A drop of water is denser than a ping-pong ball.
Usually, water is made of particles that are firmly pressed together. In differentiation, plastic (the material ping pong balls are made of) may be a lightweight fabric and the particles are not as firmly stuffed together.
The thickness of a ping pong ball is 0.0840 g/cm³, though water’s thickness is 997 kg/m³. Subsequently, ping pong balls aren’t about as thick as water and will continuously coast and surface greatly quickly.
The ping pong ball appears to oppose gravity and coast within the air.
Ping-pong balls drift within the water since they are amazingly lightweight, empty, and filled with air. Too, the water’s surface pressure makes it simple for the ping pong ball to drift.
In expansion, water is denser than ping pong balls, making them look for the most noteworthy point of water.
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Imagine the movement of gas particles in a closed container. According to the kinetic molecular theory, which statements below are true of the gas particles? Check all that apply.
Based on the kinetic molecular theory, the statements that are true of the gas particles are as follows:
Gas particles act like tiny, solid spheres; option A.Gas particles are in constant, random motion; option BCollisions are elastic, there is no energy lost as the particle hits the sides of the container; option C.What is the kinetic molecular theory?The Kinetic Molecular Theory is the theory that molecules of a gas are in constant random motion colliding with one another and with the wall of their container.
These collisions between the gas molecules are perfectly elastic and are responsible for the pressure of gas molecules.
The higher the kinetic energy of the gas molecules, the higher the temperature of the gas.
The kinetic molecular theory explains the physical state of gases since gas molecules have negligible intermolecular forces.
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Complete question:
Imagine the movement of gas particles in a closed container. According to the kinetic molecular theory, which statements below are true of the gas particles? Check all that apply. Gas particles act like tiny, solid spheres. Gas particles are in constant, random motion. Gas particles at lower temperatures move faster. Collisions are elastic, there is no energy lost as the particle hits the sides of the container. Slower moving particles collide more often and with more force with the container.
Question 1 (3 points)
Listed in the Item Bank are some important labels for sections of the image below.
Select the label for each corresponding area it identifies in
-noble gas
-alkaline earth metals
-transition metals
-alkali metals
-nonmetal
-metalloids
-halogens
The first column represent the alkali metal, the second column represent alkaline earth metals, the 3rd horizontal middle block is d block, the right 6 columns of the periodic table represent p block and the rightest column is noble gas.
What is noble gas, transition metals and it's places?See figuring out the right place of the elements mentioned in the periodic table is easy because its a predicted and factual thing.The first column we see in the periodic table is alkali metals.The second column we see in the periodic table is alkaline earth metals.The middle horizontal slit of elements called as d block elements starting from scandium.The right six columns belongs to p block elements and the rightest column represents noble gas elements.To know more about metals visit:
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A marine biologist is preparing a deep-sea submersible for a dive. The sub stores breathing air under high pressure in aspherical air tank that measures 73.0 cm wide.The biologist estimates she will need 8200. L of air for the dive. Calculate the pressure to which this volume of air must becompressed in order to fit into the air tank. Write your answer in atmospheres. Round your answer to 3 significant digits.0atm0.0XS ?EoloPEBH
To solve the problem we will assume the following:
1. Air behaves as an ideal gas during all the process.
2. The initial air equivalent to 8200L is at atmospheric pressure. It means 1 atm.
3. The temperature remains constant.
Taking into account the above, we can apply the Boyle-Marriote Law that relates the change in pressure and volume at constant temperature. The equation that we will use will be:
[tex]P_1V_1=P_2V_2[/tex]Where,
P1 is the atmospheric pressure. 1atm
V1 is the initial volume of air required, 8200L
P2 is the final pressure we want to find
V2 is the final volume, it means the volume of the spherical air tank. We will calculate this volume using the volume equation for a sphere:
[tex]V_2=\frac{4}{3}\pi r^3[/tex]r is the radius of the sphere, r=73cm/2=36.5cm
So, the volume of the spherical air tank will be:
[tex]\begin{gathered} V_2=\frac{4}{3}\pi\times(36.5cm)^3=20.4\times10^4cm^3 \\ V_2=20.4\times10^4cm^3\times\frac{1L}{1000cm^3}=204L \end{gathered}[/tex]No, we clear P2 from the first equation and replace known data:
[tex]\begin{gathered} P_2=\frac{P_1V_1}{V_2} \\ P_2=\frac{1atm\times8200L}{204L} \\ P_2=40.3atm \end{gathered}[/tex]The pressure of the gas must be 40.3 atm
Answer: 40.3
The volume of a gas is 480 mL at 45.0 oC the temperature is increased to 60.0 oC at a constant pressure what is the new volume
Answer
The new volume = 0.503 L or 503 mL
Explanation
Given:
Initial volume, V₁ = 480 mL = (480/1000) = 0.480 L
Initial temperature, T₁ = 45.0°C = (45 + 273 K) = 318 K
Final temperature, T₂ = 60.0°C = (60 + 273 K) = 333 K
What to find:
The new volume, V₂ when the temperature increases to 60.0°C.
Step-by-step solution:
Since the pressure is constant, the new volume, V₂ can be calculated using Charle's law formula.
[tex]\begin{gathered} \frac{V_1}{T_1}=\frac{V_2}{T_2} \\ \\ \Rightarrow V_2=\frac{V_1T_2}{T_1} \end{gathered}[/tex]Plugging the values of the given parameters into the formula, we have
[tex]V_2=\frac{0.480L\times333K}{318K}=0.503\text{ }L[/tex]The new volume, V₂ when the temperature increased to 60.0°C = 0.503 L or (0.503 x 1000 mL) = 503 mL
Silver Sulfate reacts with Potassium Chloride according to the following reaction:Ag2SO4 + 2KCl -> 2AgC1 + K2SO4a. If 30.0 grams of Ag2SO4 reacts with 10.0 grams of KCl, what mass of AgCl is produced by the reaction b. the limiting reactant is ______c. how many grams ok K2SO4 can be produced.d. how many grams of excess reactant remain after the reaction e. what is the percent yield if there is 10.0g of AgCl
Answer:
Explanations:
Given the chemical reaction
[tex]Ag_2SO_4+2KCl\rightarrow2AgCl+K_2SO_4[/tex]Given the following
Mass of Ag2SO4 = 30grams
Mass of KCl = 10grams
Determine the moles of the reactants
[tex]\begin{gathered} mole\text{ of Ag}_2SO_4=\frac{mass}{molar\text{ mass}}molar\text{ mass} \\ mole\text{ of Ag}_2SO_4=\frac{30g}{311.799} \\ mole\text{ of Ag}_2SO_4=0.0962moles \end{gathered}[/tex][tex]\begin{gathered} mole\text{ of KCl}=\frac{10g}{74.5513} \\ moleof\text{ KCl}=0.1341moles \\ 1mole\text{ of KCl}=\frac{0.1341}{2}=0.06707moles \end{gathered}[/tex]B) B) Since the 1 moles of KCl is lower than the moles of Ag2SO4, hence KCl willl be the limiting reactant.
A) A) According to stoichiometry, 2 moles of KCl produces 2 moles of AgCl, the mass of AgCl produced will be given as;
[tex]\begin{gathered} mass=mole\times molar\text{ mass} \\ mass\text{ of AgCl}=0.1341\times143.32 \\ mass\text{ of AgCl}=19.22grams \end{gathered}[/tex]C) According to stoichiometry, 2 moles of KCl produces 1 moles of K2SO4, the mole of K2SO4 produced is;
[tex]\begin{gathered} mole\text{ of }K_2SO_4=\frac{1}{2}\times0.1341 \\ mole\text{ of }K_2SO_4=0.06707moles \end{gathered}[/tex][tex]\begin{gathered} mass\text{ of K}_2SO_4=mole\times molar\text{ mass} \\ mass\text{ of K}_2SO_4=0.06707\times174.259 \\ mass\text{ of K}_2SO_4=11.69grams \end{gathered}[/tex]I cant figure this one out. How do I know ΔH∘ for the preceding reaction without knowing the standard enthalpie of F2?
The enthalpy of formation (ΔH°) of the given chemical reaction is equal to -905.3 KJ/mol.
What is the standard enthalpy of formation?The standard enthalpy of formation is defined as the change in enthalpy when one mole of a substance is formed from its pure elements under the standard state (1 atm of pressure and 298.15 K).
The standard enthalpy of formation is the energy released or absorbed when one mole of a substance is formed from its constituents under standard conditions. The symbol of the standard enthalpy of formation is represented by [tex]\triangle {H^o_f}[/tex].
The equation for the standard enthalpy change of formation for a balanced chemical reaction is shown below,
ΔH° (reaction) = ∑ [ΔH° of (products)− ∑ ΔH° of (Reactants)]
The standard enthalpy of the formation of fluorine gas = 0
The standard enthalpy of the formation of iodine gas = 62.23 KJ/mol
The standard enthalpy of the formation of IF₅ = -843 KJ/mol
ΔH° (reaction) = - 843 - 62.23 = - 905 .3 KJ/mol
Therefore, the enthalpy of formation (ΔH°) of the given chemical reaction is -905.3 KJ/mol.
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For the reaction A (g) → 3 B (g), Kp = 67900 at 298 K. When ∆G = -14.2 kJ/mol, what is the partial pressure of A when the partial pressure of B is 2.00 atm for this reaction at 298 K.
The partial pressure is the individual pressure of a gas. The partial pressure of gas A is 1.11 × 10⁻⁴atm.
What is partial pressure?
In a mixture of gases, the partial pressure of a gas is the individual pressure of that gas in the mixture of gases.
Given the reaction; A (g) → 3 B (g)
Kp = P(B)³/PA
Where Kp = 67900
PB = 2.00 atm
67900 = (2)³/PA
PA = (2)³/67900
PA = 1.11 × 10⁻⁴atm
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An atom with 3 protons in the nucleus and 3 electrons in the orbitals would have what overall charge?
Ans6
wer:
b
Explanation:
yes or no ? experimentation with corn and other crops led to the development of new fuels called biofuels
Yes, the experimentation with corn and other crops led to the development of new fuels called biofuels.
What are biofuels?Biofuel is a biodegradable, inexhaustible, fuel that is produced from Biomass. Biofuel is considered the easiest available and pure fuel on the earth. Biofuels are manufactured from biomass such as wood and straw, which are liberated by direct combustion of dry matter and converted into a gaseous and liquid fuel.
The organic matter such as sludge, sewage, and vegetable oils, can be changed into biofuels by a wet process such as fermentation and digestion. Biofuel is available in all regions of the world, and mainly includes fuels such as Biodiesel, Bioethanol, and Bio methanol.
The two types of biofuels are bioethanol and biodiesel most commonly used these days. Both of these biofuels are the first generation of biofuel technology.
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When a bright light is shined on glow in the dark powder what is happening to the electrons.
a.The electrons are absorbing some of the light and are become excited.
b.The atoms become negatively charged ions.
c.Nothing, the electrons are not affected by light.
d.The electrons move to a lower energy state.
Show the conversions required to solve this problem and calculate the grams of Al2O3 .
Explain how the following reaction demonstrates that matter is neither created or destroyed in a chemical reaction: Ca(OH)2 + 2HCI-> CaCl2 + 2H20
Answer:
In this reaction, Ca(OH)2 is a reducing agent. It reacts with hydrogen chloride to form calcium chloride and water. Therefore, the following reaction shows that matter is neither created nor destroyed in a chemical reaction: Ca(OH)2 + 2HCI -> CaCl2 + 2H20. The formation of calcium chloride and water from the hydrolysis of calcium hydroxide is not an example of matter being created or destroyed in a chemical reaction because it does not involve the breaking down of any bonds between atoms.
Explanation:
what is combustion? explain to me pls
Answer:
rapid chemical combination of a substance with oxygen, involving the production of heat and light.
Explanation:
Combustion is a chemical process in which a substance reacts rapidly with oxygen and gives off heat. The original substance is called the fuel, and the source of oxygen is called the oxidizer.
if you want to learn more:
https://www.britannica.com/science/combustion
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explain Maxwell equations and maxwell thermodynamics relations; it's significance and application to ideal gases?
Answer:
In order to understand the Maxwell equations and relations derived from Maxwell relations, you will have to know Euler's Reciprocity.
It states that for any state thermodynamic quantity (let x and y) having a state function phi, it must satisfy the following condition.
[tex](\frac{ {∂}^{2}\phi }{∂x \cdot∂y}) = (\frac{ {∂}^{2}\phi }{∂y \cdot∂x})[/tex]
Maxwell equations: These are a set of equations derived from the application of Euler's Reciprocity. The four Maxwell equations are as follows:
[tex]dH = TdS + VdP[/tex]
[tex]dG = VdP - SdT[/tex]
[tex]dA = -PdV - SdT[/tex]
[tex]dU = TdS - PdV[/tex]
Let's derive each Maxwell relations step by step,
1) dH = TdS + VdP
Enthalpy is a function of Entropy & Pressure
[tex] \sf \qquad \qquad H = f(S,P)[/tex]
[tex]dH = TdS + VdP[/tex]
[tex]dH = (\frac{∂H}{∂S})_{_P} dS + (\frac{∂H}{∂P})_{_S}dP[/tex]
Comparing both the above equation,
[tex]T = (\frac{∂H}{∂S})_{_P}[/tex]
[tex]V =(\frac{∂H}{∂P})_{_S}[/tex]
now we know that Enthalpy is a state function hence applying cross reciprocality,
[tex](\frac{∂V}{∂S})_{_P}= (\frac{∂T}{∂P})_{_S}[/tex]
This is called the first Maxwell relation,
Similarly,
2) dG = -SdT + VdP
Free energy is a function of Temperature & Pressure,
[tex] \sf \qquad \qquad G = f(T,P)[/tex]
[tex]dG = -SdT + VdP[/tex]
[tex]dG = (\frac{∂G}{∂P})_{_T} dP + (\frac{∂G}{∂T})_{_P}dT[/tex]
Comparing both the above equation,
[tex](\frac{∂G}{∂P})_{_T} = V [/tex]
[tex] (\frac{∂G}{∂T})_{_P}= -S[/tex]
now we know that Free energy is a state function hence applying cross reciprocality,
[tex]-(\frac{∂S}{∂T})_{_P}= (\frac{∂V}{∂P})_{_T}[/tex]
This is called the second Maxwell relation,
3) dA = -PdV - SdT
helmholtz free energy is a function of Temperature & Volume,
[tex] \sf \qquad \qquad A = f(T,V)[/tex]
[tex]dA = -PdV - SdT[/tex]
[tex]dA = (\frac{∂A}{∂V})_{_T} dV + (\frac{∂A}{∂T})_{_V}dT[/tex]
Comparing both the above equation,
[tex](\frac{∂A}{∂V})_{_T} = -P [/tex]
[tex] (\frac{∂A}{∂T})_{_V}= -S[/tex]
now we know that helmholtz free energy is a state function hence applying cross reciprocality,
[tex]-(\frac{∂S}{∂V})_{_T}= -(\frac{∂P}{∂T})_{_V}[/tex]
[tex](\frac{∂S}{∂V})_{_T}= (\frac{∂P}{∂T})_{_V}[/tex]
This is called the third Maxwell relation,
4) dU = TdS - PdV
Internal energy is a function of Entropy & Volume,
[tex] \sf \qquad \qquad A = f(S,V)[/tex]
[tex]dU = TdS - PdV [/tex]
[tex]dU = (\frac{∂U}{∂S})_{_V} dS + (\frac{∂U}{∂V})_{_S}dV[/tex]
Comparing both the above equation,
[tex](\frac{∂U}{∂V})_{_S} = -P [/tex]
[tex] (\frac{∂U}{∂S})_{_V}= T[/tex]
now we know that Internal energy is a state function hence applying cross reciprocality,
[tex](\frac{∂T}{∂V})_{_S}= -(\frac{∂P}{∂S})_{_V}[/tex]
This is called the fourth Maxwell relation,
The main significance of the Maxwell relation is that those thermodynamic quantities which are unmeasurable can be replaced with measurable quantities with the help of the Maxwell relation.
The derivative of the extensive asset in relation to the extensive asset gives the intensive asset; with respect to the intensive asset, the derivative of the extensive asset gives the extensive asset. This is the result of the overall Maxwell relations.
The coefficient of expansion and compression of a gas in thermodynamics is the application of the Maxwell relations.
There are 3 coefficients introduced,
Coefficient of thermal expansion (expansivity) α
[tex]α= \frac{1}{V} (\frac{∂V}{∂T} )_{_P}[/tex]
Coefficient of isothermal compressibility β
[tex]β = -\frac{1}{V} (\frac{∂V}{∂P} )_{_T}[/tex]
Isochoric thermal expansion coefficient γ
[tex] γ= \frac{1}{P} (\frac{∂P}{∂T} )_{_V}[/tex]
For ideal gases,
PV = nRT
For one mole ideal gas (n=1),
PV = RT
Taking derivative with respect to T at constant pressure,
[tex]V(\frac{∂P}{∂T} )_{_P}+ P(\frac{∂V}{∂T} )_{_P}= R(\frac{∂T}{∂T} )_{_P}[/tex]
At constant pressure ∂P = 0, & R.H.S = 1, hence
[tex](\frac{∂V}{∂T} )_{_P}= \frac{R}{P}[/tex]
[tex]α= \frac{1}{V} (\frac{∂V}{∂T} )_{_P}[/tex]
[tex]α= \frac{1}{V} \cdot \frac{R}{P}[/tex]
[tex]\sf Also, PV=RT\\ \frac{R}{PV} = \frac{1}{T}[/tex]
[tex]\boxed{α= \frac{1}{T}}[/tex]
Following the same procedure, by taking derivating w.r.t. pressure at constant temperature.
[tex]V(\frac{∂P}{∂P} )_{_T}+ P(\frac{∂V}{∂P} )_{_T}= R(\frac{∂T}{∂P} )_{_T}[/tex]
[tex]V+ P(\frac{∂V}{∂P} )_{_T}= 0[/tex]
[tex](\frac{∂V}{∂P} )_{_T}= \frac{-V}{P}[/tex]
Substituting the above value in,
[tex]β = -\frac{1}{V} (\frac{∂V}{∂P} )_{_T}[/tex]
[tex]β = -\frac{1}{V} \cdot \frac{-V}{P}[/tex]
[tex] \boxed{β = \frac{1}{P}}[/tex]
Repeating the same procedure again, i.e. derivative w.r.t. T at constant volume
[tex]V(\frac{∂P}{∂T} )_{_V}+ P(\frac{∂V}{∂T} )_{_V}= R(\frac{∂T}{∂T} )_{_V}[/tex]
At constant Volume ∂V = 0, and R.H.S = 1, hence overall equation becomes,
[tex]V(\frac{∂P}{∂T} )_{_V} = R \\ (\frac{∂P}{∂T} )_{_V} = \frac{R}{V}[/tex]
Substituting above value in,
[tex]γ= \frac{1}{P} (\frac{∂P}{∂T} )_{_V}[/tex]
[tex]γ= \frac{1}{P} \cdot \frac{R}{V}[/tex]
[tex] \boxed{γ= \frac{1}{T}}[/tex]
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The Maxwell equation in thermodynamics is very important and useful because the set of relations allows the scientists to change specific unknown quantities.
What are the Maxwell's thermodynamic?The Maxwell equation in thermodynamics is very useful because this is the set of relations that allows physicists to change certain unknown quantities that are hard to measure in the real world. These quantities need to be replaced by many easily measured quantities. Maxwell's relations are a set of equations in thermodynamics that are derivable from the second derivatives and from the definitions of the thermodynamic potentials. These relations are named for the nineteenth-century physicist James Clerk Maxwell. So entropy and pressure are the natural variables of enthalpy. Maxwell relations are thermodynamic equations that establish the relations between various thermodynamic quantities in equilibrium and other fundamental quantities known as thermodynamical potentials
So we can conclude that Maxwell's thermodynamics are the set of relations allows the scientists to change unknown quantities.
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1. What is the difference between a homogenous and a heterogeneous mixture? 2. Give 3 examples of each mixture and explain its uses in our daily lives.
Answer:By combining two or more substances, a mixture is produced. A homogeneous solution tends to be identical, no matter how you sample it. Homogeneous mixtures are sources of water, saline solution, some alloys, and bitumen. Sand, oil and water, and chicken noodle soup are examples of heterogeneous mixtures.
Explanation:
A liquid in the lab has a density of 1.24 g/cm³ . What is the mass in grams of 254 mL of the liquid? I dont know where to start when to convert because of the Activ app. I need help on how to answer this.
The mass in grams of the 254 mL of the liquid, given that it has a density of 1.24 g/cm³ is 314.96 grams
How to determine the mass of the liquidWe'll begin by listing the out the parameters obtained from the question. This is given below:
Density of liquid = 1.24 g/cm³Volume of liquid = 254 mL = 254 cm³Mass of liquid =?The mass of the liquid can be obtained as follow:
Density = mass / volume
Cross multiply
Mass = Density × Volume
Mass of liquid = 1.24 × 254
Mass of liquid = 314.96 grams
Thus, the mass of the liquid is 314.96 grams
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5. During factory inspection, Just Lemons, Inc. discovered that a water valve to the
lemonade mixing station was not functioning. Once they repair it, more water will
enter the mixing station. From what you know about the limiting and excess
ingredients for current lemonade production, what advice would you give engineers
about the upcoming increase in water?
The advise which would be given to the engineers is for them to determine the limiting reactant so water will not be in excess.
What is a Limiting reagent?This is referred to as the reactant which is completely used up in a chemical reaction and it helps to determine when the reaction stops. In this scenario, either water, lemon juice or sugar is a limiting reactant.
The best option in this scenario is to tell the engineers to determine the limiting reactant so as to ensure that the quality of lemonade made is maximized and is therefore the right choice.
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PLEASE HELP ASAP, I will mark the "brainiest"
In the generic reaction: 5A + 2B --> 6C + 3D The molar mass of A is 146.70g/mol while the molar mass of C is 21.31g/mol. If 4.253 g of A reacts with an excess of B, how many grams of C can form? Remember significant figures and units.
The amount, in grams, of C that can form from the reaction, is 0.7414 g.
Stoichiometric problemFrom the equation of the reaction, the mole ratio of A to that of C is 5:6.
In other words, for every 5 moles of A that reacts, 6 moles of C is produced.
Molar mass of A = 146.70 g/mol
Molar mass of C = 21.31 g/mol
Recall that: mole = mass/molar mass
Mole of 4.253 g of A = 4.253/146.70
= 0.029 mol
From the mole ratio, the equivalent mol of C that will be produced is:
0.029 x 6/5 = 0.03479 mol
mass = mole x molar mass
Mass of 0.03479 mol of C = 0.03479 x 21.31
= 0.7414 g (to the appropriate sig. fig.)
In other words, the amount of C that will be produced from 4.253 g of A will be 0.7414 g.
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Consider the following processes: 1. A ─→ B delta H = ─100 kJ 2. B ─→ C + D delta H = ─50 kJ 3. E ─→ D delta H = ─25 kJ What is the delta H for the process: A ─→ C + E?
The enthalpy change, ΔH of the process: A → C + E is -125 kJ.
What is the enthalpy change of a reaction?The enthalpy change of a reaction is the sum of the heat changes that occur as reactants combine to form products.
The enthalpy change of the reaction is given by the following formula:
Enthalpy change = sum of energy of bonds broken - sum of energy of the bonds formedΔH = ∑H(bonds broken) - ∑H(bonds formed)The delta H for the process A → C + E is given as follows;
1. A ─→ B ΔH = ─100 kJ
2. B ─→ C + D ΔH = ─50 kJ
3. E ─→ D delta H = ─25 kJ
-3. D → E = +25 kJ
A → C + D = 1 + 2 + (- 3)
A → C + E = (- 100 - 50 + 25) kJ
A → C + E = -125 kJ
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How many molecules are in 59.73 grams of the theoretical acid borofuric acid, H2B2O2?
remember units and sig figs.
If your answer is a very large number, use scientific notation with an 'E' in this format:
120,000,000 = 1.2E8
The number of molecules in 59.73 grams of the theoretical acid, borofuric acid, would be 6.44 x [tex]10^2^3[/tex] molecules.
Avogadro's numberAccording to Avogadro, 1 mole of any substance contains 6.022 x [tex]10^2^3[/tex] molecules.
Recall that: number of moles in a substance = mass of the substance/molar mass of the substance.
Molar mass of borofuric acid, [tex]H_2B_2O_2[/tex]:
H = 1 g/mol
B = 10.8 g/mol
O = 16 g/mol
(1x2) + (10.8x2) + (16x2) = 55.6 g/mol
Number of moles of 59.73 grams [tex]H_2B_2O_2[/tex] = 59.73/55.6
= 1.07 moles
Since 1 mole = 6.022 x [tex]10^2^3[/tex] molecules
1.07 moles of borofuric acid = 1.07 x 6.022 x [tex]10^2^3[/tex] molecules.
= 6.44 x [tex]10^2^3[/tex] molecules
Thus, 59.73 grams of borofuric acid contains 6.44 x [tex]10^2^3[/tex] molecules.
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Please help me it’s due today!!!!
The following are the examples of a chemical property of a substance;
1) A banana reacts with orange and turns brown
2) Iron will rust when exposed to oxygen and water
3) Iron will react with acid to form iron chloride.
What is a chemical reaction?The term chemical reaction has to do with the change in the properties of substances that are combined together in order to yield a product. Recall that the properties of the reactant change as the product is formed.
When a chemical reaction occurs
1: A new color may appear
2) A new gas may be seen
3) The temperature may be changed
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How much work (in J ) is involved in a chemical reaction if the volume decreases from 4.25 to 1.97 L against a constant pressure of 0.838 atm ?
The amount of work (in J ) is involved in a chemical reaction if the volume decreases from 4.25 to 1.97 L against a constant pressure of 0.838 atm is: 8.78 Joules.
What is work done?The work done by an object can be calculated by multiplying the force acting on it by the perpendicular distance covered by the physical object. Also, the work done during a chemical reaction can be determined by multiplying the change in volume by the accompanying level of pressure.
Mathematically, the work done by a system is given by the formula:
Work done = PΔV = P(V₂ - V₁)
Where:
P represents the pressure.V represents the volume.Substituting the given parameters into the formula, we have;
Work done = 0.038(4.25 - 1.97)
Work done = 0.038(2.28)
Work done = 0.08664 L.atm
Next, we would convert L.atm to Joules:
Work done = 101.325 × 0.08664
Work done = 8.78 Joules.
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The Ka of a monoprotic weak acid is 0.00469. What is the percent ionization of a 0.141 M solution of this acid?
Use quadratic equation.
The percent ionization of an acidic solution can be calculated from the H+ concentration. the percent ionization of the monoprotic acid of 0.141 M is 18.23 %.
What is percent ionization?Percent ionization of an acidic solution is the percent of H+ ions in the solution. Thus, mathematically, it is the ratio of H+ ion concentration to the concentration of solution multiplied by 100.
Let HA be the monoprotic acid when it ionizes, forming equal concentration of H+ and A- let it be x. Thus ionization constant can be written as follows:
Ka = [x]² /[HA]
0.00469 =[x]²/[0.141 M]
[X] = 0.025. = [H+]
Percentage ionization = (0.025 M / 0.141 M)× 100
= 18.23 %
Therefore percentage ionization of the acid is 18.23%.
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6. In order to prepare 50.0 mL of 0.100 M NaCl you will add _____ grams of NaCl to _____ mL of wate
The first step to solve this problem is to find the number of moles of NaCl in 50.0mL of 0.100M NaCl. To do it, convert the volume of solution to liters and multiply it by the concentration of the solution (M=mol/L).
[tex]50mL\cdot\frac{1L}{1000mL}\cdot\frac{0.100mol}{L}=0.005mol[/tex]Now, use the molar mass of NaCl to find the mass of 0.005moles of it (MM=58.44g/mol):
[tex]0.005mol\cdot\frac{58.44g}{mol}=0.2922g[/tex]It means that you have to add 0.2922g to 50mL of water.
Scientists can use chemicals to clean toxins from the soil. What does this
show about chemicals?
Any substance made of matter is by definition a chemical, which includes solids, liquids, and gases. Pure substances or mixtures of substances can both be found in chemicals. Water (H2O), for example, is a pure chemical because the same molecules and molecular combinations are present throughout its whole structure.
Explain about the chemicals ?Any material with a known composition is a chemical. Some chemicals, like water, are found in nature, meaning they always consist of the same "stuff." Other chemicals, including chlorine, are produced (used for bleaching fabrics or in swimming pools).
A chemical compound is a material that contains a specific combination of atoms or ions. Chemical compounds are made up of two or more elements coming together through a chemical process. While all substances are compounds, not all substances are compounds.
On its list of substances covered by the Toxic Substances Control Act, the EPA has more than 85,000 chemicals listed.
Not all chemicals are detrimental to the economy.
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Can someone help me with this? I’m not understanding what it’s asking of me
Answer: You need to research what ever you are working on and find 2 examples.
Explanation:
The atomic radius of a boron atom is 9 times 10-11m. Give its atomic radius in nanometres?
The atomic radius of boron atom is 0.09nm
What is an atomic radius?
Atomic radius, also known as atomic radii, is the total distance between an atom's nucleus and its electron's outermost orbital.
Atomic radius is thus defined as: The distance from the nucleus's center to the edge of its surrounding electron shells.
The radius of an atom is comparable to a circle's radius. The electron's most distant orbital is likened to the circle's outside edge, and the nucleus to the circle's center. The position of the outermost electron is unclear, making it challenging to calculate the atomic radii.
According to the question,
1m = 10-9nm
So, 9*10-11m = 0.09nm
The atomic radius of boron atom is 0.09nm
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The cost of electricity to deposit 10gm of Mg is Rs. 60. How much would it cost to deposit 100gm of copper from CuSO
, ? (At. wt. of Cu = 63.5)
The cost of electricity to deposit 100g of copper from CuSO₄ will be Rs227.1
We know, m = zit
1unit current= 1kwh
Energy=Vit
m= (E. weight/F)×it
m proportional to (E. weight)×cost
m₁/m₂ = E₁/E₂ × C₁/C₂
E.weight = (Molecular weight/n-factor)
According to question,
Cu² + 2e- --- Cu
Mg² + 2e- --- Mg { n-factor of both=2}
E₁ = (molecular weight of Mg/2) = (24÷2) = 12
E₂ = (molecular weight of Cu/2) = (63.5) = 31.7
So, M₁/M₂ = E₁×C₁/E₂×C₂ (C₁=cost given)
10/100 = 12×60/31.7×C₂
C₂= Rs227.1
Therefore the cost of electricity to deposit 100g of copper from CuSO₄ will be Rs227.1
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The partial pressure of oxygen was observed To be 156 torr
When the partial pressure of oxygen is 156 torr and the atmospheric pressure is 743 torr, the mole fraction of oxygen is 0.210.
What is partial pressure?Partial pressure is the pressure exerted by an individual gas in a mixture of gases. The partial pressure of a gas depends on its mole fraction.
The relationship between the partial pressure of a gas and the total pressure is given by Dalton's law, which states that the sum of the partial pressures is equal to the total pressure.
We can calculate the partial pressure of oxygen using the mathematical expression of Dalton's law:
pO₂ = P × X(O₂)
X(O₂) = pO₂ / P
X(O₂) = 156 torr / 743 torr = 0.210
where,
pO₂ is the partial pressure of oxygen.P is the total pressure of the mixture.X(O₂) is the mole fraction of oxygen.The mole fraction of oxygen is 0.210 when its partial pressure is 156 torr and the atmospheric pressure is 743 torr.
The complete question is:
The partial pressure of oxygen was observed To be 156 torr in air with a total atmospheric pressure of 743 torr. Calculate the mole fraction of oxygen present.
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At STP, exactly 1 mole of any gas occupies 22.4 L. What size container (in Liters) do you need to hold 2.1mol O₂ gas at STP?(Show all work)A) 44.2B) 47.0C) 40.0D) 55.5
Answer
B) 47.0
Explanation
If at STP, exactly 1 mole of any gas occupies 22.4 L
Then, 2.1 moles of O₂ gas at STP will occupy x L container
To get x L, cross multiply and divide both sides by 1 mole.
[tex]x\text{ }L\text{ }container=\frac{2.1\text{ }mol}{1\text{ }mol}\times22.4\text{ }L=47.0\text{ }L[/tex]Therefore, the size of the container (in Liters) needed to hold 2.1 mol O₂ gas at STP is 47.0 L