in one experiment, 25 ml of 0.050 m sn2 is titrated with 0.10 m fe3 ; the titration is monitored using a pt working electrode (red, positive lead) and a ag/agcl reference electrode (black, negative lead). what is the potential (in v) you would expect to measure after 30.0 ml of the titrant has been added?

Answer:

try c atom i hope this helps!! : )

Explanation:

Answer:

carbon monoxide.

Explanation:

The molarity of the aqueous glucose solution is 0.716 M for a solution containing 129 g of glucose (C6H1206) in 200 mL. Option 1 is the correct answer.

To calculate the molarity (M) of the aqueous glucose solution, we need to first determine the number of moles of glucose present in the solution.

Step 1: Calculate the molar mass of glucose (C6H12O6).

The molar mass of glucose is calculated by adding the atomic masses of its constituent elements, which gives:

6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol) = 180.18 g/mol.

Step 2: Calculate the number of moles of glucose.

The number of moles of glucose can be calculated using the formula:

moles = mass/molar mass

Substituting the values, we get:

moles = 129 g / 180.18 g/mol = 0.716 mol

Step 3: Calculate the molarity of the solution.

Molarity is defined as the number of moles of solute per liter of solution. As the volume of solution given is in milliliters, we need to convert it to liters by dividing by 1000.

Molarity = moles of solute/volume of solution (in liters)

Substituting the values, we get:

Molarity = 0.716 mol / (200 mL / 1000 mL/L) = 0.716 M

Therefore, the molarity of the aqueous glucose solution is 0.716 M.

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Answer: Your answer is Chemical weathering.

Hope this helps!

Answer:

chemical

Explanation:

I hope it help.........

Peroxide is sprayed on all samples before any indicator so as to bleach the samples  before staining by the indicators

peroxide is any of a class of chemicals in which 2 oxygen atoms are linked together by a single covalent bond peroxide is useful in bleaching as they are agents bleaching.

The peroxide when applied to the sample it reduces color bodies in pulp by oxidizing the carbonyl group making the samples colorless.

When the indicator is applied to a sample that is bleached the indicator penetrates this highlights important features of the tissue as well as enhances the tissue contrast.

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The given statement, "The half-life for a first-order rate law depends on the starting concentration" is False because the starting concentration is not a factor that affects the half-life of a first-order reaction.

In a first-order reaction, the half-life is constant and does not depend on the starting concentration. The first-order rate law is represented by the equation:

rate = k[A]

where the rate is the reaction rate, k is the rate constant, and [A] is the concentration of the reactant.

The half-life of a first-order reaction (t1/2) can be calculated using the formula:

t1/2 = ln(2) / k

As you can see, the half-life depends only on the rate constant (k) and not on the starting concentration of the reactant.

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Answer:

the speed of sound in air at 0° C that is composed of 60% nitrogen, 20% oxygen, and 20% carbon is 331.5 m/s.

Explanation:

The speed of sound in a gas is determined by the temperature and the molecular weight of the gas. To calculate the speed of sound in air, we need to first determine the average molecular weight of the air. The average molecular weight of a gas mixture is given by:

M = (f1 * M1 + f2 * M2 + ... + fn * Mn) / (f1 + f2 + ... + fn)

Where M is the average molecular weight, fi is the mole fraction of each component in the gas mixture (the fraction of total moles of that component), and Mi is the molecular weight of each component.

In this case, we have:

M = (0.6 * 28 g/mol + 0.2 * 32 g/mol + 0.2 * 12 g/mol) / (0.6 + 0.2 + 0.2)

= 29.6 g/mol

The speed of sound in air at 0° C is given by:

c = 331.5 * sqrt(1 + (T / 273.15))

Where c is the speed of sound in m/s, T is the temperature in °C, and 273.15 is the temperature at which the speed of sound is 331.5 m/s.

Substituting the values and solving for c, we get:

c = 331.5 * sqrt(1 + (0 / 273.15))

= 331.5 m/s

Therefore, the speed of sound in air at 0° C that is composed of 60% nitrogen, 20% oxygen, and 20% carbon is 331.5 m/s.

