Answer:
Molar mass is the mass of one mole of a substance and is ontained by adding atomic mases of all the constituent atoms and group of atoms.
Atomic masses: Cu = 63.55 g/mol,
S = 32.07 g/mol
O x 4 = 16.00 g/mol x 4 = 64.00 g/mol
H = 1.008 g/mol,
So, H2O = (2*1.008+16.00)*5 = 90.10 g/mol
Molar mass of CuSO4,5H2O = 249.72 g/mol.
Out of that 249.72 grams, 90.10 grams is water,
So, percentage by mass of water in CuSO4,5H2O
= 90.10*100/249.72 = 36.08.
I think this is what you are asking. hope this helps
Draw a structure for an alcohol that exhibits a molecular ion at M+ = 88 and that produces fragments at m/z = 73, m/z = 70 and m/z = 59.
Answer:
3-pentanol
Explanation:
In this case, we have alcohol as the main functional group (OH) with a molecular ion at 88. If the molecular ion is 88 the molar mass is also 88 g/mol therefore the formula for the unknown molecule is [tex]C_5H_1_2O[/tex].
Additionally, if the mass spectrum shows the molecular ion peak we can not have tertiary alcohols (tertiary alcohols often do not show M+ at all). So, the structures only can be primary and secondary structures.
With this in mind, our options are:
-) 1-pentanol
-) 2-pentanol
-) 3-pentanol
Now we can analyze each structure:
-) 1-pentanol
The structure must explain all the fragments produced (73, 70, and 59). In this primary alcohol, we will have an alpha cleavage (the red bond would be broken). If this has to happen, we will have fragments at 31 and 57. These fragments dont fit with the reported ones, therefore this is not a possible structure (See figure 1).
-) 2-pentanol
On this structure, we will have also an alpha cleavage (red bond). In this rupture we will have fragments at 45 and 43, these m/z values dont fit with the reported ones, therefore this is not a possible structure (See figure 1).
-) 3-pentanol
In this structure, we have the "OH" bonded to carbon three. So, we can analyze each fragment:
-) m/z 59
This fragment, can be explained as an alpha cleavage. But, in this case we have two ruptures that can produce the same ion. The carbons on both sides of the C-OH bond.
-) m/z 71
This fragment, can be explained as a loss of water (M-18) in which we have the production of a carbocation in the carbon where we previously have the C-OH bond.
-) m/z 73
This fragment, can be explained as a beta cleavage. But, in this case, also we have two ruptures that can produce the same ion. The methyl groups on each end molecule.
See figure 2
I hope it helps!
A sample of gas is held at constant volume. If the number of moles of this sample of gas is doubled and the pressure of this sample of gas is halved, what happens to the absolute temperature of the gas?
Select one
a. The absolute temperature is doubled.
b. The absolute temperature is halved.
c. The absolute temperature is quadrupled.
d. The absolute temperature is quartered.
e. The absolute temperature stays the same.
Answer:
number of moles of gas increases the volume also increases.
What type of bond is present in NBr?
Answer:
Covalent bonding and non-covalent bonding
In the laboratory, a general chemistry student measured the pH of a 0.425 M aqueous solution of benzoic acid, C6H5COOH to be 2.270.
Use the information she obtained to determine the Ka for this acid.
Ka(experiment) = _____
Answer:
Ka = 6.87x10⁻⁵
Explanation:
The equilibrium of benzoic acid in water is:
C₆H₅COOH(aq) + H₂O(l) ⇄ C₆H₅COO⁻(aq) + H₃O⁺(aq)
The equilibrium constant, Ka, is:
Ka = [C₆H₅COO⁻] [H₃O⁺] / [C₆H₅COOH]
The initial concentration of benzoic acid is 0.425M. In equilibrium its concentration is 0.425M - X and [C₆H₅COO⁻] [H₃O⁺] = X.
X is the reaction coordinate. How many acid produce C₆H₅COO⁻ and H₃O⁺ until reach equilibrium.
