Answer:
no
Explanation:
1. A gas expands and does p-v work on the surroundings equal to
279j. At the same time it absorbs, 216j of heat from the surroundings. What
is the change in energy of the system?
Please answer if yk
The change in energy of the system : -63 J
Further explanationGiven
279 J work
216 J heat
Required
The change in energy
Solution
Laws of thermodynamics 1
ΔU=Q+W
Rules :
receives heat, Q + releases heat, Q - work is done by a system, W - work is done on a system, W +a gas work on the surrounding : W =-279 J
a gas absorb heat from surrounding : Q = +216 J
Internal energy :
= -279+216
= -63 J
John rode his bike down the street to the east. He rode 100 meters in 25
seconds. What was his velocity?
A. 0.25 m/s east
B. 25 m/s east
C. 10 m/s east
D. 4 m/s east
Answer:
The answer is D.
Explanation:
Given that velocity also applies speed formula, S = D ÷ T where D represents distance and T is time :
S = D ÷ T
S = 100m ÷ 25s
S = 4m/s
V = 4m/s east
The difference between speed and velocity is that speed has only magnitude but velocity has magnitude and direction.
Lee rode a skateboard to school. His velocity changed from 2 m/s east to 5
m/s east in 6 seconds. What was the change in Lee's velocity?
A. 5 m/s east
B. 3 m/s east
C. -3 m/s east
D. 2 m/s east
Answer:
Option B. 3 m/s east
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 2 m/s east
Final velocity (v) = 5 m/s east
Time (t) = 6 s
Change in velocity =?
We can obtain the change in Lee's velocity as illustrated below:
Initial velocity (u) = 2 m/s east
Final velocity (v) = 5 m/s east
Change in velocity =?
Change in velocity = Final velocity (v) – Initial velocity (u)
Change in velocity = v – u
Change in velocity = 5 – 2
Change in velocity = 3 m/s east
Thus, the change in Lee's velocity is 3 m/s east
Answer:
Option (B.) 3 m/s east
Explanation:
Have a good day hope it helps
Compare and contrast the thermal capacity and transfer of solids, liquids, and gases
Answer:
किसी पदार्थ की ऊष्मीय चालकता को प्रति इकाई तापमान अंतर के प्रति इकाई क्षेत्र की सामग्री की एक इकाई मोटाई के माध्यम से गर्मी हस्तांतरण की दर के रूप में परिभाषित किया गया है। किसी पदार्थ की ऊष्मीय चालकता इस बात का माप है कि उस पदार्थ में कितनी तीव्र ऊष्मा प्रवाहित होगी। थर्मल चालकता के लिए एक बड़ा मूल्य इंगित करता है कि सामग्री एक अच्छा गर्मी कंडक्टर है, और एक कम मूल्य इंगित करता है कि सामग्री एक खराब गर्मी कंडक्टर या इन्सुलेटर है। कमरे के तापमान पर शुद्ध तांबे की तापीय चालकता 401 W / m है। K, जो इंगित करता है कि एक 1m मोटी तांबे की दीवार 401 W / m 2 की दर से गर्मी का संचालन करेगीप्रति दीवार के क्षेत्र का अंतर दीवार के पार तापमान अंतर। चित्रा 2.3 सामान्य तापमान और दबाव में पदार्थ के विभिन्न राज्यों के लिए तापीय चालकता की सीमा को दर्शाता है। एक ठोस की ऊष्मीय चालकता, गैस की तुलना में चार गुना अधिक परिमाण की हो सकती है। यह प्रवृत्ति काफी हद तक दोनों राज्यों के बीच अंतर-संबंधी अंतर के कारण है।
ठोस राज्य
सामग्रियों के आधुनिक दृष्टिकोण में, एक ठोस में मुक्त इलेक्ट्रॉनों और एक आवधिक व्यवस्था में बाध्य परमाणुओं का समावेश हो सकता है जिसे जाली कहा जाता है। तदनुसार, थर्मल ऊर्जा का परिवहन दो प्रभावों के कारण होता है: मुक्त इलेक्ट्रॉनों का पलायन और जाली कंपन तरंगें। ये प्रभाव योगात्मक हैं, जैसे कि तापीय चालकता k , इलेक्ट्रॉनिक घटक k e और समरूप घटक k l का योग है
k = k e + k l
(2.7)
सामान्य तापमान और दबाव में विभिन्न राज्यों के लिए थर्मल चालकता की 2.3 रेंज चित्रा
k e विद्युत प्रतिरोधकता के व्युत्क्रमानुपाती होता है । शुद्ध धातुओं के लिए, जो कम के हैं , k e , k l की तुलना में बहुत बड़ा है । इसके विपरीत, मिश्र धातुओं के लिए, जो कि काफी बड़े होते हैं , k l से k का योगदान अब नगण्य नहीं है। गैर-धात्विक ठोस के लिए, k को मुख्य रूप से k l द्वारा निर्धारित किया जाता है , जो कि जाली के परमाणुओं के बीच परस्पर क्रिया की आवृत्ति पर निर्भर करता है। जाली व्यवस्था की नियमितता का k l पर महत्वपूर्ण प्रभाव पड़ता है , क्रिस्टलीय (सुव्यवस्थित) सामग्री की तरह, क्वार्ट्ज जैसी सामग्री में कांच जैसी अनाकार सामग्री की तुलना में अधिक ऊष्मीय चालकता होती है। वास्तव में, क्रिस्टलीय के लिए, गैर-धात्विक ठोस जैसे कि हीरा और बेरिलियम ऑक्साइड, k l काफी बड़े हो सकते हैं, जो कि अच्छे कंडक्टरों से जुड़े k के मूल्यों से अधिक होते हैं , जैसे कि एल्यूमीनियम।
इन्सुलेशन सिस्टम
थर्मल इंसुलेशन में कम तापीय चालकता वाली सामग्री शामिल होती है, जो एक कम प्रणाली वाली तापीय चालकता को प्राप्त करने के लिए संयुक्त होती है। फाइबर-, पाउडर-, फ्लेक-टाइप इंसुलेशन में, ठोस पदार्थ को पूरी तरह से एक एयर स्पेस में फैलाया जाता है। ऐसी प्रणालियों को एक प्रभावी तापीय चालकता की विशेषता होती है , जो ठोस पदार्थ की तापीय चालकता और सतह विकिरणकारी गुणों पर निर्भर करती है, साथ ही साथ हवा या शून्य स्थान की प्रकृति और मात्रात्मक अंश। प्रणाली का एक विशेष पैरामीटर इसकी थोक घनत्व (ठोस द्रव्यमान / कुल मात्रा) है, जो उस तरीके पर दृढ़ता से निर्भर करता है जिसमें ठोस सामग्री परस्पर जुड़ी हुई है।
द्रवित अवस्था
चूंकि इंटरमॉलिक्युलर स्पेसिंग बहुत बड़ी होती है और अणु की गति ठोस अवस्था की तुलना में द्रव अवस्था के लिए अधिक यादृच्छिक होती है, इसलिए थर्मल एनर्जी ट्रांसपोर्ट कम प्रभावी होता है। इसलिए गैसों और तरल पदार्थों की तापीय चालकता ठोस पदार्थों की तुलना में छोटी होती है।
ऊष्मीय विसरणशीलता
गर्मी हस्तांतरण समस्याओं के हमारे विश्लेषण में, पदार्थ के कई गुणों का उपयोग करना आवश्यक होगा। इन गुणों को आम तौर पर थर्मोफिजिकल गुणों के रूप में संदर्भित किया जाता है और इसमें दो अलग-अलग श्रेणियां, परिवहन और थर्मोडायनामिक गुण शामिल होते हैं। परिवहन गुणों में प्रसार दर गुणांक जैसे कि के, थर्मल चालकता (गर्मी हस्तांतरण के लिए), और , गतिज चिपचिपापन (गति हस्तांतरण के लिए) शामिल हैं। दूसरी ओर, थर्मोडायनामिक गुण, एक प्रणाली के संतुलन की स्थिति से संबंधित हैं। घनत्व ( ) और विशिष्ट ऊष्मा ( C p ) दो ऐसे गुण हैं जिनका उपयोग थर्मोडायनामिक विश्लेषण में बड़े पैमाने पर किया जाता है। उत्पाद सी पीआम तौर पर वॉल्यूमेट्रिक ताप क्षमता को कहा जाता है , जो थर्मल ऊर्जा को स्टोर करने के लिए एक सामग्री की क्षमता को मापता है। क्योंकि बड़े घनत्व के पदार्थों को आमतौर पर छोटे विशिष्ट हीट्स, कई ठोस और तरल पदार्थों की विशेषता होती है, जो कि बहुत अच्छा ऊर्जा भंडारण मीडिया है, इसमें तुलनीय ताप क्षमता होती है। हालांकि उनकी बहुत छोटी घनत्व के कारण, गैसें थर्मल ऊर्जा भंडारण के लिए खराब अनुकूल हैं।
ऊष्मा अंतरण विश्लेषण में, ऊष्मा चालकता के लिए तापीय चालकता का अनुपात एक महत्वपूर्ण गुण है जिसे तापीय विवर्तनशीलता कहा जाता है , जिसमें m 2 / s की इकाइयाँ होती हैं ।
(2.8)
यह तापीय ऊर्जा को संग्रहीत करने की क्षमता के सापेक्ष तापीय ऊर्जा का संचालन करने के लिए एक सामग्री की क्षमता को मापता है। बड़ी की सामग्री उनके थर्मल वातावरण में बदलाव के लिए जल्दी से प्रतिक्रिया देगी, जबकि छोटे की सामग्री अधिक सुस्त प्रतिक्रिया देगी, एक नई संतुलन स्थिति तक पहुंचने में अधिक समय लेगी।
What is produced when an acid and an alkali react?
A. Water and a base
B. Water and Carbon dioxide
C. Salt and a base
D. Salt and water
what happens when copper sulphate reacts with hydrogen
sulphide?
Answer:
When hydrogen sulphide gas is passed through a blue solution of copper sulphate, a black precipitate of copper sulphide is obtained and the sulphuric acid so formed remains in the solution. In this reaction, cupric ions (from copper sulphate) combine with sulphide ions (from hydrogen sulfide) to form cupric sulphide.
Answer:
When hydrogen sulphide gas is passed through a blue solution of copper sulphate, a black precipitate of copper sulphide is obtained and the sulphuric acid so formed remains in the solution. In this reaction, cupric ions (from copper sulphate) combine with sulphide ions (from hydrogen sulfide) to form cupric sulphide.
Explanation:
describe how atomic emission Spectra give us information about the energy levels of electrons in atoms
Answer:
Energy of atom is in lowest. The energy of electrons are given energy. The electrons absorb energy. Its converted to higher level of energy.
When excited electrons return to their ground state, atomic emission spectra are produced. When electrons return to a lower energy level, they emit light energy.
What is atomic emission Spectra?The emission spectrum of a chemical element or chemical compound is the spectrum of electromagnetic radiation frequencies emitted as a result of an electron transitioning from a high energy state to a lower energy state.
The emitted photon's photon energy equals the energy difference between the two states.
Atomic emission spectra provided further evidence of light's quantized nature, leading to the development of a new model of the atom based on quantum theory.
When excited electrons return to their ground state, atomic emission spectra are produced. The light emitted corresponds to the energies of the specific electrons.
Thus, this way atomic emission Spectra give us information about the energy levels of electrons in atoms .
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A 23.9 g sample of an unknown metal is heated from 41.2C to 68.9C. During the process the
metal absorbs 1110 Joules. Calculate the specific heat of the metal.
Answer:
1.677 J/gºC
Explanation:
From the question given above, the following data were obtained:
Mass (M) = 23.9 g
Initial temperature (T₁) = 41.2 °C
Final temperature (T₂) = 68.9 °C
Heat (Q) absorbed = 1110 J
Specific heat capacity (C) of metal =?
Next, we shall determine the change in the temperature of the metal. This can be obtained as follow:
Initial temperature (T₁) = 41.2 °C
Final temperature (T₂) = 68.9 °C
Change in temperature (ΔT) =?
ΔT = T₂ – T₁
ΔT = 68.9 – 41.2
ΔT = 27.7 °C
Finally, we shall determine the specific heat capacity of the metal as follow:
Mass (M) = 23.9 g
Heat (Q) absorbed = 1110 J
Change in temperature (ΔT) = 27.7 °C
Specific heat capacity (C) of metal =?
Q = MCΔT
1110 = 23.9 × C × 27.7
1110 = 662.03 × C
Divide both side by 662.03
C = 1110 / 662.03
C = 1.677 J/gºC
Therefore, the specific heat capacity of the metal is 1.677 J/gºC
In a sample of air at STP, the partial pressure of nitrogen gas is PN, = 0.78 atm and the partial
pressure of argon gas is Par = 0:01 atm. Calculate the partial pressure of oxygen gas Po, at STP.
Ignore units and use the proper amount of significant figures.
Answer:
0.21 atm
Explanation:
Step 1: Given data
Pressure of air (P): 1 atm (standard pressure)Partial pressure of nitrogen (pN₂): 0.78 atmPartial pressure of argon (pAr): 0.01 atmStep 2: Calculate the partial pressure of oxygen
The air is a gaseous mixture formed mainly by nitrogen, oxygen and argon. The total pressure is equal to the sum of the partial pressures of the gases.
P = pN₂ + pAr + pO₂
pO₂ = P - pN₂ - pAr
pO₂ = 1 atm - 0.78 atm - 0.01 atm
pO₂ = 0.21 atm
Find the number of grams in 3.45 x 10^21 molecules NiO
Answer:
Mass = 0.43 g
Explanation:
Given data:
Mass in gram = ?
Number of molecules of NiO = 3.45×10²¹
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ molecules
3.45×10²¹ molecules × 1 mol / 6.022 × 10²³ molecules
0.57×10⁻² mol
0.0057 mol
Mass in gram:
Mass = number of moles × molar mass
Mass = 0.0057 mol × 74.69 g/mol
Mass = 0.43 g
Examine the chemical reaction and lab scenario.
2Fe2S3+9O2→2Fe2O3+6SO2
3.17 g of iron(III) sulfide (Fe2S3) reacts with an excess amount of oxygen at standard temperature and pressure. The theoretical yield of sulfur dioxide (SO2) is 3.81 g. However, during a lab, only 3.37 g are produced.
What is the percent yield of sulfur dioxide?
94.41% SO2
81.36% SO2
44.00% SO2
88.45% SO2
Explanation:
Percentage Yield
= (3.37g/3.81g) * 100% = 88.45%.
Therefore 88.45% SO2 is the percentage yield.
Why is the atomic model the foundation for chemistry?
A. because all matter is made of atoms
B. because no particle is smaller than the atom
O c. because atoms are the invisible particles that make up light
O D. because atoms are the small forces of energy that hold together all matter
Answer:
A.
Explanation:
calculate the percentage of calcium in calcium chlorate
Answer:
19.4%
Explanation:
Calcium Chlorate is Ca(ClO3)2
Now calculate the molar mass of Ca(ClO3)2
Ca = 40.1
Cl = 35.5
O = 16.0
But they are two chlorine atoms and six oxygen atoms. So you do this:
40.1 + 35.5(2) + 16.0(6) = 207.1 grams
Now find the molar mass of just calcium.
There is only one calcium atom.
So you do this.
40.1(1) = 40.1
Now divide the molar mass of calcium by the molar mass of calcium chlorate.
40.1 / 207.1 = 0.1936
0.1936 rounds to 0.194
Now multiply 0.194 * 100 and you will get 19.4
So the final answer is 19.4%.
Hope it helped!
true or false
1. Average atomic mass is the mass of any atom based on its nuclear particles in relation to units measured in amu which is 1/12th of a carbon atom.
2. When calculating the average atomic mass of an element, the average atomic mass value falls closest to that of the least percentage abundance (least common) isotope.
Answer:
true
Explanation:
calculating the average atomic
why is water a polar covalent molecule.
Water is a Polar Covalent Molecule
The unequal sharing of electrons between the atoms and the unsymmetrical shape of the molecule means that a water molecule has two poles - a positive charge on the hydrogen pole (side) and a negative charge on the oxygen pole (side).
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Identify the metals in period 3 on the periodic table
Answer:
Aluminum (Al) is the only metal in period 3 on the periodic table.
30 POINTS HELPP PLEASE
1: selective breeding
2:traits
3: crop species
4: domestic
5: diverse
6: artificial
7: uniform
in your opinion if the answer is correct please like n mark MI s brain list
A plant takes root in a small crevice of rock, as the plant grows, its roots push on the edges of the rock, breaking it is a form of weathering?
1. True
2. False
72.40% iron,27.60% Oxygen Empirical formula
Explanation:
Fe. O
72.40/ 56 27.60/16
____________________
1.29/1.29 1.725/1.29
_____________________
1. :. 1
Emperical formula = FeO
Empirical formula is the whole number ratios of elements in a compound. It can be found from the number of moles. The empirical formula of iron and oxygen of 72.4 % and 27.6 % respectively is Fe₂O₃.
What is empirical formula?Empirical formula of a compound is the chemical formula derived from the number of moles each element present in it. Whereas, molecular formula is derived from the actual masses of the elements present in the compound.
Empirical formula can be determined from the number of moles of the elements. It is given that the mass percentages of iron and oxygen are 72.40% and 27.60% respectively. This can be taken as 72.40 g and 27.60 g in a 100 g of compound.
Thus there is 72.40 g of iron. The atomic mass of Fe = 55.85 g/mol
Thus number of moles of Fe = 72.40 / 55.85
= 1.295
The weight of oxygen in 100 g of compound is 27.60 g. Atomic mass of oxygen is 16 g/mol.
Thus number of moles of oxygen = 27.60 /16 = 1.725
Taking the ratios of these number of moles = 1.295 :1.725 = 2 :3
Hence, the ratios of iron to oxygen is 2:3 which means the empirical formula of the compound is Fe₂O₃.
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Chemistry
Definition in your own words
Will check if you got it from online.
Word: Solid
Answer:
A solid is a matter that has particles that are closely packed which as a result makes it vibrate but not flow.
Q2. Which of the following molecules may show.
Infra-red absorption
absorption Spectra, CH3CH3,
CH₃Cl3, N2.
Answer:
I think is answer
CH3cl3
five more than one third of a number is nine .
Answer:
uh
Explanation:
5+x/3=9
=x/3=9-5
=x=12
Changes in temperature can often cause a change in the state of matter of objects. A decreasein temperature causes which of the following to happen?
A liquids will boil. B Water will freeze. C Wax will melt. D Solids will turn to gas.
PLS give meh answer
Answer:
B. Water will freeze
because the latent heat of vapourization decreases.
Please select the word from the list that best fits the definition How far something moves in a specific amount of time
Answer:
Speed
Explanation:
That is what I believe the word to the definition, How far something moves in a specific amount of time, but you'd have to providea picture of the list of words for me to be sure...
if a bike travels 23m at constant speed 5289s what is the bikes speed?
Answer:
0.004 m/s
Explanation:
Given data:
Distance traveled = 23 m
Time taken = 5289 s
Speed of bike = ?
Solution:
Formula:
speed = distance / time
b y putting values,
speed = 23 m/ 5289 s
speed = 0.004 m/s
Comparin
List three characteristics of chemical reactions
and three characteristics of nuclear reactions.
DONE
Answer:
Chemical reactions involve the loss, gain, or sharing of electrons, but do not involve changes to the nucleus of an atom. Nuclear reactions involve changes to the nucleus and produce a new particle. Chemical reactions involve low energy changes. Chemical reactions never involve a change in the mass of atoms.
List three characteristics of chemical reactions, I have five:
occur when bonds are broken and formed
atoms remain unchanged though they may be rearranged
involve only valence electrons
reaction rate is influenced by, temp, pressure, concentration, and catalyst
Three characteristics of nuclear reactions, I have five:
occur when nuclei emit particles
atoms are often converted into atoms of another element
may involve protons, neutrons, and electrons
reaction rate is not normally affected by temp, pressure or catalyst
How will you process the information in the table and on the map to
make predictions? Describe how you will use the inputs to develop
an output.
Help
Answer:
By measuring through scale
Explanation:
CAN YOU TELL THE NAME OF COUNTRY BECAUSE I AM FROM NEPAL
please help. calculate the partial pressure of oxygen gas Po2 at STP.
Answer:
0.21 atm
Explanation:
From the question given above, the following data were obtained:
Partial pressure of N₂ (P₁) = 0.78 atm
Partial pressure of Ar (P₂) = 0.01 atm
Partial pressure of O₂ (P₃) =?
Since the system is at standard temperature and pressure (STP), the total pressure of the system is 1 atm.
Finally, we shall determine the partial pressure of O₂. This can be obtained as follow:
Partial pressure of N₂ (P₁) = 0.78 atm
Partial pressure of Ar (P₂) = 0.01 atm
Total pressure (Pₜ) = 1 atm
Partial pressure of O₂ (P₃) =?
Pₜ = P₁ + P₂ + P₃
1 = 0.78 + 0.01 + P₃
1 = 0.79 + P₃
Collect like terms
1 – 0.79 = P₃
P₃ = 0.21 atm
Therefore, the partial pressure of O₂ is 0.21 atm.
The electron structure of atoms in elements A and C are 2, 8, 1 and 2, 8, 18, 7, respectively. The formula for the compound formed between A and C will be AC. True or false? pls help me T^T
2. How many moles are in 2.43 x 1024 particles of Carbon Monoxide
(CO)?
Answer:
= 14.63 × [tex]10^{45}[/tex] particles.
Explanation:
According to Avogadro's priciple:
1 mole of any substance is 6.02 × [tex]10^{23}[/tex] particles
∴ 2.43 x [tex]10^{24}[/tex] moles × 6.02 × [tex]10^{23}[/tex]
= 14.63 × [tex]10^{45}[/tex] particles.
I hope this was clear and easy to understand.