3-ol are structural isomers that exhibit position isomerism.
Cyclohexane and hex-1-ene are examples of functional group structural isomers.
Consider an Al-4% Si alloy. Determine (a) if the alloy is hypoeutectic or hypereutectic, (b) the composition of the first solid to form during solidification, (c) the amounts and compositions of each phase at 578˚C, (d) the amounts and compositions of each phase at 576˚C, (e) the amounts and compositions of each microconstituent at 576˚C, and (f) the amounts and compositions of each phase at 25˚C.
Answer:
(a) Hypoeutectic
(b) Alpha solid, aluminium
(c) 70% α , 30% β
(d) 97.6% α, 2.4% β
(e) 97.6% α, 2.4% β
(f) 97% α, 3% β
Explanation:
(a) The eutectic composition for Al Si alloy is 11.7 wt% silicon, therefore, an Al-4% Si alloy is hypoeutectic
(b) For the hypoeutectic alloy, aluminium, Al, is expected to form first, such that the aluminium content is reduced till the point it gets to the eutectic proportion of 11.7 wt% silicon
(c) At 578°C we have
% α: Al (11 - 4)/(11 - 1) = 70% α
% L: Si 100 - 70 = 30% β
(d) At 576°C we have
α: 99.83% Si (99.83 - 4)/(99.83- 1.65) = 97.6% α
β: 1.65% Si (4 - 1.65)/(99.83- 1.65) = 2.4% β
(e) Primary α: 1.65% α (99.83 - 4)/(99.83 - 1.65) = 97.6% α
Eutectic 4% Si = 100 - 97.6 = 2.4% β
(f) At 25°C we have;
α%: (99.83 - 4)/(99.83 - 1) = 97% α
β%: 100 - 97 = 3% β.
g 1. An ideal gas is confined to a container with adjustable volume. The number of moles, n, and temperature, T, are constant. By what factor will the volume change if pressure decreases by a factor of 8.5 Hint: pV
Answer:
The volume will increase by a factor of 8.5
Explanation:
In an ideal gas, the volume is inversely proportional to the pressure. This means that the volume will decrease when the pressure increases and vice versa.
Let the initial volume be v1, and the initial pressure p1.
the final pressure is decreased by a factor of 8.5, i.e
p2 = p1/8.5 = 0.12p1
using p1v1 = p2v2
v2 = (p1/p2)v1
v2 = (p1/0.12p1)v1
v2 = 8.5v1
This means that the final pressure increases by a factor of 8.5
Answer:
Increases by a factor of 8.5
Explanation:
Hello,
In this case, with the given data, we can apply the Boyle's law in order to understand the pressure-volume behavior as an inversely proportional relationship:
[tex]P_1V_1=P_2V_2[/tex]
Thus, for the given pressure decrease we have:
[tex]P_2=\frac{P_1}{8.5}[/tex]
So the volume changes by:
[tex]V_2=\frac{P_1V_1}{\frac{P_1}{8.5} } \\\\V_2=8.5V_1[/tex]
Therefore, it increases by a factor of 8.5.
Regards.
Write the identity of the missing nucleus for the following nuclear decay reaction:
232 Th 228Ra+?
90
88
low.
Express your answer as a particle.
View Available Hints
Hint 1. Determine the missing mass number
1
Hint 2. Determine the missing atomic number
Hint 3. Determine the symbol of the new nucleus
?
Answer:
Alpha decay: 4 2He
Explanation:
Considering the equation given in the question, we can see that the mass number of daughter nuclei (228 88Ra) reduce by 4 and the atomic number reduce by 2 when compared with the parent nucleus (232 90Th). This simply indicates that the parent nucleus is undergoing alpha decay.
Therefore, the missing nucleus is
4 2He i.e alpha decay.
Please see attachment photo for further details.
A sample of copper absorbs 4.31E+1 kJ of heat, resulting in a temperature rise of 6.71E+1 °C. Determine the mass (in kg) of the copper sample if the specific heat capacity of copper is 0.385 J/g°C.
Answer: 1.67 kg
Explanation:
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.
[tex]Q=m\times c\times \Delta T[/tex]
Q = Heat absorbed=[tex]4.31\times 10^1kJ[/tex] = [tex]43100J[/tex] (1kJ=1000J)
m= mass of substance = ?
c = specific heat capacity = [tex]0.385J/g^0C[/tex]
Change in temperature ,[tex]\Delta T=T_f-T_i=6.71\times 10^1^0C=67.1^0C[/tex]
Putting in the values, we get:
[tex]43100J=m\times 0.385J/g^0C\times 67.1^0C[/tex]
[tex]m=1670g=1.67kg[/tex] (1kg=1000g)
Thus the mass (in kg) of the copper sample is 1.67
Complete the sentence indicating the number of protons and electrons in the given ion. Match the numbers in the left column to the appropriate blanks in the sentences on the right. The ion P3− has The ion P 3 − has blank protons and blank electrons. protons and The ion P 3 − has blank protons and blank electrons. electrons.
What isotope has 17 protons and 18 neutrons? Enter the name of the element followed by a hyphen and the mass number (e.g., uranium-234).
Which element does X represent in the following expression: 4120X? Enter the chemical symbol of the element. X is the element symbol . Provide Feedback Correct. Followup. All bromine atoms have 35 protons. A neutral atom also has 35 electrons. The sum of the protons and neutrons gives the mass number, . End of followup.
Answer: (Chlorine-35)
Element X is Ca
Explanation:
Number of proton plus the number of electron gives the mass number.
Which hormones are secreted from the posterior pituitary gland?
a. Growth hormone
b. Oxytocin
C. ACTH
d. PTH
the answer is b
The posterior pituitary secretes two important endocrine hormones—oxytocin and antidiuretic hormone
Suppose there is 0.63 g of HNO3 per 100 mL of a particular solution. What is the concentration of the HNO3 solution in moles per liter?
Answer:
There are 0.09996826 moles per liter of the solution.
Explanation:
Molar mass of HNO3: 63.02
Convert grams to moles
0.63 grams/ 63.02= 0.009996826
Convert mL to L and place under moles (mol/L)
100mL=0.1 L
0.009996826/0.1= 0.09996826 mol/L
The concentration of the HNO₃ solution is 10⁻¹ moles per liter.
What is Molar Concentration ?
Molar Concentration is known as Molarity. It is the number of moles of solute present in per liter of the solution.
Molar concentration in terms of Molecular weight
C = [tex]\frac{m}{V} \times \frac{1}{M.W}[/tex]
where,
C is the molar concentration in mol/L
m is the mass or weight of solute in grams
V is volume of solution in liters
M.W is the molecular weight in g/mol
Molar concentration of HNO₃ = [tex]\frac{\text{Mass of}\ HNO_3}{\text{Liter of solution}} \times \frac{1}{\text{Molecular weight of}\ HNO_3}[/tex]
Molecular weight of HNO₃ = Atomic weight of H + Atomic weight of N + 3 (Atomic weight of O)
= 1 + 14 + 3 (16)
= 15 + 48
= 63 g/mol
Convert mL into L
1 mL = 0.001 L
100 mL = 100 × 0.001
= 0.1 L
Now, put the value in above formula we get
Molar concentration of HNO₃ = [tex]\frac{\text{Mass of}\ HNO_3 / \text{Liter of solution}}{\text{Molar mass of}\ HNO_3}[/tex]
= [tex]\frac{0.63\ g / 0.1\ L}{63\ g/mol}[/tex]
= [tex]\frac{6.3}{63}[/tex]
= 0.1
= 10⁻¹ moles per liter
Thus, we can say that 10⁻¹ moles per liter is the concentration of the HNO₃ solution.
Learn more about Molarity here: https://brainly.com/question/14469428
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If you made a three-dimensional model of an atom and its nucleus, how would you represent the atom? 7th grade
Answer:
it shows the breakdown of the atom
Explanation:
it will show it molecularly
Automobile battery acid is 38% H2SO4 and has a destiny of 1.29g/ml. Calculate the molality and the molarity of this solution.
Answer:
[tex]M=5.0M\\\\m=6.2m[/tex]
Explanation:
Hello,
In this case, 38 % is commonly a by mass concentration, meaning that we have 38 grams of solute (sulfuric acid) per 100 grams of solution (water+sulfuric acid):
[tex]38\%=\frac{m_{H_2SO_4}}{m_{H_2SO_4}+m_{H_2O}}[/tex]
Hence, we compute the moles of sulfuric acid in 38 grams by using its molar mass (98 g/mol):
[tex]n_{H_2SO_4}=38g*\frac{1mol}{98g}=0.39mol H_2SO_4[/tex]
Next, the volume of the solution in litres by using the density of the solution:
[tex]V_{solution}=100g*\frac{1mL}{1.29g}*\frac{1L}{1000mL} =0.0775L[/tex]
This is done since the molarity is defined as the ratio of the moles of the solute to the volume of the solution in litres, thus we have:
[tex]M=\frac{n}{V}=\frac{0.38mol}{0.0775L}=5.0M[/tex]
On the other hand, the molality is defined as the ratio of the moles of the solute to the mass of the solvent in kilograms, thus, we compute the mass of water (solvent) as shown below:
[tex]m_{H_2O}=100g-38g=62g*\frac{1kg}{1000g}=0.062kg[/tex]
So compute the molality:
[tex]m=\frac{n_{solute}}{m_{solvent}}=\frac{0.39mol}{0.062kg}=6.2m[/tex]
Regards.
1. The molarity of the solution is 5 M
2. The molality of the solution is 6.26 M
Let the mass of the solution be 100 g.
Therefore, the mass of 38% of H₂SO₄ in the solution is 38 g.
Next, we shall determine the mole of 38 g of H₂SO₄.
Mass of H₂SO₄ = 38 gMolar mass of H₂SO₄ = (2×1) + 32 + (16×4) = 98 g/mol Mole of H₂SO₄ =?Mole = mass / molar mass
Mole of H₂SO₄ = 38 / 98
Mole of H₂SO₄ = 0.388 mole
Next, we shall determine the volume of the solution
Mass of solution = 100 gDensity of solution = 1.29 g/mLVolume of solution =?
Volume = mass / density
Volume of solution = 100 / 1.29
Volume of solution = 77.52 mL
1. Determination of the molarity of the solution
Mole of H₂SO₄ = 0.388 mole Volume of solution = 77.52 mL = 77.52/1000 = 0.07752 LMolarity =?
Molarity = mole / Volume
Molarity = 0.388 / 0.07752
Molarity = 5 M
2. Determination of the molality of the solution
Mole of H₂SO₄ = 0.388 mole Mass of H₂SO₄ = 38 gMass of solution = 100 gMass of water = 100 – 38 = 62 g Mass of water = 62 / 1000 = 0.062 KgMolality =?Molality = mole / mass (Kg) of water
Molality = 0.388 / 0.062
Molality = 6.26 M
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A flexible container at an initial volume of 4.11 L contains 6.51 mol of gas. More gas is then added to the container until it reaches a final volume of 11.3 L. Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.
Answer:
The added mole will be "11.388 moles".
Explanation:
The given values are:
Initial volume,
V₁ = 4.11 L
Final volume,
V₂ = 11.3 L
Number of moles,
n₁ = 6.51
On applying Avogadro's law,
⇒ [tex]\frac{V_{1}}{n_{1}}=\frac{V_{2}}{n_{2}}[/tex]
On putting the estimated values, we get
⇒ [tex]\frac{4.11}{6.51} =\frac{11.3}{n_{2}}[/tex]
On applying cross-multiplication, we get
⇒ [tex]4.11 \ n_{2}=11.3\times 6.51[/tex]
⇒ [tex]4.11 \ n_{2}=73.563[/tex]
⇒ [tex]n_{2}=\frac{73.563}{4.11}[/tex]
⇒ [tex]n_{2}=17.898 \ moles[/tex]
So that the number of moles at added gas will be:
[tex]=17.898-6.51[/tex]
[tex]=11.388 \ moles[/tex]
Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product of the instantaneous amounts of A and B not converted to chemical C. Initially, there are 40 grams of A and 50 grams of B, and for each gram of B, 2 grams of A is used. It is observed that 25 grams of C is formed in 8 minutes. How much (in grams) is formed in 16 minutes
Answer: 45. 78 g is formed in 16 minutes.
Explanation:
Let [tex]A_{o} =40 g[/tex] and [tex]B_{o} =50 g[/tex]
we know that [tex]\alpha =A_{o}\frac{M+N}{M}\, and \, \beta =B_{o}\frac{M+N}{N}[/tex]
According to the question to create [tex]x[/tex] part of the chemical C we will need 2 part of A and one part of B.
Therefore, M = 2 and N = 1.
Now, we can easily solve for the value of α and β.
[tex]\alpha =40\frac{2+1}{2}=60\\\\\, and \, \beta =50\frac{2+1}{1}=150[/tex]
Now, the differential equation must be: [tex]\frac{dX}{dt}=k\left ( \alpha -X \right )\left ( \beta -X \right )[/tex]
I separate the variable and solve the equation
[tex]\int \frac{dx}{\left ( 60-x \right )\left ( 150-x \right )}=\int k\, dt[/tex]
[tex]ln\frac{150-x}{60-x}=90kt+C_{1}[/tex]
By using [tex]X(0) =0[/tex],
[tex]\frac{150-x}{60-x}=Ce^{90k_{o}},C=\frac{5}{2}[/tex]
and using [tex]X (8)=25[/tex] and solving for the value of 'k'
[tex]\Rightarrow \frac{150-25}{60-25}=\frac{5}{2}e^{450k}\\\\\Rightarrow 3.6\times \frac{2}{5}=e^{450k}\\\\\Rightarrow 1.4=e^{450k}[/tex]
Taking 'ln' both side, we get
[tex]\Rightarrow k=7.4716\times 10^{-4}[/tex]
We obtain:[tex]X(t)=\frac{60Ce^{90kt}-150}{Ce^{90kt}-1}[/tex]
Now, for the 16 min
[tex]\Rightarrow X(16)=\frac{150e^{1.07559}-150}{2.5e^{1.0759}-1}[/tex]
[tex]\Rightarrow X(16)=\frac{439.7583-150}{7.3293-1}\\\\\Rightarrow X(16)=\frac{289.7583}{6.3293}[/tex]
[tex]\Rightarrow X(16)=45.78 g[/tex]
When of benzamide are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezing point of pure . On the other hand, when of iron(III) chloride are dissolved in the same mass of , the freezing point of the solution is lower than the freezing point of pure . Calculate the van't Hoff factor for iron(III) chloride in . Be sure your answer has a unit symbol, if necessary, and round your answer to significant digits.
The given question is incomplete, the complete question is:
When 238.g of benzamide (C7H7NO) are dissolved in 600.g of a certain mystery liquid X, the freezing point of the solution is 5.0°C lower than the freezing point of pure X. On the other hand, when 238.g of iron(III) chloride are dissolved in the same mass of X, the freezing point of the solution is 11.5°C lower than the freezing point of pure X.
Calculate the van't Hoff factor for iron(III) chloride in X. Be sure your answer has a unit symbol, if necessary, and round your answer to 2 significant digits.
Answer:
The correct answer is 3.0.
Explanation:
Based on the given question, the weight of benzamide given is 238 grams, the molecular mass of benzamide is 121.14 gram per mole. The benzamide is dissolved in 600 grams of liquid X, therefore, the solution's total weight is,
= 238 + 600 = 838 grams
In case of benzamide, ΔTf = kf × molality (It is given that the solution's freezing point is 5 degree C lesser in comparison to the pure X's freezing point). Now putting the values we get,
5 = kf × 238 / (121.14×838) × 1 (Van't Hoff factor) -------- (i)
In case of benzamide, the van't Hoff factor will be 1, as it is neither associate nor dissociate.
On the other hand, the mass of ferric chloride given is 238 grams getting dissolved in the similar mass of X, therefore again the total mass of the solution will be 838 grams. The freezing point of the solution is 11.5 degree C lesser than the pure X's freezing point. The molecular mass of ferric chloride is 162.2 gram per mol.
For FeCl3,
ΔTf = kf × molality
11.5 = kf × 238/ (162.2 × 838) × i (Van't Hoff factor)------ (ii)
Now dividing equation (ii) by (i) we get,
11.5 / 5 = (kf/kf) × (121.14 / 162.2) × i
i = 3.0
A substance is considered to have a smaller surface area when...
A. it is a fine powder
B. it is in large chunks
C. you have a small amount of
the substance
D. you have a large amount of E. the substance
Assign priorities in the following set of substituents according to Cahn-Ingold-Prelog rules. -NH2 -CH2OH -OH -Br A B C D (Provide your ranking through a string like abcd, starting with the one with the highest priority. Your answer does not need to be capitalized.)
Complete Question
The complete question is shown on the first uploaded image
Answer:
[tex]-SH \ \ A[/tex]
[tex]-NH_2 \ \ B[/tex]
[tex]-COOH \ \ C[/tex]
[tex]-CH_3 \ \ D[/tex]
Explanation:
Now to correctly assign priorities in the following set of substituents we need to consider the molecular weight of the atom that is been directly attached to the carbon bond
Now an atom with a higher molecular weight will imply that the substituent will have the the highest priority
So S (32.065 u)with a the highest molecular weight implies that -SH has the highest priority
Next is N with molecular weight(14.0067 u) implies that -[tex]NH_2[/tex] will have the next priority
Next is O with molecular weight(15.999 u) implies that [tex]-COOH[/tex] will have the next priority
The Last is H with molecular weight(1.00784 u) implies that [tex]-CH_3[/tex] will have the next priority
write briefly how oxygen gas is prepared in the industry
by fractional distillation of liquefied air or the use of zeolites to remove carbon dioxide and nitrogen from air or the electrolysis of water
Hydrogen chloride gas can be prepared by the following reaction: 2NaCl(s) + H2SO4(aq) → 2HCl(g) + Na2SO4(s) How many grams of HCl can be prepared from 2.00 mol H2SO4 and 2.56 mol NaCl?
Answer:
The correct answer is is option B
b. 93.3 g
Explanation:
SEE COMPLETE QUESTION BELOW
Hydrogen chloride gas can be prepared by the following reaction: 2NaCl(s) + H2SO4(aq) → 2HCl(g) + Na2SO4(s)
How many grams of HCl can be prepared from 2.00 mol H2SO4 and 2.56 mol NaCl?
a. 7.30 g
b. 93.3 g
c. 146 g
d. 150 g
e. 196 g
CHECK THE ATTACHMENT FOR STEP BY STEP EXPLANATION
(i) 1s 2 2s 2 2p 6 3s 1 (ii) 1s 2 2s 2 2p 6 3s 2 (iii) 1s 2 2s 2 2p 6 3s 2 3p 1 (iv) 1s 2 2s 2 2p 6 3s 2 3p 4 (v) 1s 2 2s 2 2p 6 3s 2 3p 5 The electron configuration of the atom that is expected to have a positive electron affinity is ________.
Answer:
(V) 1s 2 2s 2 2p 6 3s 2 3p 5
Explanation:
Electron Affinity can be defined as the energy associated with a neutral atom, when an electron is added to form a negative ion.
1s 2 2s 2 2p 6 3s 2 3p 5 , is the electronic configuration of Chlorine.
The electron affinity is positive because it is an exorthermic reaction, meaning that, energy was released during the addition of an electron to the atom.
Cl (g) + e- -------> Cl- (g) = -349KJ/mol
g + In a coffee-cup calorimeter, when 3.25 g of NaOH is dissolved in 50.00 g of water initially at 22.0 oC, the temperature of the solution increases to 24.8 oC. Calculate the q of the reaction in kJ/mol of NaOH. Assume that the specific heat of the solution is 4.184 J/g oC, and that there is no other heat transfer than that of the solution process itself.
Answer:
THE STANDARD HEAT OF SOLUTION OF SODIUM HYDROXIDE IN WATER IS -7.68 KJ PER MOLE.
Explanation:
Variables:
Mass of NaOH = 3.25 g
Mass of water = 50 g
Initial temperature of water = 22°C = 22 + 273 K = 295 K
Final temperature of the reaction mixture = 24.8 °C = 24.8 + 273 K = 297.8 K
Assuming that:
1. specific heat of water = 4.184 J/g °C
2. total mass of the reaction mixture = 50 g + 3.25 g = 53.25 g
3. the rise in temperature = (297.8 K - 295 K ) = 2.8 K
4. Molar mass of sodium hydroxide = ( 23 + 16 + 1) = 40 g/mol
5. number of mole of sodium hydroxide = mass / molar mass
n = 3.25 g / 40 g/mol
n = 0.08125 moles
The rise in temperature for the reaction mixture produces how much of heat:
Heat = mass * specific heat * change in temperature
Heat = 53.25 * 4.184 * 2.8
Heat = 623.8344 J of heat.
Equation of reaction:
NaOH + H2O -------> NaOH + H2O + Heat
This is not a reaction but a dissolution as sodium hydroxide is very soluble in water and this reaction is exothermic where heat is given off.
So since 3.25 g having 0.08125 moles produces 623.8344 J of heat, 1 mole of the sodium hydroxide used will produce:
0.08125 mole of sodium hydroxide = 623.8344 J of heat
1 mole of sodium hydroxide = ( 623.8344 / 0.08125 J of heat
= 7677.96 J of heat per mole of sodium hydroxide.
= 7.68 kJ of heat
So therefore, the standard heat of solution of sodium hydroxide in water is -7.68 kJ of heat since its an exothermic reaction.
A salt contains a compound containing a metal and a nonmetal. This is an
compound. *
A.molecular
B.covalent
C.Tonic
D.None of the above
Answer:
Option C. Ionic
Explanation:
Ionic bond is formed between a metal and a non metal.
Ionic bond occurs when there is a transfer of electron(s) from a metallic atom to a non metallic atom.
Calculate the final temperature of 12.0 g of Argon (considered as an ideal gas) that is expanded reversibly and adiabatically from V1 = 1.00 dm3 at T1 = 273.15 K to V2 = 3.00 dm3 . CV,m(Ar) = (3/2) * R.
Answer:
Explanation:
For adiabatic change the expression is
= [tex]PV^\gamma=constant[/tex]
[tex]=(\frac{RT}{V})V^\gamma = constant[/tex]
[tex]=TV^{\gamma-1} = constant[/tex]
[tex]=T_1V_1^{\gamma-1} =[/tex][tex]=T_2V_2^{\gamma-1}[/tex]
for Argon γ = 1.67
273.15 x 1 = T₂ x [tex]3^{1.67-1}[/tex]
T₂ = 273.15 / [tex]3^{0.67}[/tex]
= 273.15/ 2.0877
= 130.83 K.
Where does the stored energy in these cabbage leaves come from ?
Answer:
The energy in cabbage leaves comes from the light from the sun.
Explanation:
The plant keeps the light (in the leaves) as energy to help make it grow.
A titanium cube contains 3.10•10^23 atoms. The density of a titanium is 4.50g/cm^3. What is the edge length of the cube?
PLEASE HELPPP! :(
Answer:
1.76cm
Explanation:
We'll begin by calculating the mass of titanium that contain 3.10x10²³ atoms. This can be obtained as follow:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02x10²³ atoms. This implies that 1 mole of titanium also contains 6.02x10²³ atoms.
1 mole of titanium = 48g
Now, if 48g of titanium contains 6.02x10²³ atoms,
Then Xg of titanium will contain 3.10x10²³ atoms i.e
Xg of titanium = (48x3.10x10²³)/6.02x10²³
Xg of titanium = 24.72g
Next, we shall determine the volume of the titanium. This is illustrated below:
Density of titanium = 4.50g/cm³
Mass of titanium = 24.72g
Volume of titanium =..?
Density = Mass /volume
Volume = Mass /Density
Volume of titanium = 24.72g/4.50g/cm³
Volume of titanium = 5.49cm³
Finally we shall determine the edge length of the titanium cube as follow:
Volume = L³
5.49cm³ = L³
Take the cube root of both side
L = 3√(5.49cm³)
L = 1.76cm
Therefore, the edge length is 1.76cm
which of the following objects have kinetic energy? check all that apply
Answer:
what are the options?
Explanation:
do you have multiple choice for this question?
Answer:
A book inside a vehicle, a running cat, and a falling meteorite.
Explanation:
I just took it
Combustion analysis of a 13.42-g sample of an unknown organic compound (which contains only carbon, hydrogen, and oxygen) produced 36.86 g CO2 and 10.06 g H2O. The molar mass of the compound is 288.38 g/mol . Part A Find the molecular formula of the unknown compound. Express your answer as a chemical formula.
Answer:
[tex]C_{18}H_{24}O_3[/tex]
Explanation:
Hello,
In this case, combustion analyses help us to determine the empirical formula of a compound via the quantification of the released carbon dioxide and water since the law of conservation of mass is leveraged to attain it. In such a way, as 36.86 g of carbon dioxide were obtained, this directly represents the mass of carbon present in the sample, thus, we first compute the moles of carbon:
[tex]n_{C}=36.86gCO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molC}{1molCO_2} =0.838molC[/tex]
Then, into the water one could find the moles of hydrogen:
[tex]n_{H_2O}=10.06gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O} =1.12molH[/tex]
Now, we compute the moles of oxygen by firstly computing the mass of oxygen:
[tex]m_{O}=13.42g-36.86gCO_2*\frac{12gC}{44gCO_2} -10.06 gH_2O*\frac{2gH}{18gH_2O} =2.25gO[/tex]
[tex]n_O=2.25gO*\frac{1molO}{16gO} =0.141molO[/tex]
Then, we have the mole ratio:
[tex]C_{0.838}H_{1.12}O_{0.141}\rightarrow C_6H_8O[/tex]
Whose molar mass is 12x6+1x8+16=96 g/mol, but the whole compound molar mass is 288.38, the factor is 288.38/96 =3, Therefore the formula:
[tex]C_{18}H_{24}O_3[/tex]
Regards.
Answer:
Molecular mass = [tex]C_{6} H_{8} O_{16}[/tex]
Explanation:
[tex](\frac{36.86gCO_{2} }{44g/molCO_{2} })[/tex] [tex](\frac{10.06gH_{2}O}{18g/molH_{2}O} )[/tex]
(0.8377mol of CO2) (0.556 mol of H2O)
mole ratio of CO2 to H2O = 1.5 : 1 ≅ 3 : 2
[tex](CO_{2} )_{3} , (H_{2} O)_{2}[/tex]
⇒ [tex](C_{3} H_{4} O_{8} )_{x}[/tex] = 288.38
(12 × 3 + 1 × 4 + 16 × 8)x = 288.38
168x = 288.38
x = 1.71 ≅ 2
∴ Molecular mass = [tex]C_{6} H_{8} O_{16}[/tex]
The H 2 produced in a chemical reaction is collected through water in a eudiometer. If the pressure in the eudiometer is 760.0 torr and the vapor pressure of water under the experimental conditions is 23.8 torr, what is the pressure (torr) of the H 2 gas
Answer:
the pressure (torr) of theH₂ gas is 736.2 torr
Explanation:
Given that
The H₂ gas produced in a chemical reaction is collected through water in a eudiometer; during this process, the gas collected contains some droplets of water vapor along with these gas.
So; the total pressure in the eudiometer = Pressure in the H₂ gas - Pressure of the water vapor
Where;
[tex]P_{Totsl}[/tex] = total pressure in the eudiometer = 760.0 torr
[tex]P_{H_2}[/tex] = Pressure in the H₂ gas = ???
[tex]P_{H_2O}[/tex] = Pressure in the water vapor = 23.8 torr
Now:
[tex]P_{Totsl}[/tex] = [tex]P_{H_2}[/tex] + [tex]P_{H_2O}[/tex]
- [tex]P_{H_2}[/tex] = + [tex]P_{H_2O}[/tex] - [tex]P_{Totsl}[/tex]
[tex]P_{H_2}[/tex] = - [tex]P_{H_2O}[/tex] + [tex]P_{Totsl}[/tex]
[tex]P_{H_2}[/tex] = (- 23.8 + 760) torr
[tex]P_{H_2}[/tex] = 736.2 torr
Thus; the pressure (torr) of theH₂ gas is 736.2 torr
If the pressure in a eudiometer is 760.0 torr and the vapor pressure of water is 23.8 torr, the pressure of the H₂ gas is 736.2 torr.
A eudiometer is a laboratory device that measures the change in volume of a gas mixture following a physical or chemical change. Usually, the gas is collected over water.
According to Dalton's law of partial pressures, the sum of the sum partial pressure of the hydrogen and the partial pressure of the water is equal to the total pressure in the eudiometer.
[tex]P = pH_2 + pH_2O\\\\pH_2 = P - pH_2O = 760.0 torr - 23.8 torr = 736.2 torr[/tex]
If the pressure in a eudiometer is 760.0 torr and the vapor pressure of water is 23.8 torr, the pressure of the H₂ gas is 736.2 torr.
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