how do we know the amount of electrons that an element has?

I need help

There is  5.8 %  of NaOCl in the sample.

The oxidation of hypochlorite is the first process, which goes as follows:

OCl^- + 2I^- + 2H^+ -----> I2 + Cl2 + H2O

Iodine and sodium thiosulfate react in the second stage as follows:

I2 + 2NaS2O3 -----> Na2S4O6 + 2NaI

Thiosulfate moles = 0.150 M * 0.890/1000 L = 1.335 * 10-4 moles.

Since 2 molecules of thiosulfate react with 1 mole of I2,

1.335 * 10-4 moles of thiosulfate and x moles of I2 react.

x = 1.335 * 10-4 moles/2 moles times one mole.

Iodine equivalent to 6.675 x 10-5 moles

6.675 * 10-5 moles of hypochlorite were reacted because 1 mole of hypochlorite yields 1 mole of iodine.

Therefore, the mass of NaOCl is equal to 4.97 * 10-3 g (6.675 * 10-5 moles * 74.44 g/mole).

5.8% of NaOCl is equal to 4.97 * 10-3 g/0.0854 g * 100/1.

NaOCl makes up 5.8% of the sample.

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Using the molarity of the final solution and the molar mass of silver nitrate, we can calculate that the student used 8.0015 g of silver nitrate to make the stock solution.

First, we calculate the number of moles in the 500.00 mL flask using the molarity (c) of the final solution and its volume (V = 500.00 mL = 0.50000 L).

c = n/V ⇒ n = c*V

n = 9.443 * 10⁻³ M * 0.50000 L

n = 4.7215 * 10⁻³ mol

This same number of moles was in the 25.059 mL solution that was subsequently diluted to 500.00 mL. Knowing this, we can make a simple proportion to calculate the number of moles present in the original 250.00 mL:

4.7215 * 10⁻³ mol : 25.059 mL = X : 250.00 mL

X = 4.7215 * 10⁻³ mol * 250.00 mL / 25.059 mL

X = 0.047104 mol

That's the amount (n) of silver nitrate used to make the stock solution. Now we can convert it to mass (m) by using the molar mass of silver nitrate (M = 169.87 g/mol):

n = m/M ⇒ m = n*M

m = 0.047104 mol * 169.87 g/mol

m = 8.0015 g

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Five liters of hydrogen gas are stored at a pressure of 400 torrs. If the temperature does not change, then the volume of the gas would be 8 liters when the pressure is decreased to 250 torrs.

It is an imaginary gas for which the volume occupies by it is negligible, this gas does not exist in a practical situation and the concept of an ideal gas is only the theoretical one,

Using the Boyle's law for the given problem,

P₁V₁=P₂V₂

400 × 5 = 250 × V₂

V₂ = 400 × 5 / 250

    = 8 Litres

Thus, if the temperature does not change the volume of the gas would be  8 Litres.

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not gon find it here cuh

Explanation:

The concentration of the acetic acid solution is approximately 0.0128 mol/L, given the information provided.

To determine the concentration of the acetic acid solution, we need to use the given enthalpy change and the heat evolved in the reaction.

The enthalpy change (ΔH) is given as 55.7 kJ/mol, which represents the heat released or absorbed per mole of acetic acid involved in the reaction.

The heat evolved in the reaction is given as 716.12 J. However, we need to convert it to kilojoules to match the unit of the enthalpy change. There are 1000 J in 1 kJ, so 716.12 J is equal to 0.71612 kJ.

Now, we can set up a proportion to find the concentration of the acetic acid solution. The heat evolved is directly proportional to the moles of acetic acid involved in the reaction, which is in turn proportional to the concentration of the solution.

Let's assume the concentration of the acetic acid solution is c mol/L.

The proportion can be set up as:

ΔH (kJ/mol) / Heat evolved (kJ) = 1 mol / c L

Substituting the values we have:

55.7 kJ/mol / 0.71612 kJ = 1 mol / c L

Simplifying:

c = 1 mol / (55.7 kJ/0.71612 kJ)

c ≈ 0.0128 mol/L

Therefore, the concentration of the acetic acid solution is approximately 0.0128 mol/L, given the information provided.

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The molarity of the Vitamin B1 Hydrochloride solution is 0.71M.

We know,

Molarity of a solution is defined as the ratio of the moles of the solute with respect to the volume of the solution in liters.

We know,

Moles = mass of solute dissolved/Molar mass of solute

Given mass of solute is dissolved is 53 grams.

Molar mass of Vitamin B1 Hydrochloride is 337 grams/mol

Moles of Vitamin B1 hydrochloride = 53/337

The volume of the solution is 220 ml.

We know,

1 ml = 1/1000 litres

220 ml = 220/1000 litres

So, putting all the values in the formula of molarity,

Molarity of solution = (53/337)/(220/1000)

Molarity of solution = 0.71

So, the concentration of the Vitamin B1 Hydrochloride is 0.71M.

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The element that is able to gain electrons and form an ionic bond is Iodine.

The term ionic bond has to do with the kind of bond that is formed between a metal and a non metal. In this type of bond, there is a transfer of electrons from the metal to the non metal. s such the metal would become positively charged while the non metal would become negatively charged as we can see in the crystal of the ionic compound.

We just have to look at the string of elements that we have been given because, this is how we can be able to make a decision about which of the elements would be able to gain an electron. While doing this, we must keep in kind that it is only the non metal that are able to gain electron within the context of forming an ionic bond.

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Missing parts;

Based on its electron configuration which element will most likely gain electrons from another element when forming an ionic compound? Potassium , Vanadium, Iodine, Xenon

The two elements that do not have a second energy levels are Hydrogen (H) and Helium (He)

Energy levels are nothing but the 3-D spaces around the nucleus of an atom. The electrons revolve around the nucleus in their respective energy levels. Each energy levels can accommodate only a particular number of electrons.

For example, first energy level can accommodate only a maximum of 2 electrons whereas second energy level can accommodate 8 electrons.

Hydrogen has 1 electron and helium has 2 electrons around their respective nucleus. In both hydrogen and helium, electrons are accommodated in the first energy level.

Hence, the two elements that do not have a second energy levels are Hydrogen (H) and Helium (He)

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Aldol condensation is an easy way of making carbon-carbon bond.

In the field of chemistry, we can describe aldol condensation as a type of bond that occurs due to electrophilic substitution at the alpha carbons of the enol.

Aldol condensation is known to be the easiest way for one carbon to form bonds with another carbon.

Such an aldol reaction that takes place between carbonyl compunds instead of aldehydes or enols is referred to as crossed aldol condensation.

Aldol reactions are known to be an important part of synthesizing organic compounds as carbon-carbon bonds are formed due to it.

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Answer: The part I think you messed up on is the atoms not all of them have 0 and 1. Sodium Chloride has 2 atoms. Other than that it looks amazing! Hope this helps!

Using the distribution coefficient, we can calculate the masses removed from water: a) 45.8 g; b) 48.8 g.

If the distribution coefficient between DCM (methylene chloride) and water is 11, that means that the concentration in DCM is 11 times greater than the concentration in water.

a) When equal volumes of the solvents are used, the amount of the solute in DCM (A) will be 11 times greater than the amount of solute remaining in water (B).

A + B = 50.0 g

A = 11 B

11 B + B = 50.0 g

12 B = 50.0 g

B = 50.0 g / 12

B = 4.17 g

A = 11 * 4.2 g

A = 45.8 g

b) Because the solvent volumes are not the same, we have to use concentrations. We can still label the mass of Y in DCM as A, and the mass of Y in water as B. After a single extraction with 125 mL of DCM we get:

A₁ + B₁ = 50.0 g

(A₁ / 125 mL) / (B₁ / 250 mL) = 11

2 A₁ / B₁ = 11

A₁ = 11 B₁ / 2

A₁ = 5.5 B₁

5.5 B₁ + B₁ = 50.0 g

6.5 B₁ = 50.0 g

B₁ = 50.0 g / 6.5

B₁ = 7.7 g

So, after the first extraction, 7.7 g of Y has remained in water. That means that for the second extraction:

A₂ + B₂ = 7.7 g

A₂= 5.5 B₂

5.5 B₂ + B₂ = 7.7 g

6.5 B₂ = 7.7 g

B₁ = 7.7 g / 6.5

B₁ = 1.2 g

After the second extraction, only 1.2 g of Y has remained in water, while 50.0 g - 1.2 g = 48.8 g of Y was removed from water.

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Answer:

Quickly

Explanation:

The system should be heated gently so that the solvent boils and the vapors extend no more than halfway up the condenser.

Boiling is the fast vaporization of a fluid, which happens when fluid is warmed to its bubbling point, the temperature at which the vapor weight of the fluid is rise to the weight applied on the fluid by the encompassing air.

There are two fundamental sorts of bubbling: nucleate bubbling where little bubbles of vapor shape at discrete focuses, and basic warm flux bubbling where the bubbling surface is warmed over a certain basic temperature, and a film of vapor shapes on the surface.

In thermodynamics, heating a system, the prerequisite for the bubbling of unadulterated substances to happen is that T divider = T sat. But in genuine tests, bubbling does not happen until the fluid is warmed some degrees over the immersion temperature.

The surface temperature must be to some degree over the immersion temperature in arrange to maintain vapor arrangement.

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The heating curve is a graphical representation of how various phases of substances change as temperature or heat increases. The graph's C section depicts the liquid phase of the substance gaining kinetic energy.

Kinetic energy is the energy that an object has as a result of its movement. If we want to accelerate an object, we must apply force to it. Using force requires us to put in effort. When the work is finished, energy is transferred to the object, and it moves at a new constant speed.

Therefore,

Section A represents the solid phase

Section B the solid-liquid phase

Section C the liquid phase

Section D the liquid-gas phase

Section E the gas phase

Section C of the graph depicts the substance's liquid phase gaining kinetic energy. Heat energy is absorbed during Phase C, and liquid particles gain kinetic energy to change into gas form in Section E.

As a result, C.C. is the correct answer.

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Because boiling point is independent of liquid volume, the boiling point would not vary.

What is boiling point  ?

A liquid's boiling point changes depending on the pressure being applied; the normal boiling point is the temperature at which the vapor pressure is equal to the average atmospheric pressure at sea level (760 mm [29.92 inches] of mercury). Water boils at sea level at 100° C.

The temperature at which a liquid's vapor pressure equals the pressure around it and the liquid turns into a vapor is known as the boiling point of a substance

A liquid's boiling point varies depending on the atmospheric pressure in the area. The boiling point of a liquid is lower in a partial vacuum than it is at atmospheric pressure. A liquid's boiling point is lower under low pressure than it is under air pressure. As a result, water boils at 99.97 °C (211.95 °F) at sea level but at 93.4 °C (200.1 °F) at 1,905 m (6,250 ft) altitude under standard pressure. Various liquids will boil at different temperatures at a given pressure.

Because boiling point is independent of liquid volume, the boiling point would not vary.

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Metallic solids,

the atoms within such a metallic solid are held together by a unique force known as metallic bonding that gives rise to many useful and varied bulk properties. All exhibit high thermal and electrical conductivity, metallic luster, and malleability.

Answer: c. metallic

A molecule with strong intermolecular forces will have a weak or low vapor pressure.

Molecules generally with a high intermolecular force of attraction will have a low vapor pressure. This is because since the force of attraction between molecules is high as they are tightly packed, only fewer molecules will get a chance to move with kinetic energy at a given temperature.

Whereas, molecules with weak intermolecular force can travel easily and escape and so will have a greater vapor pressure. But it is not the case with strong intermolecular forces. Molecules with high vapor pressure are called volatile. Therefore, the molecules with strong intermolecular forces will have a low vapor pressure, and only that is expected to happen in such cases.

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The strongest oxidizing agent is chromium (vi); option E.

Oxidizing agents are compounds that readily accept electrons and are reduced.

Oxidizing agents will oxidize other substances, resulting in an increase in the oxidation state of the compounds. However, they are themselves reduced as their oxidation states are reduced when they accept electrons.

Strong oxidizing agents are weak reducing agents.

Reducing agents readily give up electrons to other compounds, and hence are oxidized.

Transition elements are known for their variable oxidation states due to their ability to lose different numbers of electrons.

The higher the oxidation state of a transition element in a compound, the more oxidizing the compound will be.

Hence, a compound of a transition element in the highest oxidation state will be the most oxidizing.

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Acetyl-co A containing radioactively labeled carbon atoms when fed to cells produce Co2 which are not radioactive but molecules of citrate and oxalo acetate are radioactive because citrate and oxalo acetate are part of the Krebs Cycle, which is responsible for generating energy in cells.

The Krebs Cycle involves a series of reactions in which acetyl-co A is broken down to generate energy. The carbon atoms in acetyl-co A are gradually transferred to other molecules in the cycle, including citrate and oxalo acetate. As the carbon atoms are transferred, they become radioactive. Eventually, the carbon atoms are transferred to Co2, which is then released from the cells.

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Answer:

ngl ts ngl ts ts ngl...imma keep it a buck.96 ion got the answers kanye do.

Explanation:

Low pressure systems associated with high temperatures because warm air rises :)

The rate of the reaction if the concentration of ethanol was doubled will remain the same.

The rate of the reaction tert-butyl ethyl ether, (ch3)3coch2ch3, in acidic conditions, will undergo the SN1 mechanism so the rate of reaction will remain the same.

The rate of reaction is the rate at which a chemical reaction takes region, defined as proportional to the increase inside the concentration of a product in step with unit time and to the lower within the awareness of a reactant in line with unit time. response prices can vary dramatically.

The speed with which a reaction takes area is called the fee of a response. as instance, timber combustion has a high response rate since the manner is fast, and rusting of iron has a low response rate as the technique is slow. Rate is most customarily calculated by the usage of the equation The devices of rate calculated in this manner are s -1.

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The empirical formula of the compound is CH₂O, and the molecular formula of the compound is C₂H₄O₂.

All of the carbon in the compound is converted into carbon dioxide (14.35 g) and all the hydrogen in the compound is converted to water (5.876 g). This means that we can use these masses to find the masses of carbon, hydrogen, and oxygen in the original compound.

First, we calculate the mass of carbon in carbon dioxide and the mass of hydrogen in water, using their respective molar masses (M(CO₂) = 44 g/mol; M(H₂O) = 18 g/mol):

12 g of carbon : 44 g of carbon dioxide = X : 14.35 g of carbon dioxide

X = 12 g of carbon * 14.35 g of carbon dioxide / 44 g of carbon dioxide

X = 3.914 g of carbon

2 g of hydrogen : 18 g of water = X : 5.876 g of water

X = 2 g of hydrogen : 5.876 g of water / 18 g of water

X = 0.6529 g of hydrogen

Because the compound contains carbon, hydrogen and oxygen, we can now calculate the mass of oxygen in the compound:

m(O) = m(sample) - m(C) - m(H)

m(O) = 9.791 g - 3.914 g - 0.6529 g

m(O) = 5.224 g

Now we use these masses to calculate the number of moles (n) for each of the elements, using their respective molar masses (M(C) = 12 g/mol, M(H) = 1 g/mol, M(O) = 16 g/mol):

n = m/M

n(C) = m(C) / M(C)

n(C) = 3.914 g / 12 g/mol

n(C) = 0.3284 mol

n(H) = m(H) / M(H)

n(H) = 0.6529 g / 1 g/mol

n(H) = 0.6529 mol

n(O) = m(O) / M(O)

n(O) = 5.224 g / 16 g/mol

n(O) = 0.3265 mol

We divide each of these amounts by the smallest of them (0.3265 mol) to obtain the molar ratio of the elements:

carbon: 0.3284 mol / 0.3265 mol = 1.006

hydrogen: 0.6529 mol / 0.3265 mol = 2.000

oxygen: 0.3265 mol / 0.3265 mol = 1.000

These integers are indexes in the empirical formula of the compound - CH₂O

The molar mass of this empirical formula is:

12 g/mol + 2 * 1 g/mol + 16 g/mol = 30 g/mol

We divide the molar mass of the compound by the molar mass of the empirical formula:

60.05 g/mol / 30 g/mol = 2.002

So, we multiply the indexes by 2 to obtain the molecular formula - C₂H₄O₂

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As a result, it is a natural bodily alteration. The chemical shape of the material is affected by physical changes, but not its makeup.

Since natural water undergoes a phase transformation cycle from liquid to gas and back to liquid, the production of clouds is a physical shift.

The physical conversion of water from vapour to liquid during cloud formation is known as condensation. When water evaporates during the evaporation process and condenses in the air to form clouds, there is a physical shift that takes place.

Condensation is the process by which water transforms from a gas to a liquid, whereas deposition is the process by which a gas transforms straight into a solid. Clouds are created by these two processes.

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Answer:

Atomic mass of Boron: 10.8 amu

Explanation:

Boron-10: (0.198)(10.013) = 1.98 amu

Boron-11: (0.802)(11.009) =  8.83 amu

Add them up: 1.98+8.83= 10.8 amu

Answer:

[tex] \huge{ 3.935 \times {10}^{ - 16} J}[/tex]

Explanation:

The energy of the photon can be found by using the formula

E = hf

where

E is the energy

f is the frequency

h is the Planck's constant which is

6.626 × 10-³⁴ Js

From the question the frequency is

5.94x10^17Hz

We have

[tex]E = 6.626 \times {10}^{ - 34} \times 5.94 \times {10}^{17} \\ = 3.935 \times {10}^{ - 16} J[/tex]

Hope this helps you

Answer: Homogenous mixture

Mixing honey with water will give you a homogenous mixture. It is uniform and even, as the honey particles are distributed equally throughout the water.

hope it helps, mark as brainliest.

Answer:

sp³-s sigma bond.

Hope that was the answer

By using the data given in the text and the specific heat capacity of water and its density, we can calculate that the amount of heat produced by hydrocarbon combustion was 233.6 kJ.

The heat released by hydrocarbon combustion is absorbed by the calorimeter and the water in it. To calculate the amount of heat (Q) produced by the combustion, we are going to use the specific heat capacity of water (c = 4.186 J/g°C), the heat capacity of the calorimeter (q = 68.5 J/°C) and the difference in temperature (ΔT = 54.91 °C):

heat absorbed by the calorimeter = ΔT*q

heat absorbed by water = ΔT*c*m

Q = ΔT*q + ΔT*c*m

m - the mass of water

To obtain the mass of water, we are going to use the volume (V = 1000 mL) of water and its density (d = 1.00 g/mL):

d = m/V ⇒ m = d * V

m = 1.00 g/mL * 1000 mL

m = 1000 g

Now we can plug all the known values in the formula for Q:

Q = 54.91 °C * 68.5 J/°C + 54.91 °C * 4.186 J/g°C * 1000 g

Q = 233,614.595 J

Q = 233.6 kJ

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In Niger State, Nigeria, there is a deposit of clay similar to kaolin that has a lot of promise for use in pollution control processes including wastewater treatment.

This study looks into the ideal settings for Kpautagi clay modification for use in treating the effluent from refineries. The clay was modified using sulphuric acid, and the modification factors taken into account were acid concentration, activation time, and temperature.

The sulphuric acid-modified Kpautagi clay was used in the treatment of refinery wastewater in column mode at a fixed flow rate and mass of adsorbent to establish the best conditions for the modification variables.

The results show that 4M acid concentration, 120min of activation time, and 100°C activation temperature are the best conditions for modifying Kpautagi clay for use in the treatment of petroleum refinery wastewater.

As a result, the best conditions discovered in this work for Kpautagi clay modification could be used to treat wastewater from petroleum refineries more effectively.

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