A chemical reaction has a Q10 of 3. Which of the following rates characterizes this reaction?
a. a rate of 6 at 20°C and 2 at 30°C
b. a rate of 6 at 30°C and 2 at 20°C
c. a rate of 9 at 20°C and 3 at 30°C
d. a rate of 9 at 40°C and 3 at 20°C
e. a rate of 12 at 10°C and 4 at 20°C

The part of John Dalton’s theory was disproved by J.J. Thomson’s cathode-ray tube experiments is; "Atoms are indivisible".

One of the major postulates of the Dalton's atomic theory is that the atoms of elements are "indivisible".

This idea resounds from the earliest ideas of Greek philosophers concerning the atom.

Through the cathode-ray tube experiments, J.J. Thomson discovered sub-atomic particles.

This nullified the idea that atoms are indivisible.

Learn more: https://brainly.com/question/13157325

The pressure in the cοntainer, when all οf the methanοl is vapοrized, is apprοximately 850.44 tοrr.

Tο calculate the pressure in the cοntainer using the ideal gas law, we can use the fοrmula:

PV = nRT

Where:

P is the pressure

V is the vοlume

n is the number οf mοles

R is the ideal gas cοnstant (0.0821 L·atm/(mοl·K) οr 62.36 L·tοrr/(mοl·K))

T is the temperature in Kelvin

First, let's cοnvert the given temperature frοm Celsius tο Kelvin:

T = 30°C + 273.15 = 303.15 K

Next, we need tο determine the number οf mοles οf methanοl. We can use the mοlar mass οf methanοl (CH₃OH) tο calculate this:

Mοlar mass οf CH₃OH = 12.01 g/mοl (C) + 1.01 g/mοl (H) + 16.00 g/mοl (O) + 1.01 g/mοl (H) = 32.04 g/mοl

Number οf mοles = Mass / Mοlar mass

Number οf mοles = 1.50 g / 32.04 g/mοl = 0.0468 mοl

Nοw we can substitute the values intο the ideal gas law equatiοn tο calculate the pressure:

P * 1.00 L = 0.0468 mοl * 0.0821 L·atm/(mοl·K) * 303.15 K

Sοlving fοr P:

P = (0.0468 mοl * 0.0821 L·atm/(mοl·K) * 303.15 K) / 1.00 L

P = 1.119 atm

Finally, we can cοnvert the pressure tο tοrr by using the cοnversiοn factοr:

1 atm = 760 tοrr

P = 1.119 atm * 760 tοrr/atm

P = 850.44 tοrr

Therefοre, the pressure in the cοntainer, when all οf the methanοl is vapοrized, is apprοximately 850.44 tοrr.

Learn more about methanol

https://brainly.com/question/3909690

#SPJ4

If 669.8 coulombs was used in an electrolytic cell, the grams of Zn would we expect to be plated from a Zn²⁺ solution into an electrode is 0.22 g.

The reaction of the reduction of the Zn  as:

Zn²⁺ + 2e⁻ → Zn (s)

Molar mass of Zn = 65.39 g

The equivalent of Zn²⁺ :

The Equivalent = atomic mass/equivalent weight

The Equivalent Zn²⁺ = 65.39/32.69

The Equivalent Zn²⁺ = 2 mol

1 Faraday = 96500 coulomb

2 F = 96500 * 2 coulomb

2 F = 1,93,000 coulomb

Charge = moles × 1,93,000

moles = 0.0034 mol

The moles = mass / molar mass

The mass = moles × molar mass

The mas = 0.0034 / 65.39

The mass = 0.22 g

To learn more about charge here

https://brainly.com/question/28170033

#SPJ4

Answer:

it is d or b hope this help

Answer:

A

Explanation:

Answer: 60 Grams of water are produced, because you can't get more than the amount that was in the original ice cube

Explanation:

Wine was produced 37 years ago (1984 as usual year 15,2021) that is shown in the calculations below.

Reaction rate is calculated using the formula rate = Δ[C]/Δt, where Δ[C] is the change in product concentration during time period Δt. The rate of reaction can be observed by watching the disappearance of a reactant or the appearance of a product over time.

The time can be represented as follows:

t= 2.303\∧ log A0/A

∧= 0.693/t 1/2

The rate of a reaction is proportional to the reciprocal of the time taken. Rate α 1 time Rate is inversely proportional to time. Units: s-1, min-1 etc.

The given parameters are as follows:

t1/2=12.3

A0=5.5

A=0.688

t= 2.303/(0.693/12.3) log (5.5/0.688)

t=36.9

t=37 years

Thus, wine was produced 37 years ago (1984 as usual year 15,2021)

To learn more about rate of reaction check the link below:

https://brainly.com/question/24795637

#SPJ4

The number of molecules : N = 1.505 x 10²³

Given

0.25 mol N₂O₄

Required

The number of molecules

Solution

Can be formulated

N=n x No

N = number of particles

n = mol

No = Avogadro's = 6.02.10²³

N = 0.25 x 6.02.10²³

N = 1.505 x 10²³

The mass of (NH4) 2S in the solution is : Mass = 0.0600 mol × 60.08 g/mol = 3.60 g.

The given molarity and volume of the solution can be used to calculate the number of moles of ammonium sulfide (NH4)2S.Then, the number of moles can be converted to mass using the molar mass of (NH4)2S.Mass of ammonium sulfide (NH4)2S in 3.00 L of a 0.0200 M solution is given by : Mass = moles × molar mass.The number of moles of (NH4)2S can be found using the equation:Molarity = Number of moles / Volume.Rearranging this equation, we get:Number of moles = Molarity × Volume Number of moles of (NH4)2S = 0.0200 M × 3.00 L.Number of moles of (NH4)2S = 0.0600 mol.The molar mass of (NH4)2S can be calculated by summing the molar masses of ammonium (NH4) and sulfide (S) ions.Molar mass of (NH4)2S = (2 × Molar mass of NH4) + Molar mass of S= (2 × 14.01 g/mol) + 32.06 g/mol= 60.08 g/mol.

For more question on mass

https://brainly.com/question/1838164

#SPJ8

Answer:

A chemical bond formed when two atoms share two pairs of electrons is a double bond; it is best described as a covalent bond.

Explanation:

A chemical bond in which two pairs of electrons are shared has to be defined as covalent since ionic bonds don't involve electron sharing. They consist only of electrostatic attraction between ions.

The relationship between volume is inversely proportional. This means that when there are an increase in pressure of molecules that are moving, pushes the molecules closer together, reducing the volume. If the pressure is decreased, the gases are free to move about in a larger volume.

You can see in the graph that the behaviour is: the volume is increasing, so this causes that the pressure increase. The answer would be the third one: pressure increases

Answer:

Limitations of Paper Chromatography.

1.Large quantity of sample cannot be applied on paper chromatography.

2.In quantitative analysis paper chromatography is not effective.

3.Complex mixture cannot be separated by paper chromatography.

4.Less Accurate compared to HPLC or HPTLC.

Answer:

Limitations of paper chromatography​ is described below in complete details.

Explanation:

Limitations of Paper Chromatography

Answer:

C Tympanic membrane

Explanation:

due to the nerved gathered inside the membrane can cause a nauseous feeling.

Answer: Clouds form when the invisible water vapor in the air condenses into visible water droplets or ice crystals.

Explanation:

From Q = mcΔT, we can rearrange the equation to solve for mass, m = Q/cΔT. The specific heat capacity, c, of solid gold is 0.129 J/g °C. I'm assuming that the energy is given in joules, as it's not specified in the question as written.

m = Q/cΔT = (35.73 J)/(0.129 J/g °C)(40.85 °C - 0.85°C)

m = 6.92 g of gold was present  

The heat of the substance depends on the mass, specific heat, and temperature change of the system. The mass of gold that was present was, 6.92 g.

Mass of the substance in the thermodynamic system is the ratio of the heat to the product of the specific heat and the temperature change. It is given as,

\(\rm m = \rm \dfrac{Q}{c \Delta T}\)

Given,

Heat (Q) = 35.73 J

Specific heat (c) = 0.129 J/g °C

Temperature change = 40 degrees celsius

Mass of gold is calculated as:

35.73 ÷ (0.129) (40) = 6.92 gm

Therefore, 6.92 gms of gold is present.

Learn more about mass here:

https://brainly.com/question/15528502

#SPJ2

Answer:

"Forensic science is a critical element of the criminal justice system. Forensic scientists examine and analyze evidence from crime scenes and elsewhere to develop objective findings that can assist in the investigation and prosecution of perpetrators of crime or absolve an innocent person from suspicion."

Explanation:

Strictly from: https://www.justice.gov/olp/forensic-science#:~:text=Forensic%20science%20is%20a%20critical,an%20innocent%20person%20from%20suspicion.

Forensic science is any science used to aid in a criminal/legal investigation

The volume of 3M from making stock solution is 7.5 liters.

To make a 3M solution from a 6M stock solution of AgNO3, we need to add water to the stock solution. Let us assume x liters of 3M solution from 5 liters of the 6M stock solution.

The amount of AgNO3 in the 5 liters of the stock solution is:

5 L * 6 mol/L = 30 mol AgNO3

To make a 3M solution, we need:

3 mol/L * x L = 3x mol AgNO3

We can use the formula:

C1V1 = C2V2

where,

C1 = concentration of the stock solution

C2 = concentration of the final solution

V1 = volume of stock solution

V2 = volume of the final solution

In this case, we know that:

C1 = 6 M

C2 = 3 M

V1 = x

V2 = 5 L + x

In this case, we know that:

C1 = 6 M

C2 = 3 M

V1 = x

V2 = 5 L + x

let us substitute the above values in the formula

6 M * V1 = 3 M * (5 L + x)

Solving for x, we get:

x = 2.5 L

Therefore, to make a 3M solution from a 6M stock solution, a final volume of 5 + 2.5 = 7.5 liters to get a 3M solution.

To learn more about, Volume of the molecule

https://brainly.com/question/21976792

The structure of 3-ethyl-2 fluroheptane is attached. The chemical formula of the compound is C9H19F.

It is an organic compound which has one pair of hydrogen and acceptor and zero hydrogen bond donor.

The chemical formula of the compound is \(C_9H_1_9_F.\)

Thus, the structure of 3-ethyl-2 fluroheptane is attached. The chemical formula of the compound is C9H19F.

Learn more about 3-ethyl-2 fluroheptane

https://brainly.com/question/14754676

#SPJ1

Answer:

the 2 means there are 2 al and the 3 means there are 3 o

Explanation:

I hope this helps

Answer:

The correct option is A Hydrogen

Explanation:

The correct option is A Hydrogen

Iron on reaction with hydrochloric acid, replaces the hydrogen from the acid and forms ferrous chloride. Hydrogen gas is evolved during the reaction in the form of bubbles.

Fe (s) + 2HCl (aq) → FeCl2 (aq) +H2 (g)

Answer:

He is best known for introducing the atomic theory into chemistry, and for his research into colour blindness, sometimes referred to as Daltonism in his honour.

Explanation:

Answer:

Law of conservation of mass and the law of definite proportions could be explained using the idea of atoms.

Answer:

A biology investigation usually starts with an observation—that is, something that catches the biologist’s attention. For instance, a cancer biologist might notice that a certain kind of cancer can't be treated with chemotherapy and wonder why this is the case. A marine ecologist, seeing that the coral reefs of her field sites are bleaching—turning white—might set out to understand why.

How do biologists follow up on these observations? How can you follow up on your own observations of the natural world? In this article, we’ll walk through the scientific method, a logical problem-solving approach used by biologists and many other scientists.

Explanation:

Answer:

Therefore the answer is d) The pressure exerted by a gas in the balloon results from collisions between the gas molecules and the balloon walls.

Explanation:

Answer:

No

Explanation:

No. The demonstration does not violate the conservation of mass.

The law of conservation of mass states that mass can neither be created nor destroyed in a reaction. However, mass can be converted from one form to another during the reaction.

In this case, even though the remaining bits of paper weigh 0.5 g while the original paper weighed 2.5 g, the ashes and smoke/gas from the burning will all add up to the lost weight of the paper.

The burned part has been converted into other forms. If the smoke/gas and the ashes are properly captured, they will mark up with the weight of the remaining paper to give the weight of the original paper.

The answer to the question on whether the burning violates the law of conservation of mass, we say that;

No, the demonstration does not violate the conservation of mass because the mass formed still adds up to the original mass.

The law of conservation of mass states that mass in an isolated system, matter can neither be created nor destroyed by chemical reactions or physical transformations. However, it could involve change in mass from one form to another.  

Now, we are told that the remaining bits of paper has a mass of 0.5 g while the original paper has a mass 2.5 g. The reason why the mass left is 0.5 kg is because the ashes and smoke from the burning will add up to give the original weight of the paper.

Thus, since when the ashes, and smoke add up to still give the original mass of 3.5 g, it means that the mass is neither created nor destroyed

Read more about law of conservation of mass at; https://brainly.com/question/7041730

Answer:

E = 345.6  J

Explanation:

Given that,

Mass of a soccer ball, m = 1.2 kg

Speed of the soccer ball, v = 24 m/s

We need to find the mechanical kinetic energy of the soccer ball. The formula for the kinetic energy of an object is given by :

\(E=\dfrac{1}{2}mv^2\)

Put all the values in the above formula

\(E=\dfrac{1}{2}\times 1.2\times (24)^2\\\\E=345.6\ J\)

So, the kinetic energy of the soccer ball is 345.6  J.

Answer:

they use echos to communicate

Explanation: dolphins make echos in order to signal there partners, and others in there pod. or they can use it to signal for prey, or signal that they are in trouble.

Answer: A good example for this one is bats.

Explanation: While a couple of bats are known to emanate the beats through their nose, most bats will in general express them. The recurrence is estimated to be around 200 heartbeats each second, nonetheless, the time frame can fluctuate with the types of a bat. Credits: Adi Ciurea/Shutterstock A bat's squeaks are more rewarding than Ben's on the grounds that with higher recurrence comes higher-goal or more distinctive detail – they are equipped for evading deterrents no more extensive than a human hair! They use it to find and explore towards prey, for example, a hurrying bug.   As the distinguished sound waves that structure the pictures are basically echoes, they should adjust to a standard — the recovered train of heartbeats should be sufficiently noisy to get back to the sender, yet short sufficient with the end goal that the reverberation of the sender returns before the following one withdraw.

The total volumes of solution for the 1:5, 1:10, 1:20, and 1:40 dilutions are 10.4 mL, 11 mL, 14 mL, and 19 mL, respectively.

In the case study, four dilutions of an albumin stock solution were prepared: 1:5, 1:10, 1:20, and 1:40. The total volume of solution for each dilution can be calculated using the formula:

\( \text{Total volume} = \frac{\text{Volume of stock solution}}{\text{Dilution factor}} \)

For example, if 2 mL of the stock solution was diluted to 10 mL to create a 1:5 dilution, the calculation would be:

\( \text{Total volume} = \frac{2 \, \text{mL}}{5} = 0.4 \, \text{mL} \)

Thus, the total volume of solution for the 1:5 dilution is 10 mL + 0.4 mL = 10.4 mL.

Similarly, the total volumes for the other dilutions can be calculated:

1:10 dilution: total volume = 10 mL + 1 mL = 11 mL

1:20 dilution: total volume = 10 mL + 4 mL = 14 mL

1:40 dilution: total volume = 10 mL + 9 mL = 19 mL

The total volumes of solution for the 1:5, 1:10, 1:20, and 1:40 dilutions are 10.4 mL, 11 mL, 14 mL, and 19 mL, respectively.

To know more about Dilution here: https://brainly.com/question/32814176

#SPJ11

Magnesium and Lithium have similar chemical properties.

Chemical characteristics of elements in the same group on the periodic table are comparable. This is because the number of electrons in each atoms' highest-occupied energy level is the same.

Since both Li lithium and K potassium are members of the first group of alkali metals, they share a number of chemical characteristics. Some pairs of diagonally adjacent items in the second and third periods have a diagonal relationship. The first 20 elements in the periodic table are listed here.

Examples of pairings that share characteristics include lithium (Li) and magnesium (Mg), beryllium (Be) and aluminum (Al), and boron (B) and silicon (Si). Both silicon and boron are semiconductors.

Learn more about chemical properties:

brainly.com/question/1935242

#SPJ4

1. The volume of water in the tank at time t is given by the equation

Volume(t) = 30 + t.

2.The appropriate initial condition for the ODE is S(0) = 0, as there is no salt initially in the tank.

3. It is not separable because the variables S(t) and t are not separable on opposite sides of the equation.

4. The solution can be expressed in terms of the integral as:
\(S(t) = (70 * \int e^{(t^{2/2} + 30t)} dt) / e^{(t^{2/2} + 30t)})\)

5.  the salt concentration in the tank as t→infinity is zero.

1. To determine the volume of water in the tank at time t, we need to consider the rate at which water is being added and drained. The tank is being filled at a rate of 2 liters per minute and drained at a rate of 1 liter per minute.

Since the tank starts with an initial volume of 30 liters, the volume of water in the tank at time t can be calculated using the equation:
Volume(t) = Initial volume + (Rate of filling - Rate of draining) * t

Volume(t) = 30 + (2 - 1) * t

So, the volume of water in the tank at time t is given by the equation

Volume(t) = 30 + t.

2. Let S(t) denote the amount of salt in the fish tank at time t in grams.

To show that S(t) satisfies the ODE S'(t) = 70 - (t+30)S(t),

we need to take the derivative of S(t) with respect to t and substitute it into the given ODE.

Taking the derivative of S(t), we have:

S'(t) = 0 - (1+0)S(t) + 0

S'(t) = -S(t)

Substituting this into the given ODE, we get:

-S(t) = 70 - (t+30)S(t)

Simplifying the equation, we have:

S'(t) = 70 - (t+30)S(t)

Therefore, S(t) satisfies the ODE S'(t) = 70 - (t+30)S(t).

The appropriate initial condition for the ODE is S(0) = 0,

as there is no salt initially in the tank.

3. This ODE is a first-order linear ordinary differential equation. It is not separable because the variables S(t) and t are not separable on opposite sides of the equation.

4. To solve the initial value problem for S(t) using the method of integrating factors, we first rewrite the ODE in standard form:

S'(t) + (t+30)S(t) = 70

The integrating factor is given by:
\(\mu(t) = e^{(\int (t+30) dt)} = e^{(t^2/2 + 30t)\)

Multiplying both sides of the equation by μ(t), we have:
\(e^{(t^2/2 + 30t)} * S'(t) + e^{(t^2/2 + 30t)} * (t+30)S(t) = 70 * e^{(t^2/2 + 30t)\)

Applying the product rule to the left side of the equation, we get:
\((e^{(t^{2/2} + 30t) * S(t))' = 70 * e^{(t^{2/2} + 30t)})\)

Integrating both sides of the equation with respect to t, we have:
\(\int (e^{(t^2/2 + 30t)} * S(t))' dt = \int (70 * e^{(t^2/2 + 30t))} dt\)

Using the fundamental theorem of calculus, the left side becomes:
\(e^{(t^2/2 + 30t)} * S(t) = \int (70 * e^{(t^2/2 + 30t))} dt\)

Simplifying the right side by integrating, we get:
\(e^{(t^2/2 + 30t)} * S(t) = 70 * \int e^{(t^2/2 + 30t)} dt\)

At this point, the integration of \(e^{(t^2/2 + 30t)\) becomes difficult to express in terms of elementary functions.

Hence, the solution can be expressed in terms of the integral as:
\(S(t) = (70 * \int e^{(t^2/2 + 30t)} dt) / e^{(t^2/2 + 30t)\)

5. As t approaches infinity, the exponential term \(e^{(t^2/2 + 30t)\) becomes very large, causing the salt concentration S(t) to approach zero. Therefore, the salt concentration in the tank as t→infinity is zero.

To know more about salt concentration, visit:

https://brainly.com/question/33838518

#SPJ11

The salt concentration in the tank as t approaches infinity is 70/3.

1. To determine the volume of water in the tank at time t, we need to consider the rate at which water is being poured into the tank and the rate at which water is being drained from the bottom.

At a rate of 2 litres per minute, water is being poured into the tank. So after t minutes, the amount of water poured into the tank is 2t litres.

At a rate of 1 litre per minute, water is being drained from the tank. So after t minutes, the amount of water drained from the tank is t litres.

Since the tank was initially filled with 30 litres of fresh water, the volume of water in the tank at time t is given by:
Volume(t) = 30 + 2t - t
Volume(t) = 30 + t

2. Let S(t) denote the amount of salt in the fish tank at time t. To determine the ODE for S(t), we need to consider the salt being poured into the tank and the salt being drained from the tank.

The salt concentration in the water being poured into the tank is 35 grams per litre. So the amount of salt being poured into the tank per minute is 35 * 2 = 70 grams.

The amount of salt being drained from the tank per minute is S(t)/Volume(t) * 1.

Therefore, the ODE for S(t) is:
S'(t) = 70 - S(t)/Volume(t)

The initial condition for this ODE is S(0) = 0, since there was no salt in the tank initially.

3. The ODE S'(t) = 70 - S(t)/Volume(t) is a first-order linear ODE. It is not separable since the variables S(t) and Volume(t) are mixed together.

4. To solve the initial value problem for S(t), we can rewrite the ODE as:
Volume(t) * S'(t) + S(t) = 70 * Volume(t)

This is a linear ODE of the form y'(t) + p(t)y(t) = g(t), where p(t) = 1/Volume(t) and g(t) = 70 * Volume(t).

To solve this type of ODE, we can multiply both sides by an integrating factor, which is the exponential of the integral of p(t).

The integrating factor is exp(integral of 1/Volume(t) dt) = exp(ln(Volume(t))) = Volume(t).

Multiplying both sides of the ODE by the integrating factor, we get:
Volume(t) * S'(t) + S(t) = 70 * Volume(t)
Volume(t) * S'(t) + Volume(t) * S(t) = 70 * Volume(t)^2
( Volume(t) * S(t) )' = 70 * Volume(t)^2

Integrating both sides with respect to t, we get:
Volume(t) * S(t) = 70/3 * Volume(t)^3 + C

S(t) = 70/3 * Volume(t)^2 + C/Volume(t)

Using the initial condition S(0) = 0, we can solve for C:
0 = 70/3 * 30^2 + C/30
C = -70000

Therefore, the solution for S(t) is:
S(t) = 70/3 * Volume(t)^2 - 70000/Volume(t)

5. As t approaches infinity, the volume of water in the tank becomes very large. In this case, we can approximate the volume of the tank as t, since the rate at which water is being poured in is 2 litres per minute. So the salt concentration in the tank as t approaches infinity is given by:
S(t)/Volume(t) = (70/3 * t^2 - 70000/t) / t
As t approaches infinity, the second term (-70000/t) approaches 0, so the salt concentration in the tank as t approaches infinity is:
S(t)/Volume(t) = 70/3 * t^2 / t = 70/3 * t

learn more about salt concentration

https://brainly.com/question/33838518

#SPJ11

∂c/∂t represents the rate of change of concentration with respect to time, D is the diffusion coefficient, and ∇·(D∇c) represents the divergence of the diffusion flux.

(a) The total amount of pollutant in the region R can be found by integrating the concentration c(x,y,z,t) over the volume V of the region R. Mathematically, it can be expressed as:

Total amount of pollutant = ∫∫∫V c(x,y,z,t) dV

(b) It is reasonable to assume that the flow J of the pollutant is proportional to the gradient of the concentration, as this relationship is based on Fick's law of diffusion. According to Fick's law, the flow of a substance is proportional to the concentration gradient. Mathematically, it can be expressed as:

J = -D ∇c

Where J is the flow of the pollutant, D is the diffusion coefficient, and ∇c is the gradient of the concentration.

(c) To derive the partial differential equation governing the diffusion of the pollutant, we can apply the continuity equation, which states that the rate of change of the concentration in a given volume is equal to the divergence of the flow. Mathematically, it can be expressed as:

∂c/∂t = -∇·J

Using the relationship from part (b), we can substitute it into the continuity equation:

∂c/∂t = ∇·(D∇c)

This is the partial differential equation governing the diffusion of the pollutant, where ∂c/∂t represents the rate of change of concentration with respect to time, D is the diffusion coefficient, and ∇·(D∇c) represents the divergence of the diffusion flux.

Learn more about Fick's law from given link: https://brainly.com/question/33379962

#SPJ11