Concentrations in equilibrium are:
[C₆H₅COOH] = 0.425M - X[C₆H₅COO⁻] = X [H₃O⁺] = XpH is defined as -log [H₃O⁺]. As pH = 2.270
2.270 = -log [H₃O⁺]
10^-2.270 = [H₃O⁺]
5.37x10⁻³M = [H₃O⁺] = X.
Replacing, concentrations in equilibrium are:
[C₆H₅COOH] = 0.425M - 5.37x10⁻³M = 0.4196M
[C₆H₅COO⁻] = 5.37x10⁻³M
[H₃O⁺] = 5.37x10⁻³M
Ka = [5.37x10⁻³M] [5.37x10⁻³M] / [0.4196M]
Ka = 6.87x10⁻⁵A closed-end manometer was attached to a vessel containing argon. The difference in the mercury levels in the two arms of the manometer was 9.60 cm. Atmospheric pressure was 783 mm Hg. The pressure of the argon in the container was ________ mm Hg.
Answer:
96 mmHg
[tex]h=96mmHg[/tex]
Explanation:
From this question,manometer end is closedw, So we can deduced that the height of the column will not be affected by the atmospheric pressure .
The difference of height of the mercury level is given as,
h=9.60cm
h=9.60(10mm/1cm)
[tex]h=96mm[/tex]
But it is obvious that in this closed end manometer.the pressure of the gas is equal to the height
P(gas)=h
P(gas)=96mmHg
This pressure is as a result of the presence of gas.
Therefore, the pressure of the argon gas in the container is 96mmHg.
The pressure of the argon in the container was 96mmHg.
We were told that the manometer has closed ends which means that the
height will not be affected by atmospheric pressure.
The height which is the difference in mercury level is
h=9.60cm
We can convert it to millimeter by multiplying it by 10
h=9.60 × 10 = 96mm
The pressure of the closed end manometer will be equal to the height
P(gas)=h
P(gas)=96mmHg
The pressure of the argon gas in the container is 96mmHg.
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The direction of the functional group is called?
Explanation:
they are called hydrocarbyls
pls mark me brainliest
Answer:
The first carbon atom that attaches to the functional group is referred to as the alpha carbon.
Methanol liquid burns readily in air. One way to represent this equilibrium is: 2 CO2(g) + 4 H2O(g)2 CH3OH(l) + 3 O2(g) We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above. 1) CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(g) K1 = 2) CO2(g) + 2 H2O(g) CH3OH(l) + 3/2 O2(g) K2 = 3) 2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(g)
Answer:
Answers are in the explanation
Explanation:
It is possible to obtain K of equilibrium of related reactions knowing the laws:
A + B ⇄ C K₁
C ⇄ A + B K = 1 /K₁
The inverse reaction has the inverse K equilibrium
2A + 2B ⇄ 2C K = K₁²
The multiplication of the coefficients of reaction produce a k powered to the number you are multiplying the coefficients
For the reaction:
2 CO2(g) + 4 H2O(g) ⇄ 2 CH3OH(l) + 3 O2(g) K
1) CH3OH(l) + 3/2 O2(g) ⇄ CO2(g) + 2 H2O(g)
This is the inverse reaction but also the coefficients are dividing in the half, that means:
[tex]K_1 = \frac{1}{k^{1/2}} = (1/K)^{1/2}[/tex]
2) CO2(g) + 2 H2O(g) ⇄ CH3OH(l) + 3/2 O2(g)
Here,the only change is the coefficients are the half of the original reaction:
[tex]K_2 = K^{1/2}[/tex]
3) 2CH3OH(l) + 3 O2(g) ⇄ 2 CO2(g) + 4 H2O(g)
This is the inverse reaction. Thus, you have the inverse K of equilibrium:
[tex]K_3 = \frac{1}{K}[/tex]
Notice
10 January 2018
Gift A Tree
This is to inform students of Class XII to assemble in the ground on Saturday, 31 January
2018 at 8:00 in the morning to participate in a Tree Plantation Ceremony being organised
by the Environment Club as a part of the Farewell Programme.
Amita/Amit
Secretary, Environment Club)
Exercises
As the Head Boy/ Head Girl, Central Public School, draft a notice informing all students about
a wrist watch that was found near the school canteen.answar
Answer:
gahwidsuacsgsuacayau1joagavahiq8wtw8quavakiafabajozyavqhaigavayquata
Explanation:
vahaiqgahiavavqugafayqigqvsbjsiagwyeiwvvs
For each of the processes, determine whether the entropy of the system is increasing or decreasing.
a. A snowman melts on a spring day.
b. A document goes through a paper shredder.
c. A water bottle cools down in a refrigerator.
d. Silver tarnishes
e. Dissolved sugar precipitates out of water to form rock candy.
Explanation:
Entropy refers to the degree of disorderliness of a system.
a. A snowman melts on a spring day.
Entropy is increasing because there is a change in state of matter from solid to liquid. Liquid particles have more freedom f movement compared to solids.
b. A document goes through a paper shredder.
Entropy increases because random, disorganized bits of paper are left.
c. A water bottle cools down in a refrigerator.
Entropy decreases because temperature is directly proportional to entropy.
d. Silver tarnishes
Entropy increases because random bits of the sliver particles are formed.
e. Dissolved sugar precipitates out of water to form rock candy.
Entropy decreases because the random dissolved sugar precipitates are ordered into a rock candy.
Select the correct answer.
Which state of matter is highly compressible, is made of particles moving independently of each other, and is present in large quantities near Earth’s surface?
A.
solid
B.
liquid
C.
gas
D.
plasma
Answer:
C. Gas
Explanation:
A chemist prepares a solution of silver(I) nitrate(AgNO3) by measuring out 269. mu mol of silver(I) nitrate into a 300. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mmol/L of the chemist's silver(I) nitrate solution.
Answer:
concentration in mmol/L = 8.97 × 10⁻¹ mmol/L
Explanation:
Given that:
the number of moles of silver(I) nitrate(AgNO3) the chemist used in preparing a solution = 269 mmol = 269 × 10⁻³ mmol
The volume of the volumetric flask = 300 mL = 300 × 10⁻³ L
In order to calculate the concentration in mmol/L of the chemist's silver(I) nitrate (AgNO3) solution , we used the formula which can be expressed as;
[tex]concentration \ in \ mmol/L = \dfrac{ number \ of \ mmol}{vol. \ of \ the \ solution}[/tex]
[tex]concentration \ in \ mmol/L = \dfrac{ 269 * 10^{-3 } \ mmol }{300 * 10^{-3} \ L }[/tex]
concentration in mmol/L = 0.8966 mmol/L
concentration in mmol/L = 8.97 × 10⁻¹ mmol/L
If you add a solution of NaOH to a solution of H₂CO₃, two reactions occur, one after the other. Write the chemical equations for these two reactions. (Hint: NaOH dissociate into Na+ and OH-, and the hydroxide ion is the actual base).
We have a solution of NaOH and H₂CO₃
First, NaOH will dissociate into Na⁺ and OH⁻ ions
The Na⁺ ion will substitute one of the Hydrogen atoms on H₂CO₃ to form NaHCO₃
The H⁺ released from the substitution will bond with the OH⁻ ion to form a water molecule
If there were to be another NaOH molecule, a similar substitution will take place, substituting the second hydrogen from H₂CO₃ as well to form Na₂CO₃
In the reaction A + B + C + D, what are the reactants?
O A. Just B
B. Cand D
O c. A and B
O D. A and C
Answer:
C.
Explanation:
I believe that it should be A and B.
A 135 g sample of H20 at 85°C is cooled. The water loses a total of 15 kJ of energy in the cooling
process. What is the final temperature of the water? The specific heat of water is 4.184 J/g.°C.
A. 112°C
B. 58°C
C. 70°C
D. 84°C
E. 27°C
Answer:
B. 58°C
Explanation:
Hello,
In this case, the relationship among heat, mass, specific heat and temperature for water is mathematically by:
[tex]Q=mCp\Delta T=mCp(T_2-T_1)[/tex]
In such a way, solving for the final temperature [tex]T_2[/tex] we obtain:
[tex]T_2=T_1+\frac{Q}{mCp}[/tex]
Therefore, we final temperature is computed as follows, considering that the involved heat is negative as it is lost for water:
[tex]T_2=85\°C+\frac{-15kJ*\frac{1000J}{1kJ} }{135g*4.184\frac{J}{g\°C} }\\\\T_2=58\°C[/tex]
Thereby, answer is B. 58°C .
Regards.
In the experiment, you heated a sample of metal in a water bath and quickly removed and dried the sample prior to inserting it into the cold water "system". Why did you need to dry the sample prior to placing it into the cold water "system"
a. The extra water might react with the metal which would ruin the sample.
b. Any residual hot water would have added to the heat transferred from the metal to the cold water "system".
c. The wet metal would not transfer any heat thus causing inaccuracy in you measurements.
d. The metal would oxidize in the presence of water thus ruining the sample
Answer:
b
Explanation:
Provided the experiment is about the transfer of heat from one medium to another, any residual hot water on the metal would have added to the heat transferred from the metal to the cold water system.
The residual hot water on the metal posses its own heat and when such is transferred along with the metal into the cold water, the heat from the residual hot water will interfere with the measurement of the actual heat transferred to the cold water by the metal. Hence, the accuracy of the result would be impacted.
The correct option is b.
I need to name an ionic compound containing a transition metal cation and a halogen anion. Below are the rules I should follow to write the correct name for such compound, but one of the options is incorrect: identify and select it.
a. Identify the metal and write its name first
b. Use the periodic table to work out the charge (oxidation number) on the transition metal according to the group in the periodic table
c. From the charge of the anion work out the charge of cation as Roman number in parenthesis: specify this charge in the name as a Roman number in parenthesis.
d. Write the number of the anion after the name of the metal
Answer:
b. Use the periodic table to work out the charge (oxidation number) on the transition metal according to the group in the periodic table
Explanation:
The keyword in this problem us "transition metal". Transition metals are found between the group 2 and group 3 elements. They have the d sub shells and also exhibit variable oxidation numbers (valency).
Among the options, the incorrect option is option B.
This is because transition metals d not have a fixed oxidation number and they cannot be obtained by looking up the group in the periodic table.
The iconic compounds obtain a transition of metal caution and a halon anon. As per the rules the correct name of the compounds should be written as to identify the incorrect one.
Option B use the ability to check and to work out the charges (oxidation number) of the transition metal as per the group given in the table. The problem with the keyword is transition metal.Learn more about the ionic compound containing a transition metal.
brainly.com/question/21578354.
Draw a picture of what you imagine solid sodium chloride looks like at the atomic level. (Do NOT draw Lewis structures.) Make sure to include a key. Then describe what you've drawn and any assumptions you are making.
Answer:
Kindly check the explanation section.
Explanation:
PS: kindly check the attachment below for the required diagram that is the diagram showing solid sodium chloride looks like at the atomic level.
The chemical compound known as sodium chloride, NaCl has Molar mass: 58.44 g/mol, Melting point: 801 °C and
Boiling point: 1,465 °C. The structure of the solid sodium chloride is FACE CENTRED CUBIC STRUCTURE. Also, solid sodium chloride has a coordination number of 6: 6.
In the diagram below, the positive sign shows the sodium ion while the thick full stop sign represent the chlorine ion.
The NaCl has been the ionic structure with an equal number of sodium and chlorine ions bonded.
In the structure, there has been each Na ion bonded with the Cl ions. There has been the transfer of electrons between the structure in order to attain a stable configuration.
The expected structure of the NaCl would be the image attached below.
The image has been the cubic structure of NaCl. With the presence of Na ions at the vertex of the structure, there has been the presence of the Cl ion with every Na ion for the electron transfer.
For more information about the structure of NaCl, refer to the link:
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Q1: A stock solution containing Mn+2 ions was prepared by dissolving 1.584 g pure manganese metal in nitric acid and diluting to a final volume of 1.000 L. The following solutions were then prepared by dilution: For solution A, 50.00 mL of stock solution was diluted to 1000.0 mL. For solution B, 10.00 mL of solution A was diluted to 250.0 mL. For solution C, 10.00 mL of solution B was diluted to 500.0 mL. Calculate the concentrations of the stock solution and solutions A, B, and C.