Answer:

18 ml.

Explanation:

The bird will accelerate with an acceleration of 1 m/s². Therefore, option (4) is correct.

Acceleration of a body can be explained as the change of velocity of a body with respect to time. The acceleration of a body is a vector parameter with both magnitude and direction. Acceleration can be defined as the 2nd derivative of position with respect to time.

According to Newton's 2nd law of motion, force is the product of the mass and acceleration of an object.

F = ma

And,   a = F/m

Therefore, the acceleration and mass of a body have an inverse relationship.

Given, the force acting on the bird, F =  2 N

Given the mass of the bird, m = 2Kg

The acceleration  in the speed of the bird can be calculated as:

a = F/m

a = 2N/2 Kg

a = 1 m/s²

Therefore, the bird accelerates at 1 m/s² because of the wind.

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The solubility of oxygen at 10m depth below sea level, 25 degrees Celsius, and 30 g/L salinity is approximately 6.59 mg/L.

To calculate the solubility of oxygen at a specific depth below sea level, temperature, and salinity, we can refer to the oxygen solubility tables. The solubility of oxygen can vary depending on these factors.

1. Begin by identifying the given parameters:
  - Depth: 10m below sea level
  - Temperature: 25 degrees Celsius
  - Salinity: 30 g/L

2. Use the given parameters to locate the corresponding values in the oxygen solubility table.

3. The solubility of oxygen at a depth of 10m below sea level, 25 degrees Celsius, and 30 g/L salinity is typically around 6.59 mg/L.

Therefore, the solubility of oxygen at 10m depth below sea level, 25 degrees Celsius, and 30 g/L salinity is approximately 6.59 mg/L.

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The oxygen solubility at 10m depth below sea level, 25°C, and 30 g/L salinity is approximately 1538 mol/L.

To calculate the oxygen solubility at a specific depth below sea level, temperature, and salinity, we can use the solubility formula.

The solubility of a gas decreases with increasing temperature and salinity, and increases with increasing pressure.

Here's how you can calculate the oxygen solubility at 10m depth below sea level, 25°C, and 30 g/L salinity:

1. Determine the pressure at 10m depth below sea level: -

The pressure at sea level is approximately 1 atmosphere (atm).

The pressure increases by approximately 1 atm for every 10 meters of depth.

Therefore, at 10m depth, the pressure is approximately 2 atm.

2. Convert the temperature to Kelvin: -

To convert from Celsius to Kelvin, add 273 to the temperature.

25°C + 273 = 298 K.

3. Use the solubility formula:

The solubility of oxygen in water can be calculated using Henry's law:

S = k * P * C.

S is the solubility of oxygen in moles per liter (mol/L).

k is the Henry's law constant for oxygen in water at a specific temperature and salinity.  

P is the partial pressure of oxygen in atmospheres (atm).

C is the concentration of oxygen in moles per liter (mol/L).

4. Look up the Henry's law constant for oxygen at 25°C and 30 g/L salinity:

The Henry's law constant for oxygen at 25°C and 30 g/L salinity is approximately 769 L*atm/mol.

5. Calculate the solubility:  

S = (769 L*atm/mol) * (2 atm) * (1 mol/L). - S ≈ 1538 mol/L.

Therefore, the oxygen solubility at 10m depth below sea level, 25°C, and 30 g/L salinity is approximately 1538 mol/L.

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If 100 mL of a gas at 27°C is cooled to -3°C at constant pressure, thus the new pressure of the gas comes out to be  89.94 cm³. The combined gas law, which connects the starting and end states of a gas under constant pressure, can be used to resolve this issue.

The combined gas law can be expressed as follows: P₁ * V₁/ T₁ equals P₂ * V₂ / T₂. Where: The initial and final pressures (assumed to be constant) are P₁ and P₂, respectively. The first volume is V₁.The initial temperature, T₁, is given in Kelvin.

The second volume is the one we're looking for, or V₂. The final temperature, T₂, is given in Kelvin.Let's use the information provided to solve for V₂: Volume at the start: V₁ = 100 mL = 100 cm³. Temperature at initialization: T₁= 27°C = 27 + 273.15 K = 300.15 K

T₂ = -3°C = -3 + 273.15 K = 270.15 K Final temperature. Inputting the values into the equation for the combined gas law: P₁ * V₁ / T₁ equals P₂ * V₂ / T₂. We can eliminate the pressure (P) because it is constant:(V₁ / T₁) = (V₂ / T₂)

To find V₂ by rearranging the equation: V₂ = (V₁ * T₂) / T₁, replacing the specified values: V₂ = (100 cm³ * 270.15 K) / 300.15 K. Calculating: V₂ ≈ 89.94 cm³. As a result, the gas's new volume will be roughly 89.94 cm3 when it is cooled from 100 mL at 27°C to -3°C at constant pressure.

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Answer:

emprical weight=ch2o

=12*12*1*1*16 =30

Molecular mass=n*emprical weight

60=n*30

N=60/30=2

Molecular formula=n*emprical formula=2*30

=60

C2h4o2

Hope this helps :)

When placed at the same temperature, the entropy of 10 mole of Ar(g) at 10.0 atm and the 10 mole of Ar(g) at 0.5 atm. The system has a higher entropy is 10 mole of Ar(g) at 0.5 atm.

The entropy is the measure the randomness of the of the system.  the measure of the system's thermal energy per unit the temperature that is not available for doing the useful work.

The Entropy will increases with temperature at the constant pressure. The  pressure increases leads to the higher degree of the order in the molecular arrangement. so, the entropy decreases with the increasing pressure.

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Answer: 3 or 4

Explanation:it would be air because it’s structurally atom is 3

The number of moles in a sample of neon containing 8.6 × 10²⁴ atoms is 14.3 moles.

The number of moles in a chemical compound can be calculated by dividing the number of atoms in the substance by Avogadro's number as follows:

no of moles = no of atoms ÷ 6.02 × 10²³

According to this question, there are 8.6 × 10²⁴ atoms in a sample of neon gas. The number of moles in the substance can be calculated as follows:

no of moles = 8.6 × 10²⁴ atoms ÷ 6.02 × 10²³

no of moles = 1.43 × 10¹

no of moles = 14.3 moles

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Answer:

14 moles

Explanation:

Answer:

Gene Therapy is a medical field which focuses on the utilization of the therapeutic delivery of nucleic acids into a patients cells as a drug to treat disease. The first attempt was during the 1980s by Martin Cline. This can cause social issues and debates because not many people feel comfortable with this kind of treatment, also since it was just recently discovered and successful.

Answer:

I'm pretty sure it's A)Helium

Explanation:

Answer:

AIR

Explanation:

Do you have the volume of this mass? To be able to find the density of a mass, you need 2 factors. Mass(m) and Volume(V). I also need to know if the volume is measured in cubic meters (m³) or gallons or cubic foot. If you can provide these i can gladly comment the answer down.

Answer: 2.0 atm

Explanation:

PV = nRT

P- the pressure of the gas

V- the volume it occupies

n- the number of moles of gas

R- the universal gas content

T- the absolute temp. of the gas

The pressure, in atmospheres, is exerted by 0.325 mol of hydrogen gas in a 4.08 L container at 35°C is 7.03atm.

The ideal gas law relates the pressure, volume, temperature, and number of moles of a gas through the equation:

PV = nRT

where P is the pressure in atmospheres, V is the volume in liters (L), n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin (K).

T = 35°C + 273.15 = 308.15 K

Substituting the given values into the ideal gas law equation, we get:

P × 4.08 L = 0.325 mol × 0.08206 × 308.15 K

Simplifying and solving for P, we get:

P = (0.325 mol × 0.08206× 308.15 K) / 4.08 L

P = 7.03 atm

Thus, the pressure exerted by 0.325 mol of hydrogen gas in a 4.08 L container at 35°C is 7.03 atm.

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The colors of flames in experiments can vary based on several factors, including the temperature and composition of the burning material.

In general, the color of a flame is determined by the emission spectrum of the excited molecules and atoms in the flame. When a material is burned, the heat and energy generated excites the molecules and atoms, causing them to emit light. The specific colors that are emitted depend on the temperature of the flame, as well as the chemical composition of the burning material.

For example, in a very hot flame, such as the flame produced by a welder's torch, the temperature can be high enough to excite and ionize the atoms of the burning material, causing them to emit light across the entire visible spectrum and beyond. This results in a flame that appears white.

In cooler flames, such as those produced by a candle, the temperature is not high enough to ionize the atoms, but it is still high enough to excite them and cause them to emit light. The specific colors that are emitted in this case depend on the chemical composition of the burning material. For example, in a candle flame, the wax is primarily composed of hydrocarbons, which emit a yellow-orange light when burned. The blue color that is often seen in the center of the flame is due to the reaction between the hydrogen and carbon in the wax, which produces excited molecules that emit blue light.

In conclusion, the colors of flames in experiments can vary based on the temperature and composition of the burning material. The specific colors that are observed depend on the conditions within the flame and the chemical reactions that are taking place.

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Answer:

it wont let me leave I didnt mean to click sorry

The molecular mass of 3.6 moles of H₂SO₄.

So let us first calculate the molecular mass of 1 mole of H₂SO₄ and then when we calculate the molecular mass of 1 mole of H₂SO₄, we will calculate the molecular mass of 3.6 moles of H₂SO₄ by multiplying the molecular mass of 1 mole of H₂SO₄ by 3.6

Molecular mass of 1 mole of H₂SO₄ = 2*(molecular mass of Hydrogen) + (molecular mass of Sulphur ) + 4*(molecular mass of oxygen )

Molecular mass of 1 mole of H₂SO₄ = 2*1 + 32 + 4*16 = 98 grams

Mass of 3.6 moles H₂SO₄ = 3.6*98 =352.8 grams

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Answer:

d = 0.93 g/cm³

Explanation:

Given data:

Mass of object = 28 g

Volume of object = 3cm×2cm×5cm

density of object = ?

Solution:

Volume of object = 3cm × 2cm ×5cm

Volume of object = 30 cm³

Density of object:

d = m/v

by putting values,

d = 28 g/ 30 cm³

d = 0.93 g/cm³

Answer:

Explanation:

Given data:

Mass of alcohol = 400 g

Specific heat capacity of alcohol = 2.43 J/g°C

Initial temperature of alcohol = 16°C

Mass of water = 400 g

Specific heat capacity of water = 4.19 J/g°C

Initial temperature of water = 85°C

Final temperature of mixture = ?

Solution:

Equation:

m₁c₁ (T₂-T₁ ) = m₂c₂(T₂-T₁)

by putting values,

400 × 4.19  × (T₂ -  85°C) = 400 × 2.43  × (T₂ - 16°C)

1676  × (T₂ -  85°C) = 972 × (T₂ - 16°C)

Answer: d = 2.70 g/mL

Explanation: Considering we know that it weights 40.5 brains and the volume of such amount is 15 ml we can calculate such in which it comes to be 2.7 grams per mL

The net ionic equation for the reaction between aqueous sodium fluoride and hydrochloric acid is B. Na+ (aq) + F- (aq) + H+ (aq) + Cl- (aq) → Na+ (aq) + Cl- (aq) + HF (aq)

This is the net ionic equation for the reaction between aqueous sodium fluoride and hydrochloric acid. The balanced equation for this reaction is NaF (aq) + HCl (aq) → NaCl (aq) +HF (aq).

The net ionic equation is derived by canceling out the spectator ions, which are ions that don't participate in the chemical reaction. In this case, Na+ and Cl- are spectator ions, which means that they don't participate in the reaction and are present in the same form on both sides of the equation. Therefore, they are cancelled out, leaving the net ionic equation Na+ (aq) + F- (aq) + H+ (aq) + Cl- (aq) → Na+ (aq) + Cl- (aq) + HF (aq).

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Answer:

droplets of sulfuric acid

Explanation: