The melting point of water b. Decreases when pressure is applied.
The melting point of water is typically at 0 degrees Celsius (32 degrees Fahrenheit) at standard atmospheric pressure. However, unlike most substances, the melting point of water decreases as pressure is increased. This phenomenon is known as the "anomalous expansion of water." When pressure is applied to water, it compresses the molecular arrangement, making it more difficult for the water molecules to form the stable crystal lattice structure characteristic of ice. As a result, the melting point of water decreases, allowing it to remain in the liquid state at lower temperatures than would be expected under normal conditions.
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Newton's third law of motion is summarized as the law of action-reaction. For every action, there is an equal but opposite reaction. Which example best illustrates the application of Newton's law of action-reaction? Choose all that apply.
Answer:
When we stand on the floor, we apply a force on the floor surface in the downward direction and in return the floor also exerts an upward and equal force on us.
Explanation:
Newton' third law is vey famous and it states that for each and every action, there applies an equal but opposite reaction. Thus the action force and the reaction force always acts on pairs. But they does not contribute to the motion of the object.
One such example that illustrates the action and reaction force from Newton's law is when we stand on the floor we exert a force on the floor surface in downward direction. The floor surface also exerts an opposite and equal force on us in the upward direction.
What makes sound waves different from light waves?
Answer:
Light waves are electromagnetic waves while sound waves are mechanical waves. light waves are transverse while sound waves are longitudinal. sound waves require a material medium to travel, and hence, cannot travel in a vacuum. the speed of light in a medium is constant.
Explanation:
Hope this helped Mark BRAINLIEST!!
Uma peça de metal de formato irregular feita de uma liga particular foi galvanizada com zinco usando uma solução de Zn(NO₃)₂. Quando uma corrente de 2,70 A foi usada, levou exatamente 1 hora para depositar uma camada de 0,015 mm de zinco na peça. Qual é a área total de superfície da peça? A densidade do zinco é de 7,14 g /cm³. Suponha que a eficiência seja 100%. MM (Zn) = 65,4 g/mol; 1 mol e = 96500 C.
2. The density of helium is 1.78 X 104 g/cm. What is this
density in Dg/um??
Answer:
[tex]d=1.78\times 10^{-7}\ Dg/\mu m^3[/tex]
Explanation:
Given,
The density of Helium is [tex]1.78\times 10^4\ g/cm^3[/tex]
We need to find the density in Dg/μm
We know that,
1 g = 10 dg
1 cm³ = 10¹² μm³
So,
[tex]d=1.78 \times 10^4\ g/cm^3\\\\=1.78 \times 10^4\times \dfrac{10\ dg}{10^{12}\ \mu m^3}\\\\=1.78\times 10^{-7}\ Dg/\mu m^3[/tex]
So, the density of Helium is equal to [tex]1.78\times 10^{-7}\ Dg/\mu m^3[/tex].
How many grams of calcium carbonate, CaCO3, contain 48 grams of oxygen atoms?
A) 41
B) 50
C) 62
D) 88
E) 100
Answer:
(40 g O) / (15.99943 g O/mol) x (1 mol CaCO3 / 3 mol O) x (100.0875 g CaCO3/mol) =
83 g CaCO3
So answer D), although three significant digits are not justified.
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Consider a particle in a one-dimensional box. a. For a box of length 1 nm, what is the probability of finding the particle within 0.01 nm of the center of the box for the lowest-energy level? b. Answer part (a) for the first excited state. c. The longest-wavelength transition for a particle in a box (not the box in part (a)] is 200 nm. What is the wavelength if the mass of the particle is doubled? What is the wavelength if the charge of the particle is doubled? What is the wavelength if the length of the box is doubled?
The transition wavelength, λ is given as:λ = (hc)/Eλ = (6.626 x 10^-34 Js * 3 x 10^8 m/s)/(4.585 x 10^-20 J)λ = 4.325 x 10^-7 mλ = 432.5 nmTherefore, if the length of the box is doubled, the transition wavelength remains the same as 200 nm.
The length of the box, L = 1 nm.The transition wavelength, λ = 200 nm.Formula:For particle in a one-dimensional box,For the energy of a particle in a one-dimensional box,E = (n²h²)/(8mL²)where, n = quantum numberh = Planck's constanm = mass of the particleL = length of the box.1. For the lowest energy level,n = 1E₁ = (n²h²)/(8mL²)E₁ = (1²h²)/(8mL²)E₁ = (1*6.626 x 10^-34 Js)²/(8(9.11 x 10^-31 kg)(1 x 10^-9 m)²)E₁ = 9.170 x 10^-20 JFrom the above formula, the probability of finding the particle within 0.01 nm of the center of the box is given as:ψ(x) = sqrt(2/L)sin(nπx/L)Probability, P = ∫ψ²dxFrom the above formula, the probability of finding the particle within 0.01 nm of the center of the box for the lowest-energy level is given as:P = ∫[sqrt(2/L)sin(πx/L)]²dxP = ∫2/L sin²(πx/L)dxP = 2/L * L/2P = 1Hence, the probability of finding the particle within 0.01 nm of the center of the box for the lowest-energy level is 1.2. For the first excited state,n = 2E₂ = (n²h²)/(8mL²)E₂ = (2²h²)/(8mL²)E₂ = (4*6.626 x 10^-34 Js)²/(8(9.11 x 10^-31 kg)(1 x 10^-9 m)²)E₂ = 3.668 x 10^-19 JFrom the above formula, the probability of finding the particle within 0.01 nm of the center of the box for the first excited state is given as:P = ∫[sqrt(2/L)sin(2πx/L)]²dxP = ∫2/L sin²(2πx/L)dxP = 2/L * L/2P = 0.5Hence, the probability of finding the particle within 0.01 nm of the center of the box for the first excited state is 0.5.3. (i) If the mass of the particle is doubled:For the energy of a particle in a one-dimensional box,E = (n²h²)/(8mL²)The energy of a particle is inversely proportional to the mass of the particle.Hence, if the mass of the particle is doubled, the energy of the particle is halved.E = 0.5E₁ = 0.5(9.170 x 10^-20 J)E = 4.585 x 10^-20 JFrom the above formula, the transition wavelength, λ is given as:λ = (hc)/Eλ = (6.626 x 10^-34 Js * 3 x 10^8 m/s)/(4.585 x 10^-20 J)λ = 4.325 x 10^-7 mλ = 432.5 nm(ii) If the charge of the particle is doubled:The charge of the particle does not have any effect on the energy of the particle.Therefore, the transition wavelength, λ remains the same as 200 nm.(iii) If the length of the box is doubled:For the energy of a particle in a one-dimensional box,E = (n²h²)/(8mL²)The energy of a particle is inversely proportional to the length of the box.Hence, if the length of the box is doubled, the energy of the particle is halved.E = 0.5E₁ = 0.5(9.170 x 10^-20 J)E = 4.585 x 10^-20 JFrom the above formula, the transition wavelength, λ is given as:λ = (hc)/Eλ = (6.626 x 10^-34 Js * 3 x 10^8 m/s)/(4.585 x 10^-20 J)λ = 4.325 x 10^-7 mλ = 432.5 nmTherefore, if the length of the box is doubled, the transition wavelength remains the same as 200 nm.
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Can you think of something you do or a hobby you have that is physics
related? Explain your thinking as to how what you do is Physics related.
You can take physics as a hobby and also it is great that you like both maths and physics as well and also want to be an computer science engineer because physics and math
Which of the following balances has the least uncertainty? (PSS.1)
Group of answer choices
platform balance, +/- 0.1 g
analytical balance, +/- 0.0001 g
All balances have the same uncertainty
triple-beam balance, +/- 0.01 g
electronic balance, +/- 0.001 g
Which wavelength corresponds to photons with a greater energy?
a. 320 nm
b. 530 nm
c. they are equal in energy
The wavelength that corresponds to photons with greater energy is 320 nm. Hence, option a is correct.
The energy of a photon is inversely proportional to its wavelength according to the equation:
E = hc/λ
where E is the energy of the photon, h is Planck's constant (approximately 6.626 x 10^-34 J·s), c is the speed of light (approximately 3.00 x 10^8 m/s), and λ is the wavelength of the photon.
Since energy (E) is inversely proportional to wavelength (λ), shorter wavelengths correspond to photons with greater energy. Therefore, the photon with a wavelength of 320 nm has greater energy compared to the photon with a wavelength of 530 nm.
The wavelength that corresponds to photons with greater energy is 320 nm (option a).
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Which of the following is a FALSE statement? *
A. Heat moves through solids by by conduction.
B. Molecules move faster in warmer substances.
C. Warm water is denser than cold water.
D. Heat moves through liquids and gases by convection.
Answer:
C: Warm water is denser than cold water
Help pls what is the (H+) of a solution with a pH of 4?
a. 1E-4 M
b. 0.602 M
c. 1E6 M
d. 1E4 M
Answer:
A
Explanation:
[H+] = 10^-pH
[H+] = 1 x 10^-4 M
Shaun's doctor has recommended that he consume 200 calories of protein a day. Shaun consumed 176 calories of protein in one day. How much more protein does he need to consume to reach the required amount? (1 gram of protein = 4 calories)
Answer: Shaun must consume 6 more g of protein to reach the required amount.
Explanation:
Given : Recommended amount of protein = 200 calories
Now 4 calories = 1 gram of protein
Thus 200 calories = [tex]\frac{1}{4}\times 200=50[/tex] g of protein.
Calories consumed per day by Shaun = 176 calories
176 calories = [tex]\frac{1}{4}\times 176=44[/tex] g of protein.
Thus Shaun need to consume = (50-44) g = 6 of protein to reach the required amount.
Shaun must consume 6 more g of protein to reach the required amount.
Substances have a fixed arrangement of atoms. true or false (NO LINKS PLEASE)
Answer:
True
Explanation:
Write the pressure equilibrium constant expression for this reaction. 2Li3N(s) -> 6 Li(s) + N2(g)
The pressure equilibrium constant expression for the given reaction 2Li3N(s) ⇌ 6Li(s) + N2(g) is given below: P = (PN2 ) (PLi)6/ (PLi3N )2Where, PN2 is the partial pressure of nitrogen PLi is the partial pressure of lithiumPLi3N is the partial pressure of lithium nitride.
The above formula of pressure equilibrium constant expression gives the ratio of the product of partial pressures of reactants and products to the power of their stoichiometric coefficients. Here, the stoichiometric coefficients of Li3N, Li and N2 are 2, 6 and 1 respectively. The reaction 2Li3N(s) ⇌ 6Li(s) + N2(g) represents the dissociation of lithium nitride to lithium and nitrogen gases. The dissociation reaction is endothermic.
As the temperature increases, the dissociation of lithium nitride increases which results in the formation of more lithium and nitrogen gas molecules. Therefore, the value of the equilibrium constant for this reaction increases with increasing temperature. The above-pressure equilibrium constant expression helps to determine the equilibrium concentration of reactants and products when the reaction is in equilibrium.
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Our body is also a good conductor
of electricity because of the
presence of ______.
Please answer it this is my last question on my test i'll give you 20 points to answer?
acid-catalyzed addition of alcohols to alkenes proceeds in a mechanism analogous to the acid-catalyzed addition of water to yield ethers. true or false
We can see here that the acid-catalyzed addition of alcohols to alkenes proceeds in a mechanism analogous to the acid-catalyzed addition of water to yield ethers. True.
What is alcohol?Alcohol is a broad term that refers to a class of organic compounds characterized by the presence of a hydroxyl (-OH) functional group attached to a carbon atom.
The acid-catalyzed addition of alcohols to alkenes is a useful reaction for the synthesis of ethers. It is also a relatively mild reaction, which makes it a good choice for the synthesis of ethers from sensitive compounds.
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According to the Standard Reduction Potential Table in your textbook, assign the anode and cathode for thecell pairs, B1-B2, A1-B2, AND A2-B2 described in the Procedure
For the B₁-B₂ cell pair, B₂ is the cathode, and B₁ is the anode. For the A₁-B₂ cell pair, B₂ is the cathode, and A₁ is the anode. For the A₂-B₂ cell pair, B₂ is the cathode, and A₂ is the anode.
B₁-B₂:
The half-reaction at B₁ is given as A⁺(aq) + 2e⁻ → A(s) with a reduction potential of -0.78 V.
The half-reaction at B₂ is given as B⁺(aq) + e⁻ → B(s) with a reduction potential of -0.45 V.
Since the reduction potential of B₂ (-0.45 V) is more positive than B₁ (-0.78 V), B2 will be the cathode, and B1 will be the anode.
A₁-B₂:
The half-reaction at A₁ is given as C⁺(aq) + 2e⁻ → C(s) with a reduction potential of -0.95 V.
The half-reaction at B₂ is the same as mentioned before B⁺(aq) + e⁻ → B(s) with a reduction potential of -0.45 V.
Since the reduction potential of B₂ (-0.45 V) is more positive than A₁ (-0.95 V), B₂ will be the cathode, and A₁ will be the anode.
A₂-B₂:
The half-reaction at A₂ is given as D⁺(aq) + 2e⁻ → D(s) with a reduction potential of -0.70 V.
The half-reaction at B₂ is the same as mentioned before B⁺(aq) + e⁻ → B(s) with a reduction potential of -0.45 V.
Since the reduction potential of B₂ (-0.45 V) is more positive than A₂ (-0.70 V), B₂ will be the cathode, and A₂ will be the anode.
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Given the balanced chemical equation: 2 H2 + O2 → 2 H20. Choose
the description below that most accurately describes the reaction.
O 2 moles of hydrogen react with 1 mole of oxygen to form 2 moles of water
O 2 moles of water react with 1 mole of hydrogen to form 2 moles of oxygen
O equal moles of hydrogen and oxygen react to form an equal amount of
water
O 4 moles of hydrogen react with 2 moles of oxygen to form 4 moles of water
Answer:
The answer is
2 moles of hydrogen react with 1 mole of oxygen to form 2 moles of water
How many gram of ice at 0C can be melted by the condensation of 12.39g of steam at 100C, assuming a complete transfer of thermal energy?
The mass of the ice melted by a complete transfer of thermal energy is 83.84 g.
What is the mass of the ice melted?The mass of the ice melted by a complete transfer of thermal energy is calculated by applying the following equation as follows;
Q = mcΔθ
where;
m is the massc is the specific heat of waterΔθ is change in temperatureheat gained by the ice = heat loss by the steam
The mass of the ice melted by a complete transfer of thermal energy is calculated;
m x heat of fusion = 12.39g x heat of vaporization
m x 334 J/g = 12.39 g x 2260 J/g
334m = 28,001.4
m = 28,001.4 / 334
m = 83.84 g
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Sarah and Jim investigated the effect of temperature on the solubility of copper sulphate. They dissolved copper sulphate crystals in the same volume of water until no more would dissolve. They measured the mass of copper using water at different temperatures.
One of the mass readings appears to be wrong ( anomalous ) which reading is anomalous? Use the graph to help you.
Answer:
Explanation:
H m sarav
What is the mass of sulfuric acid, H2SO4, that is contained in 2.0 L of a 5.85 M solution ?
A-1147.44 grams
B- 0.12 grams
C-11.7 grams
D-98.07 grams
Hello Friends Can Yall Plz Help Me For Brainliest Don’t Comments No B.S Yu Will Get Cursed out show ya work
Answer:
it is b
Explanation:
mid ocean ridge diverges meaning it moving in two different direction horizontally - left to right
Calculate the mass in grams of 3.6 mol of iron(II) hydroxide.
Hydrogen peroxide and water both contain the same two elements. Write their chemical formulas. Compare and contrast their properties. Explain the importance of writing correct chemical formulas
Hydrogen peroxide and water both contain the same two elements. The chemical formulas of hydrogen peroxide and water are respectively H2O2 and H2O. The two compounds have distinct physical and chemical properties. Here is a long answer to the question posed.
Water is a colorless, odorless, and tasteless liquid that is vital for life. Its melting point is 0 degrees Celsius, while its boiling point is 100 degrees Celsius. Water is the most prevalent chemical substance on Earth, with about 71% of the planet's surface covered by it.
Hydrogen peroxide is a pale blue liquid that is slightly more viscous than water. It has a bitter taste, and its odor is similar to that of bleach. The boiling point of hydrogen peroxide is 150.2°C, while the melting point is -0.43°C. Hydrogen peroxide is a powerful oxidizer that is commonly used as a bleaching agent, disinfectant, and antiseptic.
Chemical formulas are utilized in a variety of scientific applications, including chemical reactions and product descriptions. An incorrect chemical formula can have significant ramifications, potentially resulting in the formation of a hazardous substance.
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Mass = 22g Volume = 2cm3 What is the Density?
Answer:
11g/cm^3
Explanation:
The density of a substance can be found By
density= Mass/volume
From the question
mass = 22 g
volume = 2 cm³
22/2=11
We have the final answer as
11 g/cm³
Answer:
11g/cm^3
Explanation:
It is because the constant formula density is
Density=Mass/ Volume.
Therefore we have
Density=22g/2cm^3
Which is = 11g/cm^3
Use the food web below to answer the following
question.
Frog
Red fox
Grasshopper
Weasel
Goldfinch
Great horned om
Clover
Deer mouse
Which group is not represerted in the food web?
A. producers
B. primary consumers
C. secondary consumers
D. decomposers
Answer:
uhm i believe its D hope i helped
Explanation:
Answer:
D. decomposers
Use the following words to fill in the text (can be used more than once).
Anode,Anode to cathode,Battery,Cathode,Chemical to electrical,Electrical to chemical,Electrolysis,Electroplating,Endothermic,Exothermic,N/A
Negative,No,Non-Spontaneous,Oxidation,Positive,Reduction,Spontaneous,Yes.
Answer:
VOLTIC CELL
Battery
Electrical to chemical
Endothermic
Electrolysis
Chemical to electrical
Anode
Anode to cathode
Battery
Endothermic
Negative
Enthomeric
Reduction
Oxidation
ELECTRYOTIC CELL
No
Non spontaneous
positive
negative
yes
Anode
Electrical to chemical
Anode to cathode
Exothermic
positive
oxidation
Electroplating
Chemical to electrical
How is activation energy represented on an energy diagram?
Energy
Reactants
Products
Reaction Progress
O A. Activation energy is the distance from the reactants to the top of
the "hill."
B. Activation energy is the distance from the top of the "hill to the
products.
O C. Activation energy is the final energy level of the product
molecules.
o
D. Activation energy is the starting energy level of the reactant
molecules.
In accordance with the diagram, activation energy is the distance from the reactants to the top of the "hill"
The activation energy represented on the energy diagram is as given as
"Activation energy is the distance from the reactant to the top of the hill"
So, the correct option is A.
What is activation energy?The minimum amount of energy required by the reactant to do the reaction is called activation energy.
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it takes 45 hours for a 6.00 mg sample of sodium-24 to decay to 0.750 mg. what is the half-life of sodium-24?
To find the half-life of sodium-24, we can use the formula for exponential decay:
N(t) = N₀ * (1/2)^(t / t₁/₂)
Where:
N(t) is the amount of the substance remaining at time t
N₀ is the initial amount of the substance
t is the time elapsed
t₁/₂ is the half-life of the substance
Given:
N₀ = 6.00 mg (initial amount)
N(t) = 0.750 mg (amount after 45 hours)
Plugging in these values, we can solve for t₁/₂:
0.750 mg = 6.00 mg * (1/2)^(45 / t₁/₂)
Dividing both sides of the equation by 6.00 mg:
(0.750 mg) / (6.00 mg) = (1/2)^(45 / t₁/₂)
0.125 = (1/2)^(45 / t₁/₂)
To eliminate the base of 1/2, we can take the logarithm of both sides:
log₂(0.125) = 45 / t₁/₂ * log₂(1/2)
Using the property log₂(a^b) = b * log₂(a):
log₂(0.125) = 45 * log₂(1/2) / t₁/₂
We know that log₂(1/2) = -1, so we can simplify the equation further:
log₂(0.125) = -45 / t₁/₂
Now, we can solve for t₁/₂:
t₁/₂ = -45 / log₂(0.125)
Using a calculator:
t₁/₂ ≈ -45 / (-3) ≈ 15
The half-life of sodium-24 is approximately 15 hours.
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calculate the temperature (in°c) at which pure water would boil at a pressure of 537.7 torr. hvap = 40.7 kj/mol
At a pressure of 537.7 torr, pure water would boil at approximately 95.8°C.
To calculate the boiling point of water at a given pressure, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = (-ΔHvap/R) × (1/T2 - 1/T1)
where:
P1 and P2 are the initial and final pressures,
ΔHvap is the molar enthalpy of vaporization,
R is the ideal gas constant (8.314 J/(mol·K)),
T1 is the initial temperature,
T2 is the final temperature.
In this case, we want to find the boiling point (T2) of water at a pressure of 537.7 torr. The normal boiling point of water is 100°C (373.15 K) at atmospheric pressure (760 torr).
We can rearrange the equation to solve for T2:
T2 = (1/(((-ΔHvap/R) × (1/T1)) + (ln(P2/P1))))
Substituting the given values:
P1 = 760 torr
P2 = 537.7 torr
ΔHvap = 40.7 kJ/mol (convert to J/mol by multiplying by 1000)
R = 8.314 J/(mol·K)
T1 = 373.15 K
T2 = (1/(((-40.7 kJ/mol × 1000 J/kJ) / (8.314 J/(mol·K)) × (1/373.15 K)) + (ln(537.7 torr/760 torr))))
Calculating this expression gives us:
T2 ≈ 95.8°C
Therefore, at a pressure of 537.7 torr, pure water would boil at approximately 95.8°C.
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