The concentration of a solution is the amount of solute in a solution.
We have from the question that he mass of manganese = 1.584 g
Hence;
Amount of [tex]Mn^2+[/tex] = 1.584 g/55g/mol = 0.0288 moles
Recall that;
Number of moles = concentration * volume
Let the concentration of the solution be C
0.0288 moles = C * 1 L
C = 0.0288 moles/ 1 L
C= 0.0288 mol/L
Hence concentration of stock solution = 0.0288 mol/L
For solution A
From the dilution formula;
C1V1 = C2V2
where;
C1 = initial concentration
C2 = final concentration
V1 = initial volume
V2= final volume
C1 = 0.0288 mol/L
V1 = 50.00 mL
C2 = ?
V2 = 1000.0 mL
C2 = 0.0288 mol/L * 50.00 mL/1000.0 mL
C2 = 0.00144 mol/L
Hence, concentration of solution A is 0.00144 mol/L
For solution B
C1V1 = C2V2
C1 = 0.00144 mol/L
V1 = 10.00 mL
C2 = ?
V2 = 250.0 mL
C2 = 0.00144 mol/L * 10.00 mL/250.0 mL
C2 = 0.0000576 mol/L
Hence, concentration of solution B is 0.0000576 mol/L
For solution C
C1 = 0.0000576 mol/L
V1 = 10.00 mL
C2 = ?
V2 = 500.0 mL
C2 = 0.0000576 mol/L * 10.00 mL/500.0 mL
C2= 0.000001152 mol/L
Hence, concentration of solution C is 0.000001152 mol/L
To learn more about concentration
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A 100.0-mL sample of 1.00 M NaOH is mixed with 50.0 mL of 1.00 M H2SO4 in a large Styrofoam coffee cup calorimeter fitted with a lid through which a thermometer passes. The acid-base reaction is as follows:
2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l)
The temperature of each solution before mixing is 22.3 °C. After mixing, the temperature of the solution mixture reaches a maximum temperature of 31.4 °C. Assume the density of the solution mixture is 1.00 g/mL, its specific heat is 4.18 J/g.°C, and no heat is lost to the surroundings. Calculate the enthalpy change, in kj, per mole of H2SO4 in the reaction.
a. +85.6 kJ/mol.
b. -85.6 kJ/mol.
c. +5.71 kJ/mol.
d. -5.71 kJ/mol.
e. -114 kJ/mol.
Answer:
THE ENTHALPY CHANGE IN KJ/MOLE IS +114 KJ/MOLE.
Explanation:
Heat = mass * specific heat capacity * temperature rise
Total volume = 100 + 50 = 150 mL
Total mass = density * volume
Total mass = 1 * 150 mL = 150 g
So therefore, the heat evolved during the reaction is:
Heat = 150 * 4.18 * ( 31.4 - 22.3)
Heat = 150 * 4.18 * 9.1
Heat = 5705.7 J
Equation for the reaction:
2 NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2 H2O(l)
From the equation, 2 moles of NaOH reacts with 1 mole of H2SO4 to produce 1 mole of Na2SO4 and 2 moles of water
50 mL of 1 M of H2SO4 contains
50 * 1 / 1000 mole of acid
= 0.05 mole of acid
The production of 1 mole of water evolved 5705.7 J of heat and hence the enthalpy changein kJ per mole will be:
0.05 mole of H2SO4 produces 5705.7 J of heat
1 mole of H2SO4 will produce 5705.7 / 0.05 J
= 114,114 J / mole
In kj/mole = 114 kJ/mole.
Hence, the enthalpy change of the reaction in kJ /mole is +114 kJ/mole.
Which of the following elements can't have an expanded octet? answers A. oxygen B. phosphorous C. chlorine d. sulfer
answer is oxygen .
oxygen is an exception in octet rule
Among the following given elements,oxygen is an element which cannot have an expanded octet.
What is an expanded octet?Expanded octet is a condition where an octet has more than 8 electrons and which is called as hyper-valency state. This concept is related to hybrid orbital theory and Lewis theory. Hyper-valent compounds are not less common and are of equal stability as the compounds which obey octet rule.
Expansion of octet is possible for elements from third period on wards only as they have low-lying empty d - orbitals which can accommodate more than eight electrons.
Expanded octet is not applicable to oxygen as it is second period of periodic table and has less than ten electrons and even does not have the 2d -orbitals due to which it does not fulfill the criteria of an element to have an expanded octet.
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15. How many moles of carbon tetrachloride (CCI) is represented by 543.2 g of carbon tetrachloride? The atomic weight of carbon is 12.01
and the atomic weight of chlorine is 35.45.
O A. 11.4 moles
O B.3.53 moles
C. 5.42 moles
D. 8.35x10 moles
Answer:
well, first off. the formula for carbon tetrachloride is CCl4
We need to find the molar mass of carbon tetrachloride
1(Mass of C) + 4(mass of chlorine)
1(12) + 4(35.5)
12 + 142
154 g/mol
Number of moles of CCl3 in 543.2g CCl3
n = given mass / molar mass
n = 543.2/153
n = 3.53 moles
always remember to brainly the questions you find helpful
Answer:
3.53 moles
Explanation:
Account for the change when NO2Cl is added using the reaction quotient Qc. Match the words in the left column to the appropriate blanks in the sentences on the right.
1. decreases
2. loss
3. Increases
4. greater
A. Disturbing the equilibrium by adding NO2Cl______Qc to a value_____than Kc.
B. To reach a new state of equilibrium, Qc therefore______which means that the denominator of the expression for Qc______.
C. To accomplish this, the concentration of reagents______, and the concentration of products_______.
Answer:
A. Disturbing the equilibrium by adding NO2Cl decreases Qc to a value less than Kc.
B. To reach a new state of equilibrium, Qc therefore increases which means that the denominator of the expression for Qc decreases.
C. To accomplish this, the concentration of reagents decreases, and the concentration of products increases.
Explanation:
Hello,
In this case, for the equilibrium reaction:
[tex]NO_2Cl(g)+NO(g)\rightleftharpoons NOCl(g)+NO_2(g)[/tex]
Whose equilibrium expression is:
[tex]Kc=\frac{[NO_2][NOCl]}{[NO_2Cl][NO]}[/tex]
The proper matching is:
A. Disturbing the equilibrium by adding NO2Cl decreases Qc to a value less than Kc, since the denominator becomes greater, therefore, Qc decreases.
B. To reach a new state of equilibrium, Qc therefore increases which means that the denominator of the expression for Qc decreases, since the lower the denominator, the higher Qc as it has the concentration of reactants.
C. To accomplish this, the concentration of reagents decreases, and the concentration of products increases, since the reactants must be consumed in order to reestablish equilibrium by shifting the reaction towards the products.
Best regards.
The half-life of Zn-71 is 2.4 minutes. If one had 100.0 g at the beginning, how many grams would be left after 7.2 minutes has elapsed? Report your answer to 1 decimal place.
Answer:
12.5g
Explanation:
Half life = 2.4 Minutes.
The half life of a compound is the time it takes to decay to half of it's original concentration or mass.
Time lapsed= 7.2 minutes. This is equivalent to 3 half lives ( 3 * 2.4)
Initial mass = 100g
First half life;
100g --> 50g
Second half life;
50g --> 25g
Third half life;
25g --> 12.5g
The amount of Zn-71 that remains after 7.2 mins has elapsed is 12.5 g
We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:
Half-life (t½) = 2.4 mins
Time (t) = 7.2 mins
Number of half-lives (n) =?[tex]n = \frac{t}{t_{1/2}} \\\\n = \frac{7.2}{2.4} \\\\[/tex]
n = 3Thus, 3 half-lives has elapsed.
Finally, we shall the amount remaining. This can be obtained as follow:Original amount (N₀) = 100 g
Number of half-lives (n) = 3
Amount remaining (N) =?[tex]N = \frac{N_{0}}{2^{n}} \\\\N = \frac{100}{2^{3 }}\\\\N = \frac{100}{8}\\\\[/tex]
N = 12.5 gThus, the amount of Zn-71 that remains after 7.2 mins is 12.5 g
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What's the concentration of hydronium ions if a water-base solution has a temperature of 25°C (Kw = 1.0×10–14), with a concentration of hydroxide ions of 2.21×10–6 M? answer options: A) 2.8×10–8 M B) 4.52 ×10–9 M C) 1.6×10–9 M D) 3.1×10–6 M
Answer:
ITS NOT D. ITS B. 4.52x10^-9 M
Explanation:
Answer:
4.52 ×10–9 M
Explanation:
what are the five main points of kinetic theory of gas?
The kinetic-molecular theory of gases assumes that ideal gas molecules
(1) are constantly moving;
(2) have negligible volume;
(3) have negligible intermolecular forces;
(4) undergo perfectly elastic collisions; and
(5) have an average kinetic energy proportional to the ideal gas's absolute temperature.
The five main postulates of the KMT are as follows:
(1) the particles in a gas are in constant, random motion,
(2) the combined volume of the particles is negligible
(3) the particles exert no forces on one another,
(4) any collisions between the particles are completely elastic.
(5) the average kinetic energy of the particles is proportional to the temperature in kelvins.
A base which can be used to relieve indigestion
Explanation:
Antacids are medications used to manage the symptoms of indigestion and heartburn. Antacids contain active ingredients that are bases. These allow antacids to neutralize any stomach acid which could be causing digestive discomfort.
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]
Compared to an atom of C-12, an atom of C-14 has
A) fewer protons
C) more neutrons
B) fewer neutrons
D) more protons
Explanation:
Carbon-12 atoms have stable nuclei because of the 1:1 ratio of protons and neutrons.
Carbon-14 atoms have nuclei which are unstable. C-14 atoms will undergo alpha decay and produce atoms of N-14. Carbon-14 dating can be used to determine the age of artifacts which are not more than 50,000 years old.
What is the change in the freezing point of a solution made by dissolving 14.7 g of
C6H1206 into 150 ml of water? Answer in 2 decimal.
The density of water is 1.00 g/ml and and Kf = 1.86 °C/m
The change in freezing point is 1.02°C
The change in the freezing point (ΔTf) is calculated from the following equation:
ΔTf = i x Kf x m
Given:
i = 1 (Van't Hoff factor, is equal to 1 because the solute is a nonelectrolyte)
Kf = 1.86°C/m (Freezing point constant of the solvent)
m = molality
We have to calculate the molality (m), which is equal to the moles of solute in 1 kg of solvent:
m = moles solute/ kg solvent
The solute is C₆H₁₂O₆. Thus, we calculate the moles of solute by dividing the mass into its molecular weight (Mw):
Mw(C₆H₁₂O) = (12 g/mol C x 6) + ( 1 g/mol H x 12) + (16 g/mol O x 6) = 180 g/mol
moles solute = mass C₆H₁₂O/Mw(C₆H₁₂O) = 14.7 g/(180 g/mol) = 0.082 mol
The solvent is water. Its mass (in kg) is obtained from the volume and the density, as follows:
kg solvent = V x density = 150 mL x 1.00 g/mL x 1 kg/1000 g = 0.15 kg
Now, we calculate the molality (m):
m = moles of solute/kg solvent = 0.082 mol/(0.15 kg) = 0.546 m
Finally, we calculate the change in freezing point:
ΔTf = i x Kf x m = 1 x 1.86°C/m x 0.546 m = 1.02°C
To learn more about freezing point depression, you can visit this content:
https://brainly.com/question/4258941
Which of the following is not the same as 1,400 mL? a. 1.4 cm³ b 1.4 L c. 1,400 cm³ d. 140 cL
answer should be 1.4 cm³
1 L = 10 and so
dL = 100 and then
cL = 1,000
mL = 0.001 m³
1 m³ = 1,000
dm³ = 1,000,000
cm³ = 1,000,000,000
mm³ = 1,000 L
So, 1 mL = 1 cm³ = 0.001 L = 0.1 cL
1,400 mL = 1,400 cm³ = 1.4 L = 140 cL
Answer:
1.4 cm^3
Explanation: