Answer:
Melting-point apparatus
A quantity of 1.435 g of naphthalene , was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.28oC to 25.95oC If the heat capacity of the bomb plus water was , calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.
Answer:
molar heat of combustion = -5156 *10³ kJ/mol
Explanation:
A quantity of 1.435 g of naphthalene , was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.28oC to 25.95oC If the heat capacity of the bomb plus water was 10.17 kJ/°C, calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.
Step 1: Data given
Mass of naphthalene = 1.435 grams
Initial temperature of water = 20.28 °C
Final temperature of water = 25.95 °C
heat capacity of the bomb plus water was 10.17 kJ/°C
Molar mass naphtalene = 128.2 g/mol
Step 2:
Qcal = Ccal * ΔT
⇒with Qcal =the heat of combustion
⇒with Ccal = heat capacity of the bomb plus water = 10.17 kJ/°C
⇒with ΔT = the difference in temperature = T2 - T1 = 25.95 - 20.28 = 5.67°C
Qcal = 10.17 kJ/°C * 5.67 °C
Qcal = 57.7 kJ
Step 3: Calculate moles
Moles naphthalene = 1.435 grams / 128.2 g/mol
Moles naphthalene = 0.01119 moles
Step 4: Calculate the molar heat of combustion
molar heat of combustion = Qcal/ moles
molar heat of combustion = -57.7 kJ/ 0.01119 moles
molar heat of combustion = -5156 *10³ kJ/mol
Which compound contains both sigma and pi bonds... HCCl3, H2CO, H2S, or HBr?
Answer:
H2CO
Explanation:
Becuase it has 2 sigma bonds plus one pi bond and one sigma bond that consitute the double bond between C and O.
Answer:
B. H2CO
Explanation:
The mass of a single tantalum atom is 3.01×10-22 grams. How many tantalum atoms would there be in 37.1 milligrams of tantalum?
Answer: There are [tex]1.23 \times 10^{22}[/tex] atoms present in 37.1 mg of tantalum.
Explanation:
Given: Mass of single tantalum atom = [tex]3.01 \times 10^{-22} g[/tex]
Mass of tantalum atoms = 37.1 mg (1 mg = 0.001 g) = 0.0371 g
Therefore, number of tantalum atoms present in 0.0371 grams is calculated as follows.
[tex]No. of atoms = \frac{0.0371 g}{3.01 \times 10^{-22}}\\= 1.23 \times 10^{20}[/tex]
Thus, we can conclude that there are [tex]1.23 \times 10^{22}[/tex] atoms present in 37.1 mg of tantalum.
There are [tex]1.23\times 10^{20}[/tex] atoms of tantalum in 37.1 mg of tantalum.
Explanation:
Given:
Mass of single atom of tantalum =[tex]3.01\times 10^{-22} g[/tex]
To find:
The number of atoms of tantalum in 37.1 milligrams.
Solution:
Mass of tantalum = 37.1 mg
[tex]1 mg = 0.001 g\\37.1 mg=37.1\times 0.001 g=0.0371 g[/tex]
The number of atoms in 0.0371 grams of tantalum = N
Mass of a single atom of tantalum = [tex]3.01\times 10^{-22} g[/tex]
Then a mass of N atoms of tantalum will be:
[tex]0.0371 g=N\times 3.01\times 10^{-22} g\\N=\frac{0.0371 g}{ 3.01\times 10^{-22} g}\\=1.23\times 10^{20}[/tex]
There are [tex]1.23\times 10^{20}[/tex] atoms of tantalum in 37.1 mg of tantalum.
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What is the best tool for making a decorative zucchini or celery twist
Answer:
don't knoe sorry
Explanation:
Plz!!!!Plz!!!!!help help
Answer:
false
Explanation:
category 1 is the worst
The graph below shows how the temperature and volume of a gas vary when
the number of moles and the pressure of the gas are held constant. How can
the volume of the gas be increased if the pressure is constant?
v
т
A. By increasing the temperature
B. By letting the gas expand over time
C. By letting the gas contract over time
D. By decreasing the temperature
Answer:
D
Explanation:
PV =nRT
So V and T are inversely proportional
an emerald can be described as...
Answer:
green gemstone
Explanation:
hope this helps someone
The products obtained from hydroboration-oxidation of cis-2-butene are identical to the products obtained from hydroboration-oxidation of trans-2-butene. Draw the products and explain why the configuration of the starting alkene is not relevant in this case.
Answer:
a) Attached below
b) The presence of racemic mixture found as product in both cases shows that products are identical ( i.e. they have same configuration
Explanation:
Diagrams of the products obtained from hydroboration-oxidation of cis-2-butene , hydroboration-oxidation of trans-2-butene.
attached below
The presence of racemic mixture found as product in both cases shows that products are identical ( i.e. they have same configuration )
The Hammond Postulate describes the relationship between the energy of the transition state and the productdistribution in a reaction that is capable of following more than one pathway.
What are the appropriate labels for the sentence?
The Hammond postulate states that reactions which are thermodynamically endothermic and kinetically ___________ have transitions states that occur ___________ in the reacton time frame, and resemble the ___________ in terms of energy and structure.
following this list of common polyatomic ions what would be the charge for aluminum nitrate?
Nitrate NO3
Phosphate PO43,
Sulfate SO4 -2
acetate C2H3O2-1
Ammonium NH4.
Chromate CrO4-2
Carbonate C03-2
Dichromate CrO7-2
Permanganate MnO4-1
What would be the correct answer for Aluminum nitrate
Answer:
Nitrate NO3
here's your answer, hope it helps you
What is oxygen's half-equation?
answer; 1/ 20_2[2-] +2e - ->0.
Me please answer as follows
Answer:
no reaction occurs .that is no product
Rock, metal, wood, glass, animals, and plants are all forms of what?
What evidence supports the law of conservation of energy?
Light energy is converted to chemical energy during photosynthesis.
Oxygen is made from the breakdown of carbon dioxide during photosynthesis.
Energy is absorbed by chlorophyll during photosynthesis.
The sun gives off light energy that is absorbed by plants.
Answer:
light energy is converted to chemical energy during photosynthesis.
Explanation:
Law of conservation of energy says: Energy can neither be created nor destroyed-only converted from one form of energy to another.
Answer:
A. Light energy is converted to chemical energy during photosynthesis.
Identify the most oxidized compound. Group of answer choices CH3CH2CHO CH3CH2CH3 CH3CH2CH2OH CH3CH2OCH3 CH3CH2COOH
Answer:
Huh!?
Explanation:
explain me please
A chemist determines by measurements that 0.0800 moles of bromine liquid participate in a chemical reaction. Calculate the mass of bromine liquid that participates. Round your answer to 3 significant digits.
Answer:
The mass of bromine liquid that participates in a chemical reaction=12.8 g
Explanation:
We are given that
Total number of moles of bromine liquid participate in chemical reaction=0.0800 moles
We have to find the mass of bromine liquid that participates.
Atomic mass of Br=79.9 g
1 mole of bromine liquid=2 atomic mass of bromine (Br)
1 mole of bromine liquid ([tex]Br_2[/tex]) =[tex]2\times 79.9=159.8 g[/tex]
0.0800 moles of bromine liquid=[tex]159.8\times 0.0800[/tex] g
0.0800 moles of bromine liquid=12.784 g
0.0800 moles of bromine liquid[tex]\approx 12.8[/tex] g
Hence, the mass of bromine liquid that participates in a chemical reaction=12.8 g
what the movement of the earth around the sun
Answer:
Yan po Ang sagot NASA pic
Explanation:
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Answer:
The movement of the earth around the sun in a fixed path or orbit is called Revolution. The axis of the earth which is an imaginary line, makes an angle of 66½° with its orbital plane. ... The earth takes about 24 hours to complete one rotation around its axis.
Classify each of the following as a strong acid or a weak acid and indicate how each should be written in aqueous solution. Classify ... In solution this acid should be written as: weak 1. hydrocyanic acid H3O CN- _______ 2. hydrobromic acid
Answer:
HCN, weak acid
H⁺, Br⁻, strong acid
Explanation:
Hydrocyanic acid is a weak acid, according to the following equation.
HCN(aq) ⇄ H⁺(aq) + CN⁻(aq)
Thus, it should be written in the undissociated form (HCN).
Hydrobromic acid is a strong acid, according to the following equation.
HBr(aq) ⇒ H⁺(aq) + Br⁻(aq)
Thus, it should be written in the ionic form (H⁺, Br⁻).
Ammonia is produced by the reaction of nitrogen and hydrogen: N2(g) + H2(g) NH3(g)
(a) Balance the chemical equation.
(b) Calculate the mass of ammonia produced when 35.0g of nitrogen reacts with hydrogen.
Answer:
a) N2 (g) + H2 = 2 NH3
b) You have to state the mass of hydrogen
In a pure metal, the electrons can be thought of as [ Select ] throughout the metal. Using molecular orbital theory, there [ Select ] an energy gap between the filled molecular orbitals and empty molecular orbitals. The [ Select ] orbitals are typically higher in energy and are mostly [ Select ] .
Answer:
Explanation:
In a pure metal, the electrons can be thought of as [concentrated] around atoms throughout the metal. Using molecular orbital theory, there [is ] an energy gap between the filled molecular orbitals and empty molecular orbitals. The [antibonding] orbitals are typically higher in energy and are mostly (filled]
Certain ketones such as fructose can be oxidized by Benedict's reagent under basic conditions to form what type of compound
Answer:
Certain ketones such as fructose can be oxidized by Benedict's reagent under basic conditions to form what type of compound
Explanation:
Benedict's solution is a mixture of copper sulfate, anhydrous sodium carbonate, and sodium citrate.
In presence of mild reducing agents, Cu(II) ion in Benedict's solution becomes the Cu(I) ion.
Fructose has an alpha-hydroxy ketone group.
It undergoes tautomerism and forms alpha-hydroxy aldehyde which gives a positive test with Benedict's reagent.
During this test, aldehydes will be converted into carboxylic acids.
The reaction of fructose with Benedict's reagent is shown below:
A sample of calcium fluoride was decomposed into the constituent elements. Write a balanced chemical equation for the decomposition reaction. If the sample produced 294 mg of calcium, how many g of fluorine were formed
Answer:
A sample of calcium fluoride was decomposed into the constituent elements. Write a balanced chemical equation for the decomposition reaction. If the sample produced 294 mg of calcium, how many g of fluorine was formed
Explanation:
The balanced chemical equation for the decomposition of calcium fluoride is shown below:
[tex]CaF_2(s)->Ca(s)+F_2(g)[/tex]
The sample produced 294 g of calcium then, how many grams of fluorine is formed?
From the balanced chemical equation,
1 mol of CaF2 forms 1mol of calcium and 1 mol of fluorine.
That is:
40g of calcium and 38.0 g of fluorine are formed.
then,
If 294 g of calcium is formed then how many grams of fluorine is formed?
[tex]294g Ca * 38g F2 / 40g Ca\\=279.3 g F_2[/tex]
Hence, 279.3 g of fluorine will be formed.
Chloride ion is a strong nucleophile and bromide is a good leaving group. The major product of treating (S)-2-bromobutane with NaCl in CH3C(O)CH3 (acetone) is _________. (1S,2R)-1-chloro-2-bromobutane cis-2-butene (1R,2S)-1-chloro-2-bromobutane (S)-2-chlorobutane trans-2-butene (R)-2-chlorobutane
Answer:
Chloride ion is a strong nucleophile and bromide is a good leaving group. The major product of treating (S)-2-bromobutane with NaCl in CH3C(O)CH3 (acetone) is _________. (1S,2R)-1-chloro-2-bromobutane cis-2-butene (1R,2S)-1-chloro-2-bromobutane (S)-2-chlorobutane trans-2-butene (R)-2-chlorobutane
Explanation:
The reaction of (S)-2-bromobutane with NaCl in CH3C(O)CH3 (acetone) forms the following product:
The answer is (R)-2-chlorobutane.
The reaction take splace through [tex]S_{N} _2[/tex] mechansim and inversion in configuration happens.
Choose the substance with the higher entropy in each pair. Assume constant temperature, except in part (5)
(i) 1 mol of SO2(g) or 1 mol of SO3(g)
(ii) 1 mol of CO2(s) or 1 mol of CO2(g)
(iii) 3 mol of O2(g) or 2 mol of O3(g)
(iv) 1 mol of KBr(s) or 1 mol of KBr(aq)
(v) Seawater at 2°C or at 23°C
(vi) 1 mol of CF4(g) or 1 mol of CCl4(g)
Answer:
I) 1 mol of SO3(g)
2) 1 mol of CO2(g)
3) 3 mol of O2(g)
4) 1 mol of KBr(aq)
5) Seawater at 23°C
6) 1 mol of CCl4(g)
Explanation:
In molecules having greater numbers of atoms, there is an increase the number of ways by which the molecule vibrates thereby leading to a higher number of possible microstates and overall increase in entropy of the system. Hence, 1 mol of SO3(g) has a higher entropy than 1 mol of SO2.
Gases have a higher entropy than liquids and liquids have a higher entropy than gases.
Also, the greater the molecular weight of a molecule, the higher the entropy. Higher number of moles of a gas as well as the increase in temperature of a substance are also factors that lead to higher entropy.
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A sample of brass, which has a specific heat capacity of , is put into a calorimeter (see sketch at right) that contains of water. The temperature of the water starts off at . When the temperature of the water stops changing it's . The pressure remains constant at . Calculate the initial temperature of the brass sample. Be sure your answer is rounded to significant digits.
Answer:
The initial temperature of the brass sample is 90.1°C
Note: The question is incomplete. A similar but complete question is given below :
A 52.9g sample of brass, which has a specific heat capacity of 0.375·J·g−1°C−1, is put into a calorimeter (see sketch at right) that contains 100.0g of water. The temperature of the water starts off at 15.0°C. When the temperature of the water stops changing it's 18.4°C. The pressure remains constant at 1 atm. Calculate the initial temperature of the brass sample. Be sure your answer is rounded to 2 significant digits.
Explanation:
Assuming that the calorimeter is an isolated system and that no heat is lost from the calorimeter. The total heat in the system is the sum of the heat content of the brass and that of water
Total heat lost by the brass = heat gained by the water
The quantity of heat lost or gained, Q = mcΔT
Where m = mass of the substance, c = specific heat capacity of substance, ΔT = temperature change
Heat gained by water is positive while heat lost by brass is negative
mass of brass = 52.9 g, specific heat capacity of brass = 0.375·J·g−1°C−1, ΔT = (18.4 - t °C; where t is the initial temperature), mass of water = 100.0 g, specific heat capacity of water = 4.186 J/g°C, ΔT = = 18.4 - 15.0 = 3.4 °C
Heat lost by brass z= - [ 52.9 × 0.375 × (18.4 - t)] = -365.01 + 19.8375t
Heat gained by water = 100 × 4.186 × 3.4 = 1423.24
Equating heat lost by brass to heat gained by water
-365.01 + 19.8375t = 1423.24
19.8375t = 1423.24 + 365.01
19.8375t = 1788.25
t = 90.1° C
Therefore, the initial temperature of the brass sample is 90.1°C
Which of the following is a physical change?
If 11g of a gas occupies 5.6dm'3at s.t.p., calculate it's vapour density (1.0mol of a gas occupies 22.4dm'3at s.t.p.).
Answer:
[tex]{ \boxed{ \bf{vapour \: density = 2 \times molecular \: mass}}} \\{ \tt{ PV= (\frac{m}{ m_{r}}) RT}} \\ { \tt{3 \times 5.6 = \frac{11}{m _{r}} \times 0.0831 \times 273}} \\ { \tt{m _{r} = 14.85 \: g}} \\ \\ { \bf{vapour \: density = 2 \times m _{r}}} \\ = 2 \times 14.85 \\ = 29.7 \: { \tt{g {dm}^{ - 3} }}[/tex]
Which term can be used to describe the process in the reaction below? 2 NaHCO3 (s) → Na2CO3 (s) + H2O (g) + CO2 (g)
Answer:
Decomposition
Explanation:
If we look at the process;
2 NaHCO3 (s) → Na2CO3 (s) + H2O (g) + CO2 (g)
We can see that NaHCO3 was broken down into Na2CO3, H2O and CO2.
The breakdown of one compound to yield other chemical compounds is known as decomposition.
Hence the NaHCO3 was decomposed in the process above.
Question 1 Points 3 23 and Louis immerses his left hand in a beaker containing cold water and immerses his right hand in a beaker containing warm water. Then, he immerses both his hands on a beaker containing water at room temperature. Which of the following statements are true? 1. The hand that was in hot water would feel cold. 2. The hand that was in cold water would feel hot. 3. His two hands will feel the same hotness. Que O2 and 3 0 1 and 2 o 1 and 3 1.2, and 3
Answer:look down below
Explanation:
The statements that are true about hands that are immersed in the water are:
1. The hand that was in hot water would feel cold.
2. The hand that was in cold water would feel hot.
The correct option is B 1. and 2.
What is temperature?
Temperature is the measurement of the hotness or coldness of any object. It is measured in Celsius or kelvin. Our body has nerves that feel the different temperatures of any object. The high temperature is called hot and the low temperature is called cold.
When Louis put his hand in the warm water and one hand in the cold water. He feels the temperature of both glasses of water. Then he put both hands in the normal water.
So the hand that is warm would feel the water as cold and the hand with cold water would feel the water as hot.
Thus, the correct option is B. 1. and 2.
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