PLEASE HELP ASP...Patsy wants to find out that it would take her to walk a 10 K (10 km) if she maintains a
pace of 6.5 km/hr. How long will it take her?

Answer:

λ = 1.1×10⁸ m

Explanation:

Given data:

Frequency of wave = 2.7 Hz

Wavelength of wave = ?

Solution:

Formula:

Speed of wave = frequency × wavelength

Speed of wave = 3×10⁸ m/s

now we will put the values in formula.

3×10⁸ m/s = 2.7 s⁻¹ × λ

λ = 3×10⁸ m/s /2.7 s⁻¹

λ = 1.1×10⁸ m

Q = Ksp = 1 × 10^(-12).

Substituting the values into the Nernst equation, we have:

0.799 V = E° - (RT/2F) * ln(1 × 10^(-12))

Now, solving for E°:

E° = 0.799 V + (RT/2F) * ln(1 × 10^(-12))

The value of R is the ideal gas constant, T is the temperature in Kelvin, and F is the Faraday constant.

To find the standard reduction potential at 25°C for the half-reaction Ag2CrO4(s) + 2e¯ → 2Ag(s) + CrO4²-(aq), we can use the Nernst equation, which relates the standard reduction potential (E°) to the equilibrium constant (K) and the reaction quotient (Q).

The Nernst equation is given as follows:

E = E° - (RT/nF) * ln(Q)

Given information:

Ksp = 1 × 10^(-12)

E = +0.799 V (standard reduction potential of Ag+ to Ag)

Since the reaction involves the dissolution of Ag2CrO4(s), the reaction quotient Q can be expressed as [Ag+]²/[CrO4²-].

Since the stoichiometry of the reaction is 2:1 for Ag2CrO4 to Ag+, we can say that [Ag+]² = Ksp.

Therefore, Q = Ksp = 1 × 10^(-12).

Substituting the values into the Nernst equation, we have:

0.799 V = E° - (RT/2F) * ln(1 × 10^(-12))

Now, solving for E°:

E° = 0.799 V + (RT/2F) * ln(1 × 10^(-12))

The value of R is the ideal gas constant, T is the temperature in Kelvin, and F is the Faraday constant.

Please note that without specific values for temperature (T) and the ideal gas constant (R), the exact standard reduction potential at 25°C cannot be determined.

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Answer:

✔ habitat destruction

Explanation:

the 3 reasons are:

hunting

habitat loss

introduced predators

summed up its habitat destruction

Hector's velocity changed because his direction changed.

Velocity refers to the distance covered per unit time in a specified direction while speed is the distance covered per unit time without taking the direction into account.

We must note that Hector's direction changed. since magnitude and direction are both considered when discussing velocity, then Hector's velocity changed because his direction changed.

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Answer:

when carbon dioxide is burned they cause pollution

At 25°C, combine 47.0 mL of ethylene glycol (d = 1.11 g/mL; molar mass = 62.07 g/mol) with 50.0 mL of water (d = 1.00 g/mL) to create an antifreeze solution. The antifreeze solution's molarity is 0.0086 M if its density is 1.07 g/mL.

Antifreeze is an additive that reduces a water-based liquid's freezing point. To achieve freezing-point depression for cold conditions, an antifreeze combination is utilized. Common antifreezes also raise the liquid's boiling point, enabling a rise in coolant temperature. However, every typical antifreeze addition also has a lower heat capacity than water, which makes it less effective as a coolant when combined with water.

The combination of a solution's freezing and boiling temperatures depends on the amount of dissolved components present. Therefore, salts cause aqueous solutions' melting points to decrease. Although salts are widely used for de-icing, salt solutions should not be utilized in cooling systems as they cause metal corrosion. Because they typically have melting values that are lower than those of water, low molecular weight organic substances can be used as antifreeze. Alcoholic organic compound solutions, in particular, are useful. Since antifreezes were first made commercially available in the 1920s, they have all been composed primarily of alcohols such methanol, ethanol, ethylene glycol, and others.

Molarity = \(\frac{mole}{volume}\)

To calculate the moles of ethylene glycol.

The mass of ethylene glycol is 47.0 mL × 1.11 g/mL = 52.17 g.

Since molar mass of ethylene glycol is 62.07 g/mol, the moles of ethylene glycol is:

\(\frac{52.17}{62.07}\) = 0.84 mol

To calculate the total volume of the antifreeze solution. We know that the volumes of ethylene glycol and H₂O are 47.0 mL and 50.0 mL, respectively.

Therefore, the total volume of the antifreeze solution is 47.0 mL + 50.0 mL = 97.0 mL.

Finally, the molarity of the antifreeze solution can be calculated using the formula

Molarity = \(\frac{mole}{volume}\)

Therefore, the molarity of the antifreeze solution is:

\(\frac{0.84}{97.0}\) = 0.0086 M

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Answer:

B. honey dissolving in tea

Explanation:

Answer:

All is Correct

Explanation:

Fossil fuels have the following properties:

Therefore, the answer is to select all

Answer:

It's A, B, and D

Explanation:

Maybe not D, because that is burning wood like charcoal. Not sure about that. Hope this helps!

Answer:

Explanation:

Robert Boyle

Answer:

B 1650

Explanation:

The pOH of the 0.001 M NaOH solution is approximately 3.

To determine the pOH of a solution, we need to know the concentration of hydroxide ions (OH-) in the solution.

In the case of a 0.001 M NaOH solution, we can assume that all of the NaOH dissociates completely in water to form Na+ and OH- ions. Therefore, the concentration of hydroxide ions in the solution is also 0.001 M.

The pOH is calculated using the equation:

pOH = -log[OH-]

Substituting the concentration of hydroxide ions, we have:

pOH = -log(0.001)

Using a calculator, we can evaluate the logarithm:

pOH ≈ 3

Therefore, the pOH of the 0.001 M NaOH solution is approximately 3.

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To graph the function g(x) = f(-x), you can start with the graph of f(x) and then reflect it about the y-axis.

To find the graph of the function g(x) = f(-x), we can start with the graph of the function f(x) and then reflect it about the y-axis.

If the graph of f(x) is symmetric with respect to the y-axis, meaning it is unchanged when reflected, then g(x) = f(-x) will have the same graph as f(x).

However, if the graph of f(x) is not symmetric with respect to the y-axis, then g(x) = f(-x) will be a reflection of f(x) about the y-axis.

In either case, the resulting graph of g(x) = f(-x) will be symmetric with respect to the y-axis.

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The concentration on which we should pipette out are 4μl.

To recap, the working solutions are 20μg/μl, 2μg/μl, and 0.2μg/μl.

1.

It’s basically the same as the other question. Set it up algebraically:

50μg/μl x = 200 μg

Note the variable x in the equation. You can see easily that x = 4μl (remember to include the units, and recall that units cancel in equations).

So if you pipet 4μl of the 50μg/μl working solution into a centrifuge tube, you’ll have 20μg of BSA in there.

2.

Again, set it up algebraically:

5μg/μl x = 20 μg

Here, x = 4μl again. So again you are pipetting out 4μl, except this time it’s from the 5μg/μl working solution.

3.

I’m sure you can see that there’s a trend going on here (it’s not any special science trend or anything like that, though).

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Answer:

• Molecular mass of Iron (III) tetraoxide

\( \dashrightarrow \: { \tt{(56 \times 3) + (16 \times 4)}} \\ = { \tt{168 + 64}} \\ = { \tt{232\:g}}\)

[ molar masses: Fe → 56, O → 16 ]

\( \dashrightarrow \:{ \rm{232 \: g \: = 1 \: mole}} \\ \\ \dashrightarrow \: { \rm{55.85 \: g = ( \frac{55.85}{232}) \: moles }} \\ \\ \dashrightarrow \: { \boxed{ \tt{ = 0.24 \: moles}}}\)

The heat that is absorbed by the system is  1363 J. Option B

We know that in a chemical reaction that there could be the absorption or the evolution of heat. We say that there is the evolution of heat when heat has been lost from the system and there is the absorption of heat when heat has been gained by the system.

Number of moles of the barium hydroxide hydrate = 3.15 g/203 g/mol

= 0.015 moles

Number of moles of the ammonium thiocyanate = 1.52/76 g/mol

= 0.02 moles

If 1 mole of barium hydroxide hydrate reacts with 2 moles of  ammonium thiocyanate

0.015 moles of barium hydroxide hydrate reacts with 0.015 * 2 moles/1 mole

= 0.03 moles

Hence the limiting reactant is the ammonium thiocyanate.

Now the heat that is absorbed is;

H = mcdT

m = mass of the water

c = Heat capacity

dT = Temperature change

H = 100 * 4.20 * 3.1

H = 1363 J

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I would say Phenolphthalein

The decomposition of aspirin (acetylsalicylic acid,\(C_{9} H_{8} O_{4}\)) can occur through the hydrolysis reaction, resulting in the formation of acetic acid (\(CH_{3} COOH\)) and salicylic acid (\(C_{7} H_{6}O_{3}\)).

The decomposition of aspirin (acetylsalicylic acid, \(C_{9} H_{8} O_{4}\)) can occur through the hydrolysis reaction, resulting in the formation of acetic acid (\(CH_{3} COOH\)) and salicylic acid (\(C_{7} H_{6}O_{3}\)). To determine the moles of each product formed, we need to consider the balanced chemical equation for the reaction:

\(C_{9} H_{8} O_{4} = > C_{7} H_{6}O_{3} +CH_{3} COOH\)

From the equation, we can see that for every 1 mole of aspirin, 1 mole of salicylic acid and 1 mole of acetic acid are produced.

Therefore, the moles of salicylic acid and acetic acid formed will be equal to the number of moles of aspirin that decomposes. If we know the amount of aspirin in moles, we can directly calculate the moles of each product based on stoichiometry.

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Answer:

0.2M H2C6H5O7 < 0.2M H2C2O4

Explanation:

A weak acid/base ionizes to a very small extent in water. Hence, if we say that a substance is a weak acid/base, its percentage of ionization in solution is very little.

More volume of a very weak acid is required to neutralize a strong base. Since NaOH is a strong base, the weaker acid among the duo will require more volume for neutralization.

Since H2C6H5O7 is a weaker acid than H2C2O4, equal concentration of the both acids will require less volume of H2C2O4 than H2C6H5O7 to neutralize 0.50 M NaOH.

H₂C₆H₅O₇ is a weaker acid than H₂C₂O₄, and will require the least volume of 0.50 M NaOH to be neutralized.

H₂C₆H₅O₇ < H₂C₂O₄

The strength of an acid is related to the value of its dissociation constant, Ka or its pKa (negative logarithm of Ka)

Strong acids have high Ka values or low pKa value, whereas weak acids have low Ka values and high pKa values.

Between two acids, the acid with a higher Ka or lower pKa values is the stronger acid.

Acids are classified as either strong or weak depending on how well it ionizes in solution to produce hydrogen ions.

Strong acids ionizes completely to produce hydrogen ions.

Weak acid ionizes partially to a varying degrees in water to produce hydrogen ions.

In neutralization reactions between acids and bases, stronger acids will require the most volume of base or alkali in order to be neutralized.

H₂C₂O₄ has a Ka value of 5.9 x 10⁻² and a pKa value of 1.23

H₂C₆H₅O₇ has a Ka value of 8.4 x 10⁻⁴ and a pKa value of 3.08

Hence H₂C₂O₄ is a stronger acid than H₂C₆H₅O₇

For equal molar concentrations of the two acids, H₂C₂O₄ will produce more hydrogen ions than H₂C₆H₅O₇, and thus, will require more volume base (0.50 M NaOH) to be neutralized.

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The gradient of a graph shows the slope or rate of change. The graph here shows the relationship between the volume (V) and the absolute temperature (T) of a gas when the gas is kept at a constant pressure.

PV = nRT,

where

P is the pressure,

V is the volume,

n is the number of moles of the gas,

R is the ideal gas constant, and

T is the absolute temperature, the formula for the ideal gas law.

The ideal gas law can be rewritten as V = (nR/P)*T if pressure (P) is constant, volume (V) is plotted on the y-axis, and absolute temperature (T) is is plotted on the x-axis. Axis.

We can see that the gradient (m) of the graph in this equation represents (nr/p) by comparing it to the equation for a straight line, y = mx + c, where y is the dependent variable (V), x is the independent variable. variable (T), m is the gradient, and c is the y-intercept.

So, the correct option is B.

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Answer:c

Explanation:c

Answer:

Explanation:

From the given information:

Let y(t) be the amount of sugar in tank A at any time.

Then, the rate of change of sugar in the tank is given by:

\(\dfrac{dy}{dt}= (rate \ in ) - ( rate \ out )\)

The rate of the sugar coming into the tank is 0

\(\text{rate of sugar going out }= \dfrac{y(t) \ pound}{100}\times \dfrac{1 }{min} \\ \\ = \dfrac{y}{100} \ pound/min\)

\(So; \\ \\ \\ \dfrac{dy}{dt} = 0 - \dfrac{y}{100} \\ \\ \implies \dfrac{dy}{y }= -\dfrac{dt}{100 } \\ \\ \implies dx|y| = -\dfrac{t}{100}+C \\ \\ \implies e*{In|y|}= e^{-\dfrac{t}{100}+C} \\ \\ \implies y = Ce^{-\dfrac{t}{100}}\)

Initial amount of sugar = 25 Pounds

Now; y(0) = 25

25 = Ce⁰

C = 25

So;  y(t) = 25 \(e^{-\dfrac{t}{100}\)

Thus, the amount of sugar at any time t = \(\mathbf{25 e^{^{-\dfrac{t}{100}}}}\)

B) For tank B :

\(\dfrac{dy}{dt}= 1 - \dfrac{y}{50 } \\ \\ \dfrac{dy}{dt}= \dfrac{50-y}{50} \\ \\ \dfrac{dy}{50-y }= \dfrac{dt}{50}\)

Answer:

Your answer is 5 x 10⁵

I hope it's helps you

Answer:

\(thank \: you\)

Answer:

0.101 M

0.1 mol/kg

0.644

Explanation:

To calculate the molarity, molality, and mole fraction of Na2SO4 in a solution, you need to know the moles of Na2SO4 in the solution and the volume or mass of the solution.

A) Molarity: Molarity (M) is defined as the number of moles of solute per liter of solution. To calculate molarity, we need to divide the number of moles of Na2SO4 by the volume of the solution in liters.

First, calculate the number of moles of Na2SO4:

mass of Na2SO4 = 755 g

molecular weight of Na2SO4 = 142 g/mol

number of moles = mass/molecular weight = 755 g / 142 g/mol = 5.34 moles

Next, calculate the volume of the solution:

mass of water = 53,100 g

density of water = 1 g/mL

volume of water = mass/density = 53,100 g / 1 g/mL = 53,100 mL = 53.1 L

Finally, calculate the molarity:

molarity = number of moles / volume of solution = 5.34 moles / 53.1 L = 0.101 M

B) Molality: Molality (m) is defined as the number of moles of solute per kilogram of solvent. To calculate molality, we need to divide the number of moles of Na2SO4 by the mass of the solvent in kilograms.

mass of water = 53,100 g

mass of solvent in kilograms = mass in grams / 1000 g/kg = 53,100 g / 1000 g/kg = 53.1 kg

Finally, calculate the molality:

molality = number of moles / mass of solvent = 5.34 moles / 53.1 kg = 0.1 mol/kg

C) Mole fraction: The mole fraction (X) is defined as the ratio of the number of moles of solute to the total number of moles of solute and solvent in a solution. To calculate the mole fraction of Na2SO4, we need to divide the number of moles of Na2SO4 by the total number of moles of Na2SO4 and water.

number of moles of Na2SO4 = 5.34 moles

number of moles of water = 53.1 kg * 1000 g/kg / 18.015 g/mol = 2.96 moles

total number of moles = number of moles of Na2SO4 + number of moles of water = 5.34 moles + 2.96 moles = 8.30 moles

Finally, calculate the mole fraction:

mole fraction of Na2SO4 = number of moles of Na2SO4 / total number of moles = 5.34 moles / 8.30 moles = 0.644

The molarity of Na2SO4 in the solution is 0.101 M, the molality is 0.1 mol/kg, and the mole fraction of Na2SO4 is 0.644.

The SI unit to the quantity it measures are:

Mass: The mass of an object is a measure of its amount of matter. The SI unit of mass is the kilogram (kg) or gram (g).

Temperature: Temperature is a measure of the average kinetic energy of the particles in a substance. The SI unit of temperature is the kelvin (K).

Time: Time is a measure of the interval between two events. The SI unit of time is the second (s).

Electric current: Electric current is a measure of the flow of electric charge. The SI unit of electric current is the ampere (A).

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Complete question:

Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.

Match each SI unit to the quantity it measures.

The energy of combustion for one mole of butane to be approximately 2888.81 kJ/mol.

To calculate the energy of combustion for one mole of butane (C4H10), we need to use the information provided and apply the principle of calorimetry.

First, we need to convert the mass of the butane sample from grams to moles. The molar mass of butane (C4H10) can be calculated as follows:

C: 12.01 g/mol

H: 1.01 g/mol

Molar mass of C4H10 = (12.01 * 4) + (1.01 * 10) = 58.12 g/mol

Next, we calculate the moles of butane in the sample:

moles of butane = mass of butane sample / molar mass of butane

moles of butane = 0.367 g / 58.12 g/mol ≈ 0.00631 mol

Now, we can calculate the heat released by the combustion of the butane sample using the equation:

q = C * ΔT

where q is the heat released, C is the heat capacity of the calorimeter, and ΔT is the change in temperature.

Given that the heat capacity of the bomb calorimeter is 2.36 kJ/°C and the change in temperature is 7.73 °C, we can substitute these values into the equation:

q = (2.36 kJ/°C) * 7.73 °C = 18.2078 kJ

Since the heat released by the combustion of the butane sample is equal to the heat absorbed by the calorimeter, we can equate this value to the energy of combustion for one mole of butane.

Energy of combustion for one mole of butane = q / moles of butane

Energy of combustion for one mole of butane = 18.2078 kJ / 0.00631 mol ≈ 2888.81 kJ/mol

Therefore, the energy of combustion for one mole of butane is approximately 2888.81 kJ/mol.

In conclusion, by applying the principles of calorimetry and using the given data, we have calculated the energy of combustion for one mole of butane to be approximately 2888.81 kJ/mol.

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Answer:

the answer is D

Answer:

From the given equation, we can see that for every 2 moles of Al, we get 3 moles of H2

So, we can say the the number of moles of H2 is 3/2 times the number of moles of Al

We are given the number of moles of Al and we have to find the number of moles of H2

We have deduced the relationship:

Moles of Al * 3 / 2 = Moles of H2

Replacing the variables with given values

3.5 * 3 / 2 = Moles of H2

Moles of H2 = 5.25 moles

By using the ideal gas law and Avogadro's number, we find that there are approximately 6.72 × 10^22 molecules of CO2 in 2.5 L at STP.

To determine the number of molecules of CO2 in 2.5 L at STP (Standard Temperature and Pressure), we can use the ideal gas law and Avogadro's number.

Avogadro's number (N_A) is a fundamental constant representing the number of particles (atoms, molecules, ions) in one mole of substance. Its value is approximately 6.022 × 10^23 particles/mol.

STP conditions are defined as a temperature of 273.15 K (0 °C) and a pressure of 1 atmosphere (1 atm).

First, we need to convert the volume from liters to moles of CO2. To do this, we use the ideal gas law equation:

PV = nRT,

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Since we have STP conditions, we can substitute the values:

(1 atm) × (2.5 L) = n × (0.0821 L·atm/(mol·K)) × (273.15 K).

Simplifying the equation:

2.5 = n × 22.4149.

Solving for n (the number of moles):

n = 2.5 / 22.4149 ≈ 0.1116 moles.

Next, we can calculate the number of molecules using Avogadro's number:

Number of molecules = n × N_A.

Number of molecules = 0.1116 moles × (6.022 × 10^23 particles/mol).

Number of molecules ≈ 6.72 × 10^22 molecules.

Therefore, there are approximately 6.72 × 10^22 molecules of CO2 in 2.5 L at STP.